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Holt Physics Problem Bank
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Problem 1A Ch. 1–1
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 1AMETRIC PREFIXES
P R O B L E MThe scanning tunneling microscope (STM) has a magnifying ability of100 million and can distinguish between two objects that are separatedby only 3.0 × 10−10 m, or about one-hundredth the diameter of an atom.Express 3.0 × 10−10 m ina. nanometers.b. picometers.
S O L U T I O N
Given: distance = 3.0 × 10−10 m
Unknown: distance = ? nm distance = ? pm
Build conversion factors from the relationships given in Table 1-3.
1 ×
1
1
n
0
m−9 m
1 ×
1
10
p−m12 m
Convert from meters to nanometers by multiplying the distance by the first con-
version factor.
distance = 3.0 × 10−10 m × 1 ×
1
1
n
0
m−9 m = 3.0 × 10−1 nm =
Convert from meters to picometers by multiplying the distance by the second
conversion factor.
distance = 3.0 × 10−10 m × 1 ×
1
1
p
0−m
12m = 3.0 × 102 pm
0.30 nm
ADDITIONAL PRACTICE
1. One of the more unusual of world wonders is the Plain of Jars in Laos.
Several hundred huge stone jars, which do not seem to be made from
local rock, are scattered across the plain. The largest of these jars has a
mass of around 6.0 × 103 kg. Express this mass in
a. milligrams.
b. megagrams.
2. The French drink about 6.4 × 104 cm3 of mineral water per person per
year. Express this volume in
a. cubic meters.
b. cubic millimeters.
Holt Physics Problem BankCh. 1–2
NAME ______________________________________ DATE _______________ CLASS ____________________
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3. The explosive energy of powerful explosives is measured in terms of
“tons.” The ton referred to is a ton of TNT (trinitrotoluene), one of the
most powerful of chemical explosives. A ton of TNT will release 4.2 ×109 J (joules). Express this energy in
a. megajoules.
b. gigajoules.
4. A parsec, a distance measurement used by astronomers, is equal to
3.262 light years, where a light year is the distance light travels in one
year. In SI units, a parsec equals 3.086 × 1016 m. Express this distance in
a. kilometers.
b. exameters.
5. An acre is a common unit used to measure the area of a portion of
land. An acre is equal to about 4.0469 × 103 m2. Express this area in
a. square kilometers.
b. square centimeters.
6. Electric charge is measured in terms of the coulomb (1 C), although
this is a very large and not extremely practical unit of measurement.
For example, the charge in a bolt of lightning is about 15 C. Express the
charge in a lightning bolt in
a. millicoulombs.
b. kilocoulombs.
7. The wettest spot on Earth is generally considered to be Mt. Waialeale,
on the island of Kauai, Hawaii. In one year, this long-extinct volcano
receives 1.168 × 103 cm of rainfall. Express this quantity in
a. meters.
b. micrometers.
8. The United States Department of Defense is housed in the Pentagon,
one of the largest office buildings in the world. The entire floor area of
the Pentagon equals 0.344 279 km2. Express this area in
a. square meters.
b. square millimeters.
9. The Earth is approximately 4.50 billion years old. Setting 1 year equal
to 365.25 days, express the age of the Earth in
a. gigaseconds.
b. petaseconds.
10. One of the isotopes with the shortest half-life (the time it takes for half
of a sample of the element to decay) is beryllium-8. Its half life is mea-
sured as 6.7 × 10–17 s. Express this time in
a. microseconds.
b. attoseconds.
1ChapterThe Science of Physics
1. mass = 6.0 × 103 kga. mass = 6.0 × 103 kg ×
1
1
0
k
3
g
g ×
1
1
0
m−3
g
g =
b. mass = 6.0 × 103 kg × 1
1
0
k
3
g
g ×
1
10
M6 g
g = 6.0 Mg
6.0 × 109 mg
Additional Practice 1A
Givens Solutions
2. Volume = 6.4 × 104 cm3
a. volume = 6.4 × 104 cm3 × 10
12m
cm
3= 6.4 × 104 cm3 ×
10
16m
cm
3
3 =
b. volume = 6.4 × 104 cm3 × 11
0
c
m
m
m
3= 6.4 × 104 cm3 ×
10
1
3
c
m
m
m3
3
= 6.4 = 107 mm3
6.4 × 10−2 m3
3. energy = 4.2 × 109 J a. energy = 4.2 × 109 J × 1
10
M6 J
J =
b. energy = 4.2 × 109 J × 1
1
0
G9
J
J = 4.2 GJ
4.2 × 103 MJ
4. distance = 1 parsec = 3.086 × 1016 m a. distance = 1 parsec = 3.086 × 1016 m ×
1
1
0
k3m
m =
b. distance = 1 parsec = 3.086 × 1016 m × 1
1
01E8m
m = 3.086 × 10–2 Em
3.086 × 1013 km
5. area = 1 acre = 4.0469 × 103 m2 a. area = 1 acre = 4.0469 × 103 m2 × 1
1
0
k3m
m2
area = 4.0469 × 103 m2 × 1
1
0
k6m
m
2
2 =
b. area = 1 acre = 4.0469 × 103 m2 × 10
1
2
m
cm
2
area = 4.0469 × 103 m2 × 10
1
4
m
cm2
2
= 4.0469 × 107 cm2
4.0469 × 10−3 km2
6. electric charge = 15 Ca. electric charge = 15 C ×
10
1
3
C
mC =
b. electric charge = 15 C × 1
1
0
k3C
C = 1.5 × 10–2 kC
1.5 × 104 mC
Section Two—Problem Workbook Solutions V Ch. 1–1
V
Holt Physics Solution ManualV Ch. 1–2
V
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.
7. depth = 1.168 × 103 cma. depth = 1.168 × 103 cm ×
10
12m
cm =
b. depth = 1.168 × 103 cm × 10
12m
cm ×
1
1
0−m6
m
m = 1.168 × 107 mm
1.168 × 101 m = 11.68 m
Givens Solutions
8. area = 0.344 279 km2
a. area = 0.344 279 km2 × 110
k
3
m
m
2= 0.344 279 × 106 m2 =
b. area = 0.344 279 km2 × 110
k
3
m
m
2× 10
1
3
m
mm
2= 0.344 279 × 1012 mm2
area = 3.442 79 × 1011 mm2
3.442 79 × 105 m3
9. time = 4.50 × 109 years ×
365
1
.2
y
5
ea
d
r
ays ×
1
24
da
h
y ×
36
1
0
h
0 s
= 1.42 × 1017 s
a. time = 1.42 × 1017 s × 1
1
0
G9
s
s =
b. time = 1.42 × 1017 s × 1
1
01P5s
s = 1.42 × 102 Ps = 142 Ps
1.42 × 108 Gs
10. time = 6.7 × 10−17 s a. time = 6.7 × 10−17 s × 10
1
6
s
ms =
b. time = 6.7 × 10−17 s × 10
1
18
s
as = 6.7 = 101 as = 67 as
6.7 × 10−11 ms
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Problem 2A Ch. 2–1
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 2AFINDING THE AVERAGE VELOCITY
P R O B L E MTo qualify for the finals in a racing event, a race car must achieve an aver-age speed of 2.50 × 102 km/h on a track with a total length of 1.60 km. If aparticular car covers the first half of the track at an average speed of 2.30× 102 km/h, what minimum average speed must it have in the second halfof the event to qualify?
S O L U T I O NGiven: vtot, avg = 2.50 × 102 km/h
v1, avg = 2.30 × 102 km/h
∆xtot = 1.60 km
∆x1 = ∆x2 = 12
∆xtot
Unknown: v2, avg = ?
Use the definition of average velocity, and rearrange it to solve for time.
vavg = ∆∆
x
t
∆t1 = time required to travel ∆x1 = v1
∆
,
x
a
1
vg
∆t2 = time required to travel ∆x2 = v2
∆
,
x
a
2
vg
∆ttot = ∆t1 + ∆t2 = v
∆
1,
x
a
1
vg +
v
∆
2,
x
a
2
vg =
vt
∆
ot
x
,
t
a
o
v
t
g
Use the last two equations for ∆ttot to solve for v2, avg.
v
∆
to
x
t,
t
a
o
v
t
g =
v
∆
1,
x
a
1
vg +
vt
∆
2,
x
a
2
vg =
1
2 ∆xtot v1,
1
avg +
v2,
1
avg
Divide by 1
2 ∆xtot on each side.
vtot
2
, avg =
v1,
1
avg +
v2,
1
avg
Rearrange the equation to calculate v2, avg.
v2,
1
avg =
vtot
2
, avg −
v1,
1
avg
Invert the equation.
v2, avg =
v2, avg =1
2.50 × 1
2
02 km/h − 2.30 × 1
2
02 km/h
1
vtot
2
, avg −
v1,
1
avg
Holt Physics Problem BankCh. 2–2
NAME ______________________________________ DATE _______________ CLASS ____________________
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v2, avg =
v2, avg = 3.65
1
×k
1
m
0−3 h
v2, avg = 274 km/h
18.00 × 10−3 h/km − 4.35 × 10−3 h/km
ADDITIONAL PRACTICE
1. The fastest helicopter, the Westland Lynx, can travel 3.33 km in the forward
direction in just 30.0 s. What is the average velocity of this helicopter? Ex-
press your answer in both meters per second and kilometers per hour.
2. The fastest airplane is the Lockheed SR-71 Blackbird, a high-altitude spy
plane first built in 1964. If an SR-71 is clocked traveling 15.0 km west in
15.3 s, what is its average velocity in kilometers per hour?
3. At its maximum speed, a typical snail moves about 4.0 m in 5.0 min.
What is the average speed of the snail?
4. The arctic tern migrates farther than any other bird. Each year, the Arctic
tern travels 3.20 × 104 km between the Arctic Ocean and the continent of
Antarctica. Most of the migration takes place within two four-month pe-
riods each year. If a tern travels 3.20 × 104 km south in 122 days, what is
its average velocity in kilometers per day?
5. Suppose the tern travels 1.70 × 104 km south, only to encounter bad
weather. Instead of trying to fly around the storm, the tern turns around
and travels 6.00 × 102 north to wait out the storm. It then turns around
again immediately and flies 1.44 × 104 km south to Antarctica. What are
the tern’s average speed and velocity if it makes this trip in 122 days?
6. Eustace drives 20.0 km to the east when he realizes he left his wallet at
home. He drives 20.0 km west to his house, takes 5.0 min to find his wal-
let, then leaves again. Eustace is 40.0 km east of his house exactly
60.0 min after he left the first time.
a. What is his average velocity?
b. What is his average speed?
7. Emily takes a trip, driving with a constant velocity of 89.5 km/h to the
north except for a 22.0 min rest stop. If Emily’s average velocity is
77.8 km/h to the north, how long does the trip take?
8. Laura is skydiving when at a certain altitude she opens her parachute and
drifts toward the ground with a constant velocity of 6.50 m/s, straight down.
What is Laura’s displacement if it takes her 34.0 s to reach the ground?
9. A tortoise can run with a speed of 10.0 cm/s, and a hare can run exactly 20
times as fast. In a race, they both start at the same time, but the hare stops to
rest for 2.00 min. The tortoise wins by 20.0 cm. How long does the race take?
10. What is the length of the race in problem 9?
Problem 2B Ch. 2–3
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 2BAVERAGE ACCELERATION
P R O B L E MA basketball rolling with a velocity of +1.1 m/s comes to a stop in 8.5 s.What is the basketball’s average acceleration?
S O L U T I O NGiven: vi = +1.1 m/s
vf = 0 m/s
∆t = 8.5 s
Unknown: aavg = ?
Use the definition of average acceleration to find aavg.
aavg = ∆∆
v
t =
vf
∆−t
vi
aavg = 0 m/s
8
−.5
1
s
.1 m/s =
−1
8
.1
.5
m
s
/s
aavg = −0.13 m/s2
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ADDITIONAL PRACTICE
1. With a time of 6.92 s, Irina Privalova of Russia holds the women’s
record for running 60 m. Suppose she ran this distance with a constant
acceleration, so that she crossed the finish line with a speed of
17.34 m/s. Assuming she started at rest, what was the magnitude of
Privalova’s average acceleration.
2. The solid-fuel rocket boosters used to launch the space shuttle are able
to lift the shuttle 45 kilometers above Earth’s surface. During the
2.00 min that the boosters operate, the shuttle accelerates from rest
to a speed of nearly 7.50 × 102 m/s. What is the magnitude of the shut-
tle’s average acceleration?
3. A type of firework consists of a cardboard tank mounted on plastic
wheels and driven forward by a small rocket. Once the rocket ignites,
the tank rolls from rest to a maximum velocity of 0.85 m/s forward, at
which point the rocket burns out. If the total time that the rocket re-
mains ignited is 3.7 s, what is the average acceleration of the tank?
4. A handball is hit toward a wall with a velocity of 13.7 m/s in the for-
ward direction. It returns with a velocity of 11.5 m/s in the backward
direction. If the time interval during which the ball is accelerated is
0.021 s, what is the handball’s average acceleration?
Holt Physics Problem BankCh. 2–4
NAME ______________________________________ DATE _______________ CLASS ____________________
5. A certain type of rocket sled is used to measure the effects of extreme
deceleration. The sled reaches a velocity of +320 km/h, then comes to a
complete stop in 0.18 s. What is the average acceleration that takes
place in this time interval?
6. In 1970, Don “Big Daddy” Garlits set what was then the world record
for drag racing. With an average acceleration of 16.5 m/s2, Garlits
started at rest and reached a speed of 386.0 km/h. How much time was
needed for Garlits to reach his final speed?
7. A freight train traveling with a velocity of −4.0 m/s begins backing into
a train yard. If the train’s average acceleration is −0.27 m/s2, what is the
train’s velocity after 17 s?
8. A student on in-line roller skates travels at a speed of 4.5 m/s along the
top of a hill. She then skates downhill with an average acceleration of
0.85 m/s2. If her final speed is 10.8 m/s, how long does it take her to
skate down the hill?
9. The Impact is the first commercial electric car to be developed in over
60 years. During tests in 1994, the car reached a top speed of nearly
296 km/h. Suppose the car started at rest and then underwent an aver-
age acceleration of 1.60 m/s2. How long did it take the Impact to reach
its top speed?
10. A bicyclist accelerates –0.87 m/s2 during a 3.8 s interval. What is the
change in the velocity of the bicyclist and bicycle?
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Problem 2C Ch. 2–5
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 2CDISPLACEMENT WITH UNIFORM ACCELERATION
P R O B L E MThe arrow on a crossbow undergoes uniform acceleration over a distanceof 38.1 cm. If the acceleration takes place over 8.93 × 10-3 s and the arrowis initially at rest, what is the arrow’s final speed?
S O L U T I O NGiven: ∆x = 38.1 cm
∆t = 8.93 × 10−3 s
vi = 0 m/s
Unknown: vf = ?
Use the equation for displacement with uniform acceleration.
∆x = 1
2 (vi + vf ) ∆t
Rearrange the equation to solve for vf.
vf = 2
∆∆t
x − vi
vf = − 0 m/s
vf = 85.3 m/s − 0 m/s
vf = 85.3 m/s
(2)(38.1 cm)10
1
0
m
cm
8.93 × 10−3 s
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ADDITIONAL PRACTICE
1. A device at Sandia Laboratories in Albuquerque, New Mexico, uses
highly compressed air to accelerate small metal disks to supersonic
speeds. Suppose the disk, which is initially at rest, undergoes a uniform
acceleration for 0.910 s, at which point it reaches its top speed. If the disk
travels 7.19 km in that time, what is its final speed?
2. Despite their size and awkward appearance, polar bears can run at re-
spectable speeds for short distances. Suppose a polar bear running with
an initial speed of 4.0 m/s accelerates uniformly for 18 s. What is the
bear’s maximum speed if the bear travels 135 m during the 18 s of accel-
eration? Give the answer in both meters per second and kilometers per
hour?
3. A hockey puck slides 55.0 m along the length of the rink in just 1.25 s.
The slight friction between the puck and the ice provides a uniform ac-
celeration. If the puck’s final speed is 43.2 m/s, what is its initial speed?
Holt Physics Problem BankCh. 2–6
NAME ______________________________________ DATE _______________ CLASS ____________________
4. A child sleds down a snow-covered hill with a uniform acceleration.
The slope of the hill is 38.5 m long. If the child starts at rest and
reaches the bottom of the hill in 5.5 s, what is the child’s final speed?
5. The longest stretch of straight railroad tracks lies across the desolate
Nullarbor Plain, between the Australian cities of Adelaide and Perth.
The tracks extend a distance of 478 km without a curve. Suppose a
train with an initial speed of 72 km/h travels along the entire length of
straight track with a uniform acceleration. The train reaches the end of
the straight track in 5 h, 39 min. What is the train’s final speed?
6. A golf ball at a miniature golf course travels 4.2 m along a carpeted
green. When the ball reaches the hole 3.0 s later, its speed is 1.3 m/s. As-
suming the ball undergoes constant uniform acceleration, what is the
ball’s initial speed?
7. A speedboat uniformly increases its velocity from 25 m/s to the west to
35 m/s to the west. How long does it take the boat to travel 250 m west
while undergoing this acceleration?
8. Airplane racing, like horse and auto racing, uses a “track” of a specific
length. Unlike the horse or auto tracks, the racing area for airplanes is
bounded on the inside by tall columns, or pylons, around which the pi-
lots must fly, and by altitude limitations that the pilots must monitor
using their instruments. Different types of races use different arrange-
ments of pylons to make the length of the race longer or shorter. In one
particular race, a pilot begins the race at a speed of 755.0 km/h and ac-
celerates at a constant uniform rate for 63.21 s. The pilot crosses the
finish line with a speed of 777.0 km/h. From this data, calculate the
length of the course.
9. A hovercraft, also known as an air-cushion vehicle, glides on a cushion
of air, allowing it to travel with equal ease on land or water. Suppose a
hovercraft undergoes constant uniform acceleration, which causes the
hovercraft to move from rest to a speed of 30.8 m/s. How long does the
hovercraft accelerate if it travels a distance of 493 m?
10. A spaceship travels 1220 km with a constant uniform acceleration.
How much time is required for the acceleration if the spaceship in-
creases its speed from 11.1 km/s to 11.7 km/s?
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Problem 2D Ch. 2–7
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 2DVELOCITY AND DISPLACEMENT WITH UNIFORM ACCELERATION
P R O B L E MA barge moving with a speed of 1.00 m/s increases speed uniformly, sothat in 30.0 s it has traveled 60.2 m. What is the magnitude of the barge’sacceleration?
S O L U T I O N
Given: vi = 1.00 m/s
∆t = 30.0 s
∆x = 60.2 m
Unknown: a = ?
Use the equation for displaement with constant uniform acceleration.
∆x = vi∆t + 12
a∆t2
Rearrange the equation to solve for a.
12
a∆t2 = ∆x − vi∆t
a = 2(∆x
∆−t2
vi∆t)
a =
a =
a = 9
(
.
2
0
)
0
(3
×0
1
.2
02m
s
)2
a = 6.71 × 10−2 m/s2
(2)(60.2 m − 30.0 m)
9.00 × 102 s2
(2)[60.2 m − (1.00 m/s)(30.0 s)]
(30.0 s)2
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ADDITIONAL PRACTICE
1. The flight speed of a small bottle rocket can vary greatly, depending on
how well its powder burns. Suppose a rocket is launched from rest so that
it travels 12.4 m upward in 2.0 s. What is the rocket’s net acceleration?
2. The shark can accelerate to a speed of 32.0 km/h in a few seconds. As-
sume that it takes a shark 1.5 s to accelerate uniformly from 2.8 km/h to
32.0 km/h. What is the magnitude of the shark’s acceleration?
3. In order for the Wright brothers’ 1903 flyer to reach launch speed, it had
to be accelerated uniformly along a track that was 18.3 m long. A system
of pulleys and falling weights provided the acceleration. If the flyer was
initially at rest and it took 2.74 s for the flyer to travel the length of the
track, what was the magnitude of its acceleration?
Holt Physics Problem BankCh. 2–8
NAME ______________________________________ DATE _______________ CLASS ____________________
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4. A certain roller coaster increases the speed of its cars as it raises them to
the top of the incline. Suppose the cars move at 2.3 m/s at the base of
the incline and are moving at 46.7 m/s at the top of the incline. What is
the magnitude of the net acceleration if it is uniform acceleration and
takes place in 7.0 s?
5. A ship with an initial speed of 6.23 m/s approaches a dock that is 255 m
away. If the ship accelerates uniformly and comes to rest in 82 s, what is
its acceleration?
6. Although tigers are not the fastest of predators, they can still reach and
briefly maintain a speed of 55 km/h. Assume that a tiger takes 4.1 s to
reach this speed from an initial speed of 11 km/h. What is the magni-
tude of the tiger’s acceleration, assuming it accelerates uniformly?
7. Assume that a catcher in a professional baseball game catches a ball that
has been pitched with an initial velocity of 42.0 m/s to the southeast. If
the catcher uniformly brings the ball to rest in 0.0090 s through a dis-
tance of 0.020 m to the southeast, what is the ball’s acceleration?
8. A crate is carried by a conveyor belt to a loading dock. The belt speed
uniformly increases slightly, so that for 28.0 s the crate accelerates by
0.035 m/s2. If the crate’s initial speed is 0.76 m/s, what is its final speed?
9. A plane starting at rest at the south end of a runway undergoes a uni-
form acceleration of 1.60 m/s2 to the north. At takeoff, the plane’s ve-
locity is 72.0 m/s to the north.
a. What is the time required for takeoff?
b. How far does the plane travel along the runway?
10. A cross-country skier with an initial forward velocity of +4.42 m/s ac-
celerates uniformly at −0.75 m/s2.
a. How long does it take the skier to come to a stop?
b. What is the skier’s displacement in this time interval?
Problem 2E Ch. 2–9
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 2EFINAL VELOCITY AFTER ANY DISPLACEMENT
P R O B L E MA radio-controlled toy car increases speed over a distance of 15.2 m. If thecar starts at rest and has a final speed of 0.76 m/s, what is the magnitudeof its acceleration?
S O L U T I O NGiven: ∆x = 15.2 m
vi = 0 m/s
vf = 0.76 m/s
Unknown: a = ?
Choose the equation(s) or situation: Use the equation for the final velocity after
any displacement.
vf2 = vi
2 + 2a∆x
Rearrange the equation(s) to isolate the unknown(s):
a = vf
2
2∆−
x
vi2
Substitute the values into the equation(s) and solve:
a =
a =
a = 0
3
.
0
5
.
8
4 m/s2
a =
The magnitude of the acceleration suggests that the car increases speed very slowly.
This is confirmed by using the difference in speeds and the acceleration to calculate
the time interval for the acceleration. The car reaches its final speed in 40 s.
1.9 × 10−2 m/s2
0.58 m2/s2 − 0 m2/s2
30.4 m
(0.76 m/s)2 − (0 m/s)2
(2)(15.2 m)
1. DEFINE
2. PLAN
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3. CALCULATE
4. EVALUATE
ADDITIONAL PRACTICE
1. A dumptruck filled with sand moves 1.8 km/h when it begins to acceler-
ate uniformly at a constant rate. After traveling 4.0 × 102 m, the truck’s
speed is 24.0 km/h. What is the magnitude of the truck’s acceleration?
2. One of the most consistent long-jumpers is Jackie Joyner-Kersee of the
United States. Her best distance in this field and track event is 7.49 m. To
achieve this distance, her speed at the point where she started the jump
was at least 8.57 m/s. Suppose the runway for the long jump was
Holt Physics Problem BankCh. 2–10
NAME ______________________________________ DATE _______________ CLASS ____________________
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19.53 m, and that Joyner-Kersee’s initial speed was 0 m/s. What was
the magnitude of her acceleration if it was uniform acceleration?
3. Although ungraceful on land, walruses are fine swimmers. They nor-
mally swim at 7 km/h, and for short periods of time are capable of
reaching speeds of nearly 35 km/h. Suppose a walrus accelerates from
7.0 km/h to 34.5 km/h over a distance of 95 m. What would be the
magnitude of the walrus’s uniform acceleration?
4. Floyd Beattie set an unofficial speed record for a unicycle in 1986. He
rode the unicycle through a 2.00 × 102 m speed trap, along which his
speed was measured as being between 9.78 m/s and 10.22 m/s. Suppose
that Beattie had accelerated at a constant rate along the speed trap, so
that his initial speed was 9.78 m/s and his final speed was 10.22 m/s.
What would the magnitude of his acceleration have been?
5. A fighter jet lands on an aircraft carrier’s flight deck. Although the deck
is 300 m long, most of the jet’s acceleration occurs within a distance of
42.0 m. If the jet’s velocity is reduced uniformly from +153.0 km/h to
0 km/h as it moves through 42.0 m, what is the jet’s acceleration?
6. Most hummingbirds can fly with speeds of nearly 50.0 km/h. Suppose
a hummingbird flying with a velocity of 50.0 km/h in the forward di-
rection accelerates uniformly at 9.20 m/s2 in the backward direction until
it comes to a hovering stop. What is the hummingbird’s displacement?
7. A thoroughbred racehorse accelerates uniformly at 7.56 m/s2, reaching
its final speed after running 19.0 m. If the horse starts at rest, what is its
final speed?
8. A soccer ball moving with an initial speed of 1.8 m/s is kicked with a
uniform acceleration of 6.1 m/s2, so that the ball’s new speed is 9.4 m/s.
How far has the soccer ball moved?
9. A dog runs with an initial velocity of 1.50 m/s to the right on a waxed
floor. It slides to a final velocity of 0.30 m/s to the right with a uniform
acceleration of 0.35 m/s2 to the left. What is the dog’s displacement?
10. A hippopotamus can run up to 30 km/h, or 8.33 m/s. Suppose a hip-
popotamus uniformly accelerates 0.678 m/s2 until it reaches a top
speed of 8.33 m/s. If the hippopotamus has run 46.3 m, what is its ini-
tial speed?
Problem 2F Ch. 2–11
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 2FFALLING OBJECT
P R O B L E MWhen it is completed in 2002, the International Financial Center inTaipei, Taiwan, will be the tallest building in the world. Suppose a con-struction worker on the top-most floor of the building accidentallyknocks a wrench off a ledge. The wrench hits the ground below 9.56 slater. What is the distance between the top-floor of the International Fi-nancial Center and the ground. Assume there is no air resistance.
S O L U T I O NGiven: ∆t = 9.56 s
a = −9.81 m/s2
vi = 0 m/s
Unknown: ∆x = ?
Choose the equation(s) or situation: Displacement is unknown, as is the final
velocity. Because time, acceleration, and initial velocity are known, the equation
for displacement with constant acceleration can be used.
∆x = vi∆t + 12
a∆t2
Substitute the values into the equation(s) and solve:
∆x = (0 m/s)(9.56 S) + 12
(−9.81 m/s2)(9.56 s)2
∆x = (0 m) + (−448 m)
∆x = −448 m
∆x =
From the value for ∆x the wrench’s final speed can be determined as 93.8 m/s, or
nearly 340 km/h.
distance from top of building to ground = 448 m
1. DEFINE
2. PLAN
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3. CALCULATE
4. EVALUATE
ADDITIONAL PRACTICE
1. Suppose a safety net at one of the floors of the International Financial
Center catches the wrench in Problem 2F. The wrench falls into the net
with a velocity of 49.5 m/s downward. How far above the ground is the
safety net located?
2. A gumdrop is released from rest at the top of the Empire State Building,
which is 381 m tall. Disregarding air resistance, calculate the displace-
ment of the gumdrop after 1.00, 2.00, and 3.00 s.
3. A small sandbag is dropped from rest from a hovering hot-air balloon.
After 2.0 s, how far below the balloon is the sand bag?
Holt Physics Problem BankCh. 2–12
NAME ______________________________________ DATE _______________ CLASS ____________________
4. A physics student throws a softball straight up into the air with a speed
of 17.5 m/s. The ball is in the air for a total of 3.60 s before it is caught
at its original position. How high does the ball rise?
5. A surface probe lands on a highland region of the planet Mercury. A
few hours later the ground beneath the probe gives way and the probe
falls, landing below its original position with a velocity of 11.2 m/s
downward. If the free-fall acceleration near Mercury’s surface is
3.70 m/s2 downward, what is the probe’s displacement?
6. A ball thrown vertically is caught by the thrower after 5.1 s. Find the
maximum height the ball reaches.
7. Find the initial velocity with which the ball in problem 6 is thrown.
8. An archer fires an arrow directly upward, then quickly runs from the
launching spot to avoid being struck by the returning arrow. If the
arrow’s initial velocity is 85.1 m/s upward how long does the archer
have to run away before the arrow lands?
9. A popular scene in recent action films shows a character in free-fall
speed up to catch a freely falling parachute. Suppose a packed para-
chute is dropped from rest from an airplane and that a daredevil is
launched straight down from the plane 3.00 s later. Neglecting air
resistance, the daredevil catches up to the parachute 4.00 s after the
daredevil leaves the plane. What are the daredevil’s initial and final
velocities?
10. The elevators in the Landmark Tower in Yokohama, Japan, are among
the fastest in the world. They accelerate upward at 3.125 m/s2 for 4.00 s
to reach their maximum speed. Suppose an empty elevator is moving
upward with its maximum speed when the cable breaks, so that the ele-
vator slows down, comes to a stop, and then begins to fall freely. What
will the elevator’s velocity be 0.00 s, 1.00 s, 2.00 s, and 3.00 s after the
cable breaks?
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Problem 3A Ch. 3–1
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 3AFINDING RESULTANT MAGNITUDE AND DIRECTION
P R O B L E MA hummingbird flies 9.0 m horizontally and then flies up for 3.0 m. Whatis the bird’s resultant displacement?
S O L U T I O NGiven: ∆x = 9.0 m ∆y = 3.0 m
Unknown: d = ? q = ?
Diagram:
∆x = 9.0 m
∆y = 3.0 m
y
d
xθ
Choose the equation(s) or situation: The Pythagorean theorem can be used to
find the magnitude of the hummingbird’s displacement. The direction of the dis-
placement can be found using the tangent function.
d2 = ∆x 2 + ∆y 2
tan q =
Rearrange the equation(s) to isolate the unknown(s):
d =√
∆x2+ ∆y2
q = tan−1 Substitute the values into the equation(s) and solve:
d =√
(9.0 m)2 + (3.0m)2 =√
81 m2+ 9.0 m2 =√
9.0× 101 m2
d =
q = tan−1 39..00
mm
q =
The resultant displacement (d) is only slightly larger than the largest component
(∆x), as is the case for small angles (q 20°).
18° above horizonal
9.5 m
∆y∆x
∆y∆x
1. DEFINE
2. PLAN
3. CALCULATE
4. EVALUATE
Holt Physics Problem BankCh. 3–2
NAME ______________________________________ DATE _______________ CLASS ____________________
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ADDITIONAL PRACTICE
1. A tiger paces back and forth along the front of its cage, which is 8 m
wide. The tiger starts from the right side of the cage, paces to the left
side, then back to the right side, and finally back to the left.
a. What total distance has the tiger paced?
b. What is the tiger’s resultant displacement?
2. A particular type of rubber ball is able to bounce to 0.90 times the height
from which it is dropped. The ball is dropped from a height of 0.91 m,
but it falls slightly away from the vertical, so that by the time it has
bounced to its new height it has undergone a horizontal displacement of
0.11 m. What is the ball’s resultant displacement from its initial height to
its maximum height after one bounce?
3. A helicopter flies 165 m horizontally and then moves downward to land
45 m below. What is the helicopter’s resultant displacement?
4. A toy parachute is dropped from an open window that is 13.0 m above
the ground. If the parachute travels 9.0 m horizontally, what is the resul-
tant displacement?
5. An octopus swims 36.0 m east, 42.0 m north, and then rises 17.0 m to-
ward the surface of the water. What is the octopus’s displacement?
(TWO-DIMENSIONAL METHOD: Visualize a horizontal and a vertical
triangle. Find the horizontal resultant; use that with the vertical distance
to calculate the final resultant. Studying this method can lead to under-
standing the easier three-dimensional solution in the solutions manual.)
6. An airplane taxis to the end of a runway before taking off. The magni-
tude of the plane’s total displacement is 599 m. If the northern compo-
nent of this displacement is 89 m, what is the displacement’s eastern
component? What is the direction of the total displacement?
7. The straightest stretch of railroad tracks in the world extends for 478 km
in southwestern Australia. A train traveling along these tracks is dis-
placed to the south by about 42 km. What is the train’s displacement to
the west? What is the direction of the total displacement?
8. Before the widespread use of steamships, sailing from Europe to North
America was accomplished by use of the “trade winds.” The trade winds
move from the northeast to the southwest between 30° and 60° latitude
in the northern hemisphere. A ship sailing from Europe to the Caribbean
Sea would first travel southward to the Canary Islands, off the coast of
North Africa, and then use the trade winds to sail west. Suppose a ship
travels south from Iceland to the Canary Islands, and then west to
Florida. The ship’s total displacement is 7400 km at 26° south of west. If
the ship sails 3200 km south from Iceland to the Canary Islands, how
large is the western component of its journey?
Problem 3A Ch. 3–3
NAME ______________________________________ DATE _______________ CLASS ____________________
9. The Palm Springs Aerial Tramway extends 3.88 km from the Valley Sta-
tion, which is located 0.8 km above sea level, to the Mountain Station
atop San Jacinto Peak. The actual path of the tramway’s cables is not
along a straight line, but if it were, the horizontal displacement of the
tramway would be 3.45 km. How far is San Jacinto Peak above sea
level?
10. The islands that form the Tristan da Cunha Group in the South At-
lantic Ocean are considered to be the most remote places in the world:
the next nearest inhabited island is 2400 km away. If you sail from
Capetown, South Africa, in a south by southwest direction, you must
travel 2.9 × 103 km before reaching the Tristan da Cunha islands. If the
western component of your displacement is 2.8 × 103 km, what is your
displacement south? In what direction is the resultant displacement?
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Holt Physics Problem BankCh. 3–4
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 3BRESOLVING VECTORS
P R O B L E MThe straight stretch of Interstate Highway 5 from Mettler, California, to apoint near Buttonwillow, California, is 53.0 km long and makes an angleof 48.7° north of west. What are the northern and western components ofthis highway segment?
S O L U T I O NGiven: d = 53.0 km q = 48.7° north of west
Unknown: ∆x = ? ∆y = ?
Diagram:
1. DEFINE
2. PLAN
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∆x
∆y
θ = 48.7°
N
d = 53.0 km
Choose the equation(s) or situation: Because the axes are perpendicular, the
sine and cosine functions can be used to find the components.
sin q = ∆dy
cos q = ∆dx
Rearrange the equation(s) to isolate the unknown(s):
∆x = d (cos q)
∆y = d (sin q)
Substitute the values into the equation(s) and solve:
∆x = (53.0 km)(cos 48.7°)
∆x =
∆y = (53.0 km)(sin 48.7°)
∆y =
Using the Pythagorean theorem to check the answers confirms the magnitudes of
the components.
d2 = ∆x2 + ∆y2
(53.0 km)2 = (35.0 km)2 + (39.8 km)2
39.8 km, north
35.0 km, west
3. CALCULATE
4. EVALUATE
Problem 3B Ch. 3–5
NAME ______________________________________ DATE _______________ CLASS ____________________
1. The distance from an observer on the plain to the top of a nearby
mountain is 5.3 km, and the angle between this line and the horizontal
is 8.4°. How tall is the mountain?
2. A bowling ball is released at the near right corner of a bowling lane and
travels 19.1 m at an angle of 3.0° with respect to the lane’s length. The
ball reaches the far left corner of the lane, where it knocks over the “7”
pin. What is the width of the lane?
3. A skyrocket travels 113 m at an angle of 82.4° with respect to the ground
and toward the south. What is the rocket’s horizontal displacement?
4. A hot-air balloon descends with a velocity of 55 km/h at an angle of
37° below the horizontal. What is the vertical velocity of the balloon?
5. A billiard ball travels 2.7 m at an angle of 13° with respect to the long
side of the table. What are the components of the ball’s displacement?
6. One hole at a certain miniature golf course extends for about 60 m. A golf
ball on this hole travels with a velocity of 1.20 m/s at 14.0° east of north.
What are the eastern and northern components of the ball’s velocity?
7. The Very Large Array in western New Mexico consists of several radio
telescopes that can be rearranged along railroad tracks. The largest of
these arrangements has the telescopes positioned in a “Y” pattern for
18 km along three separate tracks. Suppose an electrician inspects the
instruments in each antenna from the end of the northern track to the
end of the southwestern track. If the electrician’s resultant displace-
ment is 31.2 km at 30 0° west of south, what are the southern and west-
ern components of the displacement?
8. Barnard’s Star is the closest star to Earth after the sun and the triple
star Alpha Centauri. Barnard’s Star has a velocity of 165.2 km/s at an
angle of 32.7° away from its forward motion. What are the forward and
side components of this velocity?
9. A tiger leaps with an initial velocity of 55.0 km/h at an angle of 13.0°
with respect to the horizontal. What are the components of the tiger’s
velocity?
10. A certain type of balloon is designed to ascend rapidly. Suppose this
balloon has a velocity 13.9 m/s at 26.0° above the horizontal and 24.0°
east of north. What are the upward, northern, and eastern components
of the balloon’s velocity? (HINT: Draw horizontal and vertical right tri-
angles whose sides represent the velocity’s components.)
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ADDITIONAL PRACTICE
2.80 × 103 km2 = 1.22 × 103 km2 + 1.58 × 103 km2
2.80 × 103 km2 = 2.80 × 103 km2
Holt Physics Problem BankCh. 3–6
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 3CADDING VECTORS ALGEBRAICALLY
P R O B L E MThe southernmost point in the United States is called South Point, and islocated at the southern tip of the large island of Hawaii. A plane designedto take off and land in water leaves South Point and flies to Honolulu, onthe island of Oahu, in three separate stages. The plane first flies 83.0 km at22.0° west of north from South Point to Kailua Kona, Hawaii. The planethen flies 146 km at 21.0° west of north from Kailua Kona to Kahului, onthe island of Maui. Finally, the plane flies 152 km at 17.5° north of westfrom Kahului to Honolulu. What is the plane’s resultant displacement?
S O L U T I O NGiven: d1 = 83.0 km q1 = 22.0° west of north
d2 = 146 km q2 = 21.0° west of north
d3 = 152 km q3 = 17.5° north of west
Unknown: d = ? q = ?
Diagram:
1. DEFINE
2. PLAN
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θ
θ
d
d2 = 146 km
d1 = 83.0 km
1 = 22.0°
θ2 = 21.0°
θ3 = 17.5°
d3 = 152 km N
Choose the equation(s) or situation: Express the components of each vector in
terms of sine or cosine functions.
∆x1 = d1 (sin q1) ∆y1 = d1 (cos q1)
∆x2 = d2 (sin q2) ∆y2 = d2 (cos q2)
∆x3 = d3 (cos q3) ∆y3 = d3 (sin q3)
Note that the angles q1 and q2 are with respect to the y axis (north), and so the x
components are in terms of sin q . Write the equations for ∆xtot and ∆ytot , the
components of the total displacement.
Problem 3C Ch. 3–7
NAME ______________________________________ DATE _______________ CLASS ____________________
∆xtot = ∆x1 + ∆x2 + ∆x3
= d1 (sin q1) + d2(sin q2) + d3 (cos q3)
∆ytot = ∆y1 + ∆y2 + ∆y3
= d1 (cos q1) + d2(cos q2) + d3 (sin q3)
Use the components of the total displacement, the Pythagorean theorem, and the
tangent function to calculate the total displacement.
d =√
(∆xtot)2 + (∆ytot)2
q = tan−1 Substitute the values into the equation(s) and solve:
∆xtot = (83.0 km)(sin 22.0°) + (146 km)(sin 21.0°) + (152 km)
(cos 17.5°)
= 31.1 km + 52.3 km + 145 km
= 228 km
∆ytot = (83.0 km)(cos 22.0°) + (146 km)(cos 21.0°) + (152 km)
(sin 17.5°)
= 259 km
d =√
(228 km)2 + (259km)2 =√5.20 × 104 km2+ 6.71× 104 km2 =
√11.91× 104 km2
=
q = tan−1 =
If the diagram is drawn to scale, compare the calculated results to the drawing.
The length of the drawn resultant is fairly close to the scaled magnitude for d,
while the angle appears to be slightly greater than 45°.
48.6° north of west259 km228 km
345.1 km
∆ytot∆xtot
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3. CALCULATE
4. EVALUATE
1. U.S. Highway 212 extends 55 km at 37° north of east between Newell and
Mud Butte, South Dakota. It then continues for 66 km nearly due east
from Mud Butte to Faith, South Dakota. If you drive along this part of
U.S. Highway 212, what will be your total displacement?
2. Wrigley Field is one of only three original major-league baseball fields
that are still in use today. Suppose you want to drive to Wrigley Field
from the corner of 55th Street and Woodlawn Avenue, about 14 miles
south of Wrigley Field. Although not the fastest or most direct route, the
most straightforward way to reach Wrigley Field is to drive 4.1 km west
on 55th Street to Halsted Street, then turn north and drive 17.3 km on
Halsted until you reach Clark Street. Turning on Clark, you will reach
Wrigley Field after traveling 1.2 km at an angle of 24.6° west of north.
What is your resultant displacement?
ADDITIONAL PRACTICE
3. A bullet traveling 850 m ricochets from a rock. The bullet travels an-
other 640 m, but at an angle of 36° from its previous forward motion.
What is the resultant displacement of the bullet?
4. The cable car system in San Francisco is the last of its kind that is still
in use in the United States. It was originally designed to transport large
numbers of people up the steep hills on which parts of the city are
built. If you ride seven blocks on the Powell Street cable car from the
terminal at Market Street to Pine Street, you will travel 2.00 × 102 m on
level ground, then 3.00 × 102 m at an incline of 3.0° to the horizontal,
and finally 2.00 × 102 m at 8.8° to the horizontal. What will be your re-
sultant displacement?
5. An Arctic tern flying to Antarctica encounters a storm. The tern
changes direction to fly around the storm. If the tern flies 46 km at 15°
south of east, 22 km at 13° east of south, and finally 14 km at 14° west
of south, what is the tern’s resultant displacement?
6. A technique used to change the direction of space probes, as well as to
give them additional speed, is to use the gravitational pull of nearby
planet. This technique was first used with the Voyager probes. Voyager 2
had traveled about 6.3 × 108 km when it reached Jupiter. Jupiter’s grav-
ity changed Voyager’s direction by 68°. The probe then traveled about
9.4 × 108 km when it reached Saturn, and its direction was changed by
94°. Voyager 2 was now redirected; it encountered Uranus after traveling
3.4 × 109 km from Saturn. Use this information to calculate the resul-
tant displacement of Voyager 2 as it traveled from Earth to Uranus.
7. The city of Amsterdam, in the Netherlands, has several canals that con-
nect different sections of the city. Suppose you take a barge trip to the
harbor, starting at a point near the northwest corner of the Vondel-
park. You would sail 2.50 × 103 m at 58.5° north of east, 375 m at 21.8°
north of east, and 875 m at 21.5° east of north. What would be your re-
sultant displacement?
8. The elevated train, or “L,” in Chicago is a major source for mass transit
in that city. One of the lines extends from Jefferson Park, in the north-
west part of town, to the Clark Street station downtown. The route of
this line runs 5.0 km at 36.9° south of east, 1.5 km due south, 8.5 km at
42.2° south of east, and 0.8 km due east. What is the resultant displace-
ment of an “L” train from Jefferson Park to Clark Street?
9. A billiard table is positioned with its long side parallel to north. A cue
ball is then shot so that it travels 1.41 m at an angle of 45.0° west of
north, is deflected by the table’s left side, and continues to move 1.98 m
east of north at an angle of 45.0°. The ball is then deflected by the
table’s right side, so that it moves 0.42 m west of north at an angle of
45.0°. After a reflection on the north end of the table, the ball travels
1.56 m at an angle of 45.0° south of west. Determine the resultant dis-
placement of the cue ball.
Holt Physics Problem BankCh. 3–8
NAME ______________________________________ DATE _______________ CLASS ____________________
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Problem 3C Ch. 3–9
NAME ______________________________________ DATE _______________ CLASS ____________________
10. Hurricane Iniki was the most destructive cyclone to have crossed the
Hawaiian Islands in the twentieth century. It’s path was also unusual: it
moved south of the islands for 790 km at an angle of 18° north of west,
then moved due west for 150 km, turned north and continued for
470 km, and finally turned back 15° east of north and moved 240 km
to cross the island of Kauai. What was the resultant displacement of
Hurricane Iniki?
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Holt Physics Problem BankCh. 3–10
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 3DPROJECTILES LAUNCHED HORIZONTALLY
P R O B L E MAlthough not the fastest or tallest or steepest roller coaster in the world,the “High Roller” roller coaster atop the Stratosphere Tower, in Las Vegas,Nevada, is the highest. Suppose that during construction of the ride ametal bolt was accidentally knocked horizontally off the edge of theStratosphere. If the bolt’s initial speed was 0.80 m/s, it would have trav-eled 6.76 m in the horizontal direction before hitting the ground. Use thisinformation to calculate how tall the Stratosphere Tower is.
S O L U T I O NGiven: vx = 0.80 m/s
∆x = 6.76 m
g = 9.81 m/s2
Unknown: ∆y = ?
Diagram:
1. DEFINE
2. PLAN
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∆x = 6.76 m
∆y
vx
Choose the equation(s) or situation: The magnitude of the vertical displace-
ment is given by the equation for falling bodies with no ititial vertical velocity.
∆y = − 12
g∆t 2
The magnitude for horizontal displacement is given by the equation for displace-
ment at constant velocity.
∆x = vx ∆ t
Rearrange the equation(s) to isolate the unknown(s): Substitute for ∆ t in the
falling-body equation.
∆t = ∆vx
x
∆y = − 12
g ∆vx
x2
Substitute the values into the equation(s) and solve:
∆y = − 12
(9.81 m/s2) 2
= 350 m
height of building = 350 m
6.76 m0.80 m/s
3. CALCULATE
Problem 3D Ch. 3–11
NAME ______________________________________ DATE _______________ CLASS ____________________
The solution can be checked by using both equations to solve for ∆t. From the
equation for falling bodies, ∆t, is found to be 8.4 s. From the equation for horizon-
tal displacement, ∆t is 8.4 s. Both times are the same, so ∆y is correctly calculated.
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4. EVALUATE
ADDITIONAL PRACTICE
1. Lookout Mountain, which overlooks the Tennessee River Valley near
Chattanooga, Tennessee, was of great strategic importance during the
Civil War. Today, some of the artillery used in the war remain at the park
located on top of the mountain. Suppose one of these cannons fired a
projectile horizontally with a speed of 430 m/s, so that the projectile
landed at a horizontal distance of 4020 m from the cannon. How high
would the ridge of the mountain be with respect to the valley below?
2. In 1977, a helicopter at the heliport atop the 59-story Pan Am building in
New York fell over, causing the rapidly-turning rotor blades to splinter.
One of these fragments landed about 101 m away, near the corner of
Madison Avenue and 43rd Street. Suppose the fragment moved off the
building horizontally with a speed of 14.25 m/s. Use this information to
find the height of the Pan Am building.
3. The LZ N07 is a newly designed airship in the manner of the old Zep-
pelin airships built in Germany between 1908 and 1940. New technology
has made the LZ N07 more efficient and safe, as well as speedier. This
airship can travel with a horizontal speed of up to 1.30 × 102 km/h. If a
parcel is dropped from this airship, so that it lands 135 m in front of the
spot over which it was released, how far above the ground is the airship?
4. The shape of Sugarloaf mountain, in Rio de Janeiro, Brazil, is such that, if
you were to kick a soccer ball hard enough, it could land near the base of
the mountain without hitting the mountain’s side. Suppose the ball is
kicked horizontally with an initial speed of 9.37 m/s. If the ball travels a
horizontal distance of 85.0 m, how tall is the mountain?
5. Although many structures taller than 500 m have been designed, few
have been built due to practical limitations, such as cost and safety. In
light of this, the Bionic Tower in Hong Kong may never be more than a
design. If it is built, the Bionic Tower will provide working space for
100,000 people, and transport them using over 300 elevators. Suppose a
plate-glass window falls out of place from the top floor of the Bionic
Tower. Although the window’s speed is only 6.32 cm/s in the horizontal
direction, the window will still have a horizontal displacement of 1.00 m
once it hits the street below. Use this information to calculate the pro-
posed height of the Bionic Tower.
6. A squirrel on a limb near the top of a tree loses its grip on a nut, so that
the nut slips away horizontally at a speed of 10.0 cm/s. If the nut lands at a
horizontal distance of 18.6 cm, how high above the ground is the squirrel?
Holt Physics Problem BankCh. 3–12
NAME ______________________________________ DATE _______________ CLASS ____________________
7. A lunch pail is accidentally kicked off a steel beam on a building under
construction. Suppose the initial horizontal speed is 1.50 m/s. How far
does the lunch pail fall after it travels 3.50 m horizontally?
8. If the building in problem 7 is 2.50 × 102 m tall, and the lunch pail is
knocked off the top floor, what will be the horizontal displacement of
the lunch pail when it reaches the ground?
9. What is the velocity of the lunch pail in problem 8 when it reaches the
ground?
10. What is the range of an arrow shot horizontally at 85.3 m/s if it is ini-
tially 1.50 m above the ground?
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Problem 3E Ch. 3–13
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 3EPROJECTILES LAUNCHED AT AN ANGLE
P R O B L E MA flying fish leaps out of the water with a speed of 15.3 m/s. Normallythese fish use winglike fins to glide about 40 m before reentering theocean, but in this case the fish fails to use its “wings” and so only travelshorizontally about 17.5 m. At what angle with respect to the water’s sur-face does the fish leave the water? Use the trigonometric identity2(sinq)(cosq) = sin (2q) to solve for q .
S O L U T I O NGiven: vi = 15.3 m/s
∆x = 17.5 m
g = 9.81 m/s2
Unknown: q = ?
Diagram:
1. DEFINE
2. PLAN
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∆x = 17.5 m
vi θ
Choose the equation(s) or situation: The horizontal component of the fish’s
initial velocity, vx, is equal to the horizontal displacement divided by the time of
the jump.
vx = vi (cos q) =
The vertical displacement of the fish is given by the equation for falling bodies,
with the vertical component of the initial velocity, vy , used.
∆y = vy∆t − 12
g∆t 2
Because the fish lands at the same vertical position from which it started, ∆y = 0.
∆y = 0
vy = vi(sin q) = 12
g∆t
Rearrange the equation(s) to isolate the unknowns: Substitute for ∆t using the
equation for horizontal velocity.
∆t =
vi (sin q) = 12
g (sin q)(cos q) =
Using the trigonometric identity allows a solution for q to be found.
(sin q)(cos q) = 12
[sin (2q)]
sin (2q) = g∆xvi
2
g∆x2vi
2
∆xvi (cos q)
∆xvi (cos q)
∆x∆t
Holt Physics Problem BankCh. 3–14
NAME ______________________________________ DATE _______________ CLASS ____________________
q =
q =
=
Substituting the value for q into the original equations and solving for ∆t pro-
duces a time of 1.25 s for both, thus confirming the result for q .
23.6° above the horizontal
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sin−1 2
g∆xvi
2
sin−1 2
(9.81 m/s2)(17.5 m)
(15.3 m/s)2
3. CALCULATE
4. EVALUATE
ADDITIONAL PRACTICE
1. A baseball is thrown with an initial speed of 15.0 m/s. If the ball’s hori-
zontal displacement is 17.6 m, at what angle with respect to the ground is
the ball pitched? Use the trigonometric identity 2(sin q)(cos q) = sin
(2q) to solve for q .
2. A football is kicked so that its initial speed is 23.1 m/s. If the football
reaches a maximum height of 16.9 m, at what angle with respect to the
ground is the ball kicked?
3. Jackie Joyner-Kersee’s record long jump is 7.49 m. Suppose she ran
9.50 m/s to jump this horizontal distance. At what angle above the hori-
zontal did she jump? Use the trigonometric identity 2(sin q)(cos q) =sin (2q) to solve for q .
4. The small jumping spiders make up for their size by their ability to leap
relatively large distances. Some can jump fifty times the length of their
bodies. Suppose a jumping spider leaps a horizontal distance of 18.5 cm
with an initial speed of about 141 cm/s. At what angle above the horizon-
tal would a spider with this speed have to leap in order to travel a range
of 18.5 cm? Use the trigonometric identity 2(sin q)(cos q) = sin (2q) to
solve for q .
5. Olympic platform divers jump from a diving board that is 10.0 m above
the water. Suppose a diver jumps from the board with an initial speed of
6.03 m/s. The diver reaches a maximum height of 11.7 m above the
water, and lands in the water at a horizontal distance of 3.62 m from the
end of the board. At what angle with respect to the board does the diver
leave the board?
6. A ball is thrown from a roof with a speed of 10.0 m/s and an angle of
37.0° with respect to the horizontal. What are the vertical and horizontal
components of the ball’s displacement 2.5 s after it is thrown?
7. A downed pilot fires a flare from a flare gun. The flare has an initial
speed of 250 m/s and is fired at an angle of 35° to the ground. How long
does it take for the flare to reach its maximum altitude?
Problem 3E Ch. 3–15
NAME ______________________________________ DATE _______________ CLASS ____________________
8. In the sport of ski jumping, a skier travels down the slope of a hill until
he or she reaches the takeoff. The takeoff is slanted slightly below the
horizontal, so that the skier is able to travel in the air just above the
ground. Suppose a skier leaves the takeoff and lands 73.0 m horizon-
tally beyond the takeoff and −52.8 m below the takeoff. If the takeoff
angle is −8.00° below the horizontal, what is the skier’s initial speed?
9. A shingle slides down a roof having a 30.0° pitch and falls off with a
speed of 2.0 m/s. How long will it take to hit the ground 45 m below?
10. A hole at a miniature golf course requires the ball to roll up a ramp, fly
over a small stream, and then land on the green beyond the stream.
The stream is 0.46 m wide, and the cup is 4.00 m beyond the stream’s
edge. The ramp makes an angle of 41.0° with the horizontal, and its
upper edge is 0.35 m above the green. What must the ball’s initial speed
be in order for the ball to fly over the water and land directly in the
cup?
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Holt Physics Problem BankCh. 3–16
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 3FRELATIVE VELOCITY
P R O B L E MA polar bear swims 2.60 m/s south relative to the water. The bear is swim-ming against a current that moves 0.78 m/s at an angle of 40.0° north ofwest, relative to Earth. How long will it take the polar bear to reach theshore, which is 5.50 km to the south?
S O L U T I O NGiven: vbc = 2.60 m/s due south (velocity of the bear, b, with respect to
the current, c)
vce = 0.78 m/s at 40.0° north of west (velocity of the current, c,
with respect to Earth, e)
∆y = 5.50 km, south
Unknown: ∆t = ?
Diagram:
1. DEFINE
2. PLAN
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N
θce = 40.0°vce
vbcvbe
Choose the equation(s) or situation: To find vbe , write the equation so that the
subscripts of the vectors on the right begin with b and end with e.
vbe = vbc + vce
Because vectors vbc and vce are not perpendicular, their x and y components
must be calculated. Aligning the positive y axis with north and treating west as
the positive x direction for convenience, the following equations apply for the
magnitude of the components of vbe .
vx,be = vx,bc + vx,ce = vx,ce = vce (cos q ce)
vy,be = vy,bc + vy,ce = −vbc + vce (sin q ce)
From these components the magnitude and direction of vbe could be found
from the Pythagorean theorem and the tangent function, respectively. However,
only the component vy,be is needed to calculate the time required for the bear to
swim in the negative y direction.
∆t = −∆yvy,be
Problem 3F Ch. 3–17
NAME ______________________________________ DATE _______________ CLASS ____________________
Rearrange the equation(s) to isolate the unknown(s):
∆t = =
Substitute the values into the equation(s) and solve:
∆t =
= =
∆t =
Without the current, the polar bear would arrive about 500 s or 8.3 min sooner.
The 500 s delay is about one fourth (25%) of the bear’s swimming time without
the current. This proportion is equal to the ratio of the current’s northern com-
ponent to the bear’s velocity to the south.
2.62 × 103 s, or 43 min 40 s
−5.50 × 103 m−2.10 m/s
−5.50 km(−2.60 m/s + 0.50 m/s)
−5.50 km(−2.60 m/s + 0.78 m/s)(sin 40.0°)]
−∆y[−vbc + vce (sin q ce)]
−∆yvy,be
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3. CALCULATE
4. EVALUATE
ADDITIONAL PRACTICE
1. A bird flies directly into a wind. If the bird’s forward speed relative to the
wind is 58.0 km/h and the wind’s speed in the opposite direction is
55.0 km/h, relative to Earth, how long will it take the bird to fly 1.4 km?
2. A moving walkway at an airport has a velocity of 1.50 m/s to the west. A
man rushing to catch his flight runs down the walkway with a velocity of
4.20 m/s to the west relative to the walkway. If the walkway if 8.50 ×102 m long, how much time does the man save by running on the walk-
way as opposed to running on a non-moving surface?
3. The greatest average speed for a race car in the Daytona 500 is 286 km/h,
which was achieved in 1980. Suppose a race car moving at this speed is in
second place, being 0.750 km behind a car that is moving at a speed of
252 km/h. How long will it take the second-place car to catch up to the
first-place car?
4. A mosquito can fly with a speed of 1.10 m/s with respect to the air. Sup-
pose a mosquito flies east at this speed across a swamp. The mosquito is
flying into a breeze that has a velocity of 5.0 km/h with respect to Earth
and moves 35° west of south. If the swamp is 540 m across, how long will
it take the mosquito to cross the swamp?
5. A glider descends with a velocity relative to the air of 150 km/h at an
angle of 7.0° below the horizontal. Suppose that the glider encounters an
updraft with a velocity relative to Earth of 15 km/h upward. How long
will it take the glider to reach the ground if it encounters the updraft at
166 m? How long would it take for the glider to land without the updraft?
6. A flare gun is mounted on an automobile and fired perpendicular to the
car’s motion. The car’s velocity with respect to Earth is 145 km/h to the
north. The flare’s velocity with respect to the car is 87 km/h to the west.
What are the components of the flare’s displacement with respect to
Earth 0.45 s after the flare is launched?
Holt Physics Problem BankCh. 3–18
NAME ______________________________________ DATE _______________ CLASS ____________________
7. An airship moving north at 55.0 km/h with respect to the air encoun-
ters a wind from 17.0° north of west. If the wind’s speed with respect to
Earth is 40.0 km/h, what is the airship’s velocity with respect to Earth?
8. How far to the north and west does the airship in problem 7 travel after
15.0 minutes?
9. A torpedo fired at an anchored target moves against a current. Suppose
the torpedo’s velocity with respect to the current is 51 km/h east, and
the current’s velocity with respect to the target is 4.0 km/h south. If the
torpedo hits the target in 14 s, how far away is the target from the point
where the torpedo is launched? How far north of the target must the
torpedo be launched in order to hit the target?
10. A sailboat travels south with a speed of 12.0 km/h with respect to the
water. Suppose the boat encounters a current that has a velocity with
respect to Earth of 4.0 km/h at 15.0° south of east. What is the sail-
boat’s resultant velocity with respect to Earth?
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Problem 4A Ch. 4–1
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 4ANET EXTERNAL FORCE
P R O B L E MTwo soccer players kick a ball at the same instant. One player kicks with aforce of 65 N to the north, while the other player kicks with a force of88 N to the east. In what direction does the ball travel?
S O L U T I O NGiven: F1 = 65 N north
F2 = 88 N east
Unknown: q = ?
Diagram: F1 = 65 Ν
F2 = 88 Ν
Ν
Select a coordinate system and apply it to the free-body diagram. Choose the
positive x-axis to align with east and the positive y-axis to align with north.
Find the x and y components of all vectors.
F1,x = 0 N F1,y = 65 N
F2,x = 88 N F2,y = 0 N
Find the net external force in both the x and y directions.
Fx,net = ΣFx = F1,x + F2,x = 0 N + 88 N = 88 N
Fy,net = ΣFy = F1,y + F2,y = 65 N + 0 N = 65 N
Find the direction of the net external force. Use the tangent function to find
the angle q of Fnet.
q = tan−1FFx
y
,
,n
n
e
e
t
t = tan−1685
8
N
N = 36°
q =
The direction is about three-fourths of the way to the midpoint (45°) between
north and east. This corresponds closely to the ratio of 65 N to 88 N (0.74).
1. Two tugboats pull a barge across the harbor. One boat exerts a force of
7.5 × 104 N north, while the second boat exerts a force of 9.5 × 104 N at
15.0° north of west. Precisely, in what direction does the barge move?
2. Three workers move a car by pulling on three ropes. The first worker ex-
erts a force of 6.00 × 102 N to the north, the second a force of 7.50 × 102 N
to the east, and the third 6.75 × 102 N at 30.0° south of east. In what pre-
cise direction does the car move?
36° north of east
1. DEFINE
2. PLAN
3. CALCULATE
4. EVALUATE
ADDITIONAL PRACTICE
Holt Physics Problem BankCh. 4–2
NAME ______________________________________ DATE _______________ CLASS ____________________
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3. Four forces are acting on a hot-air balloon: F1 = 2280.0 N up, F2 = 2250.0 N
down, F3 = 85.0 N west, and F4 = 12.0 N east. What is the precise direction
of the net external force on the balloon?
4. What is the magnitude of the largest net force that can be produced by com-
bining a force of 6.0 N and a force of 8.0 N? What is the magnitude of the
smallest such force?
5. Two friends grab different sides of a videotape cartridge and pull with forces
of 3.0 N to the east and 4.0 N to the south, respectively. What force would a
third friend need to exert on the cartridge in order to balance the other two
forces? What would be that force’s precise direction?
6. A four-way tug-of-war has four ropes attached to a metal ring. The forces on
the ring are as follows: F1 = 4.00 × 103 N east, F2 = 5.00 × 103 N north,
F3 = 7.00 × 103 N west, and F4 = 9.00 × 103 N south. What is the net force on
the ring? What would be that force’s precise direction?
7. A child pulls a toy by exerting a force of 15.0 N on a string that makes an
angle of 55.0° with respect to the floor. What are the vertical and horizontal
components of the force?
8. A shopper pushes a grocery cart by exerting a force on the handle. If the
force equals 76 N at an angle of 40.0° below the horizontal, how much force
is pushing the cart in the forward direction? What is the component of force
pushing the cart against the floor?
9. Two paramedics are carrying a person on a stretcher. One paramedic exerts a
force of 350 N at 58° above the horizontal and the other paramedic exerts a
force of 410 N at 43° above the horizontal. What is the magnitude of the net
upward force exerted by the paramedics?
10. A traffic signal is supported by two cables, each of which makes an angle
of 40.0° with the vertical. If each cable can exert a maximum force of
7.50 × 102 N, what is the largest weight they can support?
Problem 2A Ch. 4–3
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 4BNEWTON’S SECOND LAW
P R O B L E MTwo students reach for a jar of mustard at the same time. One studentpulls to the left with a force of 13.2 N, while the other student pulls to theright with a force of 12.9 N. If the jar has a net acceleration of 0.44 m/s2 tothe left, what is the mass of the jar?
S O L U T I O NGiven: F1 = 13.2 N to the left
F2 = 12.9 N to the right
anet = 0.44 m/s2 to the left
Unknown: m = ?
Use Newton’s second law and solve for m.
ΣF = m a = m anet
ΣF = F1 + F2 = F1 − F2 to the left
F1 − F2 = 13.2 N − 12.9 N = 0.30 N
m = a
Σ
n
F
et =
0.
0
4
.
4
30
m
N
/s2
m =
1. A house is lifted from its foundations onto a truck for relocation. The
house is pulled upward by a net force of 2850 N. This force causes the
house to move from rest to an upward speed of 15 cm/s in 5.0 s. What
is the mass of the house?
2. Suppose an empty grocery cart rolls downhill in a parking lot. The cart
undergoes a constant increase in speed of 1.0 m/s over a 5.0 s interval.
If the downhill force acting on the cart is 18.0 N and the uphill force
due to friction is 15.0 N, what is the cart’s mass?
3. A certain cable of an elevator is designed to exert a force of 4.5 × 104 N. If
the maximum acceleration that a loaded car can withstand is 3.5 m/s2
(the current fastest elevators in the world undergo an acceleration of less
than 3.2 m/s2), what is the combined mass of the car and its contents?
4. An 2.0-kg fish pulled upward by a fisherman rises 1.9 m in 2.4 s, start-
ing from rest. Assuming the acceleration is constant, find the magnitude
and direction of the net force acting on the fish during this interval.
0.68 kg
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ADDITIONAL PRACTICE
Holt Physics Problem BankCh. 4–4
NAME ______________________________________ DATE _______________ CLASS ____________________
5. An 8.0-kg bag of coins is being pulled upward by a rope rises 20.0 cm
in 0.50 s, starting from rest. Assuming the acceleration is constant, cal-
culate the net force on the bag. What is the upward force on the bag ex-
erted by the rope?
6. A pedestrian with a mass of 75 kg accelerates at 0.15 m/s2 to the west.
A high wind comes up, blowing toward the east. The wind is capable of
giving the pedestrian an acceleration of 2 × 10−2 m/s2. What are the
magnitude and direction of the net force acting on the pedestrian?
7. Assume that a catcher in a professional baseball game exerts a force of
−65.0 N to stop the ball. If the baseball has a mass of 0.145 kg, what is
its net acceleration as it is being caught?
8. A 214 kg boat is sinking in the ocean. The boat’s weight is partially off-
set by the 790 N buoyant force of the water. What is the net accelera-
tion of the boat?
9. The Goliath beetle, which is found in Africa, can reach a mass of
0.080 kg. Suppose a Goliath beetle is placed on a slope that makes an
angle of 37.0° with the horizontal. Find the acceleration of the beetle
along the slope, assuming the slope to be frictionless.
10. If an force of 1.40 N upward along the slope is applied to the beetle in
problem 9, what is the beetle’s acceleration?
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Problem 4C Ch. 4–5
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 4CCOEFFICIENTS OF FRICTION
P R O B L E MA cabinet initially at rest on a horizontal surface requires a 115 N hori-zontal force to set it in motion. If the coefficient of static friction betweenthe cabinet and the floor is 0.38, what is the normal force exerted on thecabinet? What is the mass of the cabinet?
S O L U T I O NGiven: Fs,max = 115 N
ms = 0.38
g = 9.81 m/s2
Unknown: Fn = ? m = ?
Use the equation for the coefficient of static friction to find Fn.
ms = Fs
F, m
n
ax
Fn = Fs,
mm
s
ax = 1
0
1
.
5
38
N = 3.0 × 102 N
Fn =
Use the definition for the normal force to find m.
Fn = mg, for a horizontal surface
m = F
gn =
3
9
.0
.8
×1
1
m
0
/
2
s2N
m =
1. A ship launched from a dry-dock slides into the water at a constant ve-
locity. Suppose the force of gravity that pulls the ship downward along
the dry-dock is 4.26 × 107 N. If the coefficient of kinetic friction be-
tween the ship’s hull and the dry-dock is 0.25, what is the magnitude of
the normal force that the dry-dock exerts on the ship’s hull?
2. If the incline of the dry-dock in problem 1 is 10.0°, what is the ship’s
mass?
3. A frictional force of 2400 N keeps a crate of machine parts from sliding
down a ramp. If the coefficient of static friction between the box and
the ramp is 0.20 and the incline of the ramp is 30.0°, what is the nor-
mal force of the ramp on the box? What is the mass of the crate of ma-
chine parts?
31 kg
3.0 × 102 N upward
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ADDITIONAL PRACTICE
Holt Physics Problem BankCh. 4–6
NAME ______________________________________ DATE _______________ CLASS ____________________
4. A passenger with a mass of 60.0 kg is standing in a subway car that is
accelerating at 3.70 m/s2. If the coefficient of static friction between the
passenger’s shoes and the car floor is 0.455, will the passenger be able
to stand without sliding?
5. A 90.0 kg skier glides down a slope with an incline of 17.0°. What fric-
tional force is needed for the skier to move at a constant velocity?
6. A dogsled with a mass of 47 kg is loaded with 33 kg of supplies. If the
coefficient of kinetic friction between the sled’s runners and the snow
is 0.075, what is the magnitude of the frictional force on the sled as it
moves across flat ground? What is the magnitude of the frictional force
on the sled as it moves up a hill with a 15° incline?
7. A car with a mass of 1.8 × 103 kg is parked on a hill in San Francisco. Sup-
pose the hill makes a 15.0° incline with the horizontal. If the frictional
force required to keep the car from sliding down the hill is 1.25 × 104 N,
what is the coefficient of static friction between the pavement and the
car’s tires?
8. A metal disk with a mass of 15.0 g slides along a metal sheet. Both the
disk and sheet have been coated with a substance that reduces fric-
tional forces. If the sheet needs to be tilted only 2.3° for the disk to slide
down the sheet with a constant velocity, what is the coefficient of ki-
netic friction between the disk and sheet?
9. In 1994, a commercial automobile accelerated from rest to 88.0 km/h
in 3.07 s. Cars accelerate because of traction, which in turn depends on
the force of static friction between the rubber of their tires and the
road. If the force of acceleration is entirely provided by static friction
between the tires and pavement (an overly simplified assumption), cal-
culate the coefficient of static friction between the tires and the road.
10. A snowboarder slides down a 5.0° slope at a constant speed. What is
the coefficient of kinetic friction between the snow and the board?
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Problem 4D Ch. 4–7
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 4DFINAL VELOCITY AFTER ANY DISPLACEMENT
P R O B L E MA bicyclist riding in the rain suddenly applies the brakes and slides to astop. If the acceleration is −9.5 m/s2, what is the coefficient of kinetic fric-tion between the bicycle’s rubber tires and the wet concrete?
S O L U T I O N
Given: anet = −9.5 m/s2
g = 9.81 m/s2
Unknown: mk = ?
Choose the equation(s) or situation: Use Newton’s second law to describe the
forces acting on the bicycle.
Fnet = m anet = −Fk
Use the definition of frictional force to express Fk in terms of the coefficient of
friction.
Fk = mk Fn = mk (mg)
Rearrange the equation(s) to isolate the unknown(s):
m anet = −mkmg
mk = − an
get
Substitute the values into the equation(s) and solve:
mk = −(
9
−.8
9
1
.5
m
m
/s
/2s2)
mk =
The coefficient of static friction for rubber and most surfaces is high. This is indi-
cated by the value for rubber and wet concrete. Even under these conditions, ms is
nearly 1.
1. Blocks of ice are slid down a metal chute with an incline of 12.0° above
the horizontal. The blocks undergo a constant acceleration of 1.22 m/s2.
What is the coefficient of kinetic friction between the ice and the chute?
2. A force of 1760 N is required to start moving a bundle of wooden
planks up a ramp. If the ramp’s incline is 17° and the mass of the
planks is 266 kg, what is the coefficient of static friction between the
planks and the ramp?
0.97
1. DEFINE
2. PLAN
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3. CALCULATE
4. EVALUATE
ADDITIONAL PRACTICE
Holt Physics Problem BankCh. 4–8
NAME ______________________________________ DATE _______________ CLASS ____________________
3. A bundle of bricks is pulled up a ramp to a construction site. The bun-
dle has a mass of 5.1 × 102 kg, and the incline of the ramp is 14°. If the
minimum force needed to move the bricks up the ramp is 4.1 × 103 N,
what is the coefficient of static friction between the bricks and the
ramp?
4. A force of 5.00 N to the left causes a 1.35 kg book to have a net acceler-
ation of 0.76 m/s2 to the left. What is the frictional force acting on the
book?
5. A jar is slid horizontally across a smooth table. If the coefficient of ki-
netic friction between the jar and the table is 0.20, what is the magni-
tude of the jar’s acceleration?
6. A skier is pulled by an applied force of 2.50 × 102 N up a slope with an
incline of 18.0°. If the combined mass of the skier and skis is 65.0 kg
and the net acceleration uphill is 0.44 m/s2, what is the frictional force
between the skis and the snow?
7. If the skier in problem 6 skis down the same hill, what will the skier’s
acceleration be?
8. A crate is pushed across a level floor by a force of 3.00 × 102 N exerted
at an angle of 20.0° below the horizontal. The coefficient of kinetic
friction between the crate and floor is 0.250. If the crate’s velocity is
constant, what is the magnitude of the normal force exerted on the
crate by the floor? What is the mass of the crate?
9. A horse must exert a force of 590 N just to keep a sleigh from sliding
down a snowcovered hill. The component of the sleigh’s weight down
the slope of the hill is 950 N, and the coefficient of static friction be-
tween the sleigh’s runners and the snow is 0.095. What is the normal
force exerted by the ground on the sleigh? What is the sleigh’s mass if
the slope of the hill is 14.0°?
10. A freight elevator accelerates upward at 1.20 m/s2. A crate is lifted in-
side the elevator. In order to move the crate along the floor of the ele-
vator, a worker must exert a force of 1.50 × 103 N at an angle of 10.0°
below the horizontal on the upper corner of the crate. If the coefficient
of static friction is 0.650, what is the normal force that the elevator
floor exerts on the crate? What is the crate’s mass?
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Problem 5A Ch. 5–1
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 5AWORK
P R O B L E MA girl playing tug-of-war with her dog pulls the dog a distance of 8.0 m byexerting a force at an angle of 18° with the horizontal. If the amount ofwork the girl does in pulling the dog is 190 J, what is the magnitude of theforce?
S O L U T I O NGiven: W = 190 J
d = 8.0 m
q = 18°
Unknown: F = ?
Use the equation for work done by a constant force, and rearrange it to solve for F.
W = Fd (cos q)
F = d(c
W
os q) =
(8.0 m
1
)
9
(
0
co
J
s 18°)
F = 25 N
1. A roller coaster must do work raising its cars to the highest point on the
ride. From there, the cars coast at varying speed until they return to the
starting point. Suppose a loaded roller coaster car must be pulled 3.00 ×102 m from the ride’s starting point to the top of the first rise. If 2.13 ×106 J of work must be done on the car during this stage of the ride, how
large is the force exerted on the car by the raising mechanism?
2. A building under construction requires building materials to be raised
to the upper floors by cranes or elevators. An amount of cement is
lifted 76.2 m by a crane, which exerts a force on the cement that is
slightly larger than the weight of the cement. If the net work done on
the cement is 1.31 × 103 J, what is the magnitude of the net force
exerted on the cement?
3. Two workers load identical refrigerators into identical trucks by differ-
ent methods. One worker has the refrigerator lifted upward onto the
back of the truck, which is 1.5 m above the ground. The other worker
uses a ramp to slide the refrigerator onto the back of the truck. The
ramp is 5.0 m long, and raises the refrigerator 1.5 m above the ground.
The amount of work done by both workers is the same: 1800 J. What
are the magnitudes of the forces each worker must exert to load the
refrigerators?
ADDITIONAL PRACTICE
Holt Physics Problems BankCh. 5–2
NAME ______________________________________ DATE _______________ CLASS ____________________
4. A sunken treasure has a mass of 2140 kg, most of which is due to silver
and gold coins. In order to make it easier to raise the treasure, a diver
descends 17 m to where the treasure is located and attaches balloon-
like bladders to each corner of the treasure chest. The diver then in-
flates these bladders, so they provide buoyancy to the chest. The chest
is still too heavy to float upward, but its weight has been largely
counteracted by the inflated bladders, so that now it can be easily lifted
by 4.27 × 103 J of work. What is the magnitude of the net force that is
exerted on the treasure in order to raise it to the water’s surface?
5. A wrench slides off a tilted shelf, although if a force of 1.6 N is applied
opposite the wrench’s motion the wrench will slide down the shelf with
a constant velocity. If the shelf is 1.2 m long, what is the work done by
the applied force on the wrench?
6. A girl pulls a wagon along a level path for a distance of 15.0 m. The
handle of the wagon makes an angle of 20.0° with the horizontal, and
the girl exerts a force of 35.0 N on the handle. Friction provides a force
of 24.0 N. Find the net work that is done on the wagon.
7. In 1947, a deceleration sled was built to test the effects of extreme
forces on humans and equipment. In this sled, a test pilot undergoes a
sudden negative acceleration of about 50.0 times free-fall acceleration
(g). In 0.181 s, through a distance of 8.05 m, the pilot’s speed decreases
from 88.9 m/s to 0 m/s. If the pilot’s mass is 70.0 kg, how much work is
done against the pilot’s body during the deceleration?
8. A car has run out of gas. Fortunately, there is a gas station nearby. You
must exert a force of 715 N on the car in order to move it. By the time
you reach the station, you have done 2.72 × 104 J of work. How far have
you pushed the car?
9. A catcher picks up a baseball from the ground. If the net upward force
on the ball is 7.25 × 10–2 N and the net work done lifting the ball is
4.35 × 10–2 J, how far is the ball lifted?
10. At the 1996 Summer Olympics in Atlanta, Georgia, a mass of 260 kg
was lifted for the first time ever in a clean-and-jerk lift. The lift, per-
formed by Russian weightlifter Andrei Chemerkin, earned him the un-
official title as “the world’s strongest man.” If Chemerkin did 6210 J of
work in exerting a force of 2590 N, how high did he lift the mass?
Problem 5B Ch. 5–3
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 5BKINETIC ENERGY
P R O B L E M
A 2.00 g projectile has a speed of 3.00 102 m/s. What is its kinetic energy?
S O L U T I O NGiven: m = 2.00 g
v = 3.00 × 102 m/s
Unknown: KE = ?
Use the kinetic energy equation to solve for KE.
KE = 12
mv 2 = 12
(2.00 × 10–3 kg)(3.00 × 102 m/s)2
KE = 90.0 J
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ADDITIONAL PRACTICE
1. The Queen Mary was one of the largest ocean liners of the mid-twentieth
century, crossing the Atlantic Ocean 1000 times. The ship is now a
tourist attraction at Long Beach, California. Given that the mass of the
Queen Mary is 7.5 × 107 kg and her maximum cruising speed was
57 km/h, what would be the kinetic energy of the ship at maximum speed?
2. The fastest speed achieved on Earth for any object, with the exception
of sub-atomic particles in particle accelerators, is 15.8 km/s. A device at
Sandia Laboratories in Albuquerque, New Mexico, uses highly com-
pressed air to accelerate a small metal disk to supersonic speeds. Sup-
pose the disk has a mass of 0.20 g. What is the maximum kinetic
energy of the disk?
3. Although ungraceful on land, walruses are fine swimmers. They nor-
mally swim at 7 km/h, and for short periods of time are capable of
reaching speeds of nearly 35 km/h. If a walrus swimming at a speed of
35.0 km/h has a mass of 9.00 × 102 kg, what is its kinetic energy?
4. The Shinkansen, Japan’s high-speed trains, have been in service since
1964. Since that time, several train designs have been developed. Most
of these trains travel between 240 km/h and 285 km/h. The exceptions
are the “0” series, which began service in 1964, and the “500” series,
which began service in 1997. Series 0 trains travel up to 220.0 km/h
and have a total mass of about 8.84 × 105 kg. The lighter, streamlined
series 500 trains travel up to 320.0 km/h, and have an estimated total
mass of about 4.80 × 105 kg. What are the maximum kinetic energies
that can be achieved by each of these trains?
Holt Physics Problems BankCh. 5–4
NAME ______________________________________ DATE _______________ CLASS ____________________
5. The most massive of the Shinkansen are the series 200 trains, yet they
are among the fastest. Series 200 trains can reach speeds of 275 km/h. If
a 16-car series 200 train has a maximum kinetic energy of 2.78 × 109 J,
what is its mass?
6. The largest airplane built that has flown more than once is the Ukrainian-
built Antonov-225 Mriya. With a length of 85 m and a wingspan of 88 m,
the Mriya (Dream) was designed to carry the space shuttle of the Soviet
Union’s space program. Unloaded, the top speed of Mriya is 850 km/h, at
which point its kinetic energy is 9.76 × 109 J. What is its mass?
7. Though slow on land, the leatherback turtle holds the record for the
fastest water speed of any reptile: 9.78 m/s. It is also among the largest
of reptiles. Suppose the largest leatherback yet discovered were to swim
at the top leatherback speed. If its kinetic energy was 6.08 × 104 J, what
was its mass?
8. At the time a 55.0 kg skydiver jumps from a plane, her speed steadily
increases until air resistance provides a force that balances that due to
free-fall. How fast is the skydiver falling if her kinetic energy at the
moment is 7.81 × 104 J?
9. The kinetic energy of a golf ball is measured to be 1433 J. If the golf
ball has a mass of about 47.0 g, what is the ball’s speed?
10. A running student has half the kinetic energy that his younger brother
has. The student speeds up by 1.3 m/s, at which point he has the same
kinetic energy as his brother. If the student’s mass is twice as large as
his brother’s mass, what were the original speeds of both the student
and his brother? (See Appendix A of the text for hints on solving
quadratic equations.)
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Problem 5C Ch. 5–5
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 5CWORK-KINETIC ENERGY THEOREM
P R O B L E MA forward force of 11.0 N is applied to a loaded cart over a distance of15.0 m. If the cart, which is initially at rest, has a final speed of 1.98 m/s,what is the combined mass of the cart and its contents?
S O L U T I O NGiven: Fapplied = 11.0 N
d = 15.0 m
q = 0°vi = 0 m/s
vf = 1.98 m/s
Unknown: m = ?
Diagram:
Choose the equation (s) or situation: The net work done on the cart can be ex-
pressed by using the definition of work in terms of net force. Because the force is
in the same direction as the cart’s displacement (q = 0°), the net work is simply
the product of the net force and the distance the cart is pushed. The net work can
also be explained in terms of changing kinetic energy by using the work-kinetic
energy theorem.
Wnet = Fnetd(cos q) = Fnetd
Wnet = ∆KE = KEf − KEi = 12
mvf2 − 1
2mvi
2
The net force on the cart is equal to the applied force. By inserting Fapplied into
the equation for Wnet and using the work-kinetic energy equivalence, the follow-
ing equation is obtained.
Fappliedd = 12
m(vf2 − vi
2)
Rearrange the equation(s) to isolate the unknown (s):
m = 2
v
F
f
a2p
−pli
ve
i
d2d
Substitute the values into the equation(s) and solve:
m =
m =
m = 84.2 kg
(2)(11.0 N)(15.0 m)
(1.98 m/s)2
(2)(11.0 N)(15.0 m)(1.98 m/s)2 − (0 m/s)2C
opyr
ight
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Hol
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ineh
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nd W
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.
1. DEFINE
2. PLAN
3. CALCULATE
Fapplied
v
d
Holt Physics Problems BankCh. 5–6
NAME ______________________________________ DATE _______________ CLASS ____________________
1. A hockey puck with an initial speed of 8.0 m/s coasts 45 m to a stop
across the ice. If the force of friction on the puck has a magnitude of
0.12 N, what is the puck’s mass?
2. A meteoroid is a small fragment of rock that orbits a planet or the sun.
When a meteoroid enters a planet’s atmosphere, it most likely will burn
up entirely, glowing brilliantly as it does so. It is then referred to as a
meteor. Consider a meteoroid that has an initial speed of 15.00 km/s
when it enters the thin upper region of Earth’s atmosphere. Suppose
this meteoroid encounters a force of resistance with a magnitude of
9.00 × 10−2 N, so that after it travels 500.0 km parallel to Earth’s surface
the meteoroid’s speed is 14.97 km/s. Assume that the meteoroid does
not lose any mass as its temperature increases, and that the change in
the gravitational potential energy is negligible. What is the mass of the
meteoroid?
3. A car moving at a speed of 48.0 km/h accelerates 100.0 m up a steep
hill, so that at the top of the hill its speed is 59.0 km/h. If the car’s mass
is 1100 kg, what is the magnitude of the net force acting on it?
4. A 450 kg compressor slides down a loading ramp that is 7.0 m long. Ini-
tially at rest, the compressor’s speed at the bottom of the ramp is 1.1 m/s.
What is the magnitude of the net force acting on the compressor?
5. The force that stops a fighter jet as it lands on the flight deck of an
aircraft carrier is provided by a series of arresting cables. These cables
are attached to large springs that stretch enough to keep the plane from
slowing down too suddenly. Suppose a Hornet jet traveling with an
initial speed of 2.40 × 102 km/h lands on the flight deck, where it is
brought to rest by a net acceleration of magnitude 30.8 m/s2. If the jet’s
mass is 1.30 × 104 kg, how far does the jet travel during its deceleration?
6. A 50.0 kg parachutist falls at a speed of 47.00 m/s when the parachute
opens. The parachutist’s speed upon landing is 5.00 m/s. How much
work is done by the air to reduce the parachutist’s speed?
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ADDITIONAL PRACTICE
Note that the form of the equation is equivalent to Newton’s second law, with ac-
celeration given by the kinematic equation.
a = vf
2
2
−d
vi2
4. EVALUATE
Problem 5C Ch. 5–7
NAME ______________________________________ DATE _______________ CLASS ____________________
7. The giant sequoia redwood trees of the Sierra Nevada in California are
said never to die from old age. Instead, an old tree dies when its shallow
roots become loosened and the tree falls over. Removing a dead mature
redwood from a forest is no easy feat, as the tree can have a mass of
nearly 2.0 × 106 kg. Suppose a redwood with this mass is lifted 7.5 m
with a net upward acceleration of 7.5 × 10–2 m/s2. If the tree’s initial ki-
netic energy is zero, what is the final kinetic energy?
8. An applied force of 92 N is exerted horizontally on an 18 kg box of
books. The coefficient of kinetic friction between the floor and the box
is 0.35. If the box is initially at rest with zero kinetic energy, what is the
final kinetic energy after it has been moved 7.6 m across the floor?
9. A 2.00 × 102 kg iceboat is propelled across the horizontal surface of a
frozen lake by the wind. The wind exerts a constant force of 4.00 × 102 N while the boat moves 0.90 km. Assume that frictional forces are
negligible and that the boat starts from rest. What is the boat’s final
speed?
10. A certain firework is made of a small cardboard tube with a mass of
about 20.0 g. When lit, the tube slides 2.5 m across a smooth surface. If
the forward force on the tube is 7.3 × 10−2 N and the coefficient of fric-
tion between the tube and the ground is 0.20, what is the tube’s final
speed? Assume the tube is initially at rest.
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Holt Physics Problems BankCh. 5–8
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 5DPOTENTIAL ENERGY
P R O B L E MA 70.0 kg stuntman jumps from a bridge that is 50.0 m above the water.Fortunately, a bungee cord with an unstretched length of 15.0 m is at-tached to the stuntman, so that he breaks his fall 12.0 m above the water’ssurface. If the total potential energy associated with the stuntman andcord is 3.43 104 J, what is the force constant of the cord?
S O L U T I O NGiven: m = 70.0 kg h = 12.0 m
x = 50.0 m − 12.0 m − 15.0 m = 23.0 m
PEg = 0 J at river level
PEtot = 3.43 × 104 J
g = 9.81 m/s2
Unknown: k = ?
Diagram:
Choose the equation(s) or situation: The zero level for gravitational potential
energy is chosen to be at the water’s surface. The total potential energy is the sum
of the gravitational and elastic potential energies.
PEtot = PEg + PEelastic = mgh + 12
kx2
Rearrange the equation(s) to isolate the unknown(s):
12
kx2 = PEtot − mgh
k = 2(PEto
xt2− mgh)]
Substitute the values into the equation(s) and solve:
k =
k = =
k =
The situation in this problem is basically the same as that in Sample Problem 5D
in the textbook. In order for the stuntman to have a greater height above the water,
the bungee cord must store more elastic potential energy with less stretching.
This occurs when the spring constant is larger (98.7 N/m > 71.8 N/m).
98.7 N/m
(2)(2.61 × 104 J)
(23.0 m)2(2)(3.43 × 104 J − 8.24 × 103 J)
(23.0 m)2
(2)[3.43 × 104 J − (70.0 kg)(9.81 m/s2)(12.0 m)]
(23.0 m)2
1. DEFINE
2. PLAN
3. CALCULATE
4. EVALUATE
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50.0 m
12.0 m
Relaxed length = 15.0 m
Stretched length = 38.0 m
Problem 5D Ch. 5–9
NAME ______________________________________ DATE _______________ CLASS ____________________
1. A highway guardrail is designed so that it can be distorted as much as
5.00 cm when struck by an automobile. What is the minimum force
constant of the guardrail if it is to withstand the impact of a car with
1.09 × 104 J of energy?
2. An arresting cable helps to slow jet planes as they land on an aircraft
carrier. This is accomplished by two springs, each of which is attached
to one end of the cable. Suppose the elastic potential energy stored in
the springs while a jet is landing is 5.78 × 107 J. If each spring is
stretched 102 m, what is the force constant of each spring?
3. A produce scale at a supermarket uses a stretched spring to indicate the
weight of fruits and vegetables. If five oranges with a total mass of
0.76 kg are placed in the scale, the spring will be stretched 2.3 cm.
What is the force constant of the spring?
4. A 5.0 kg stone is slid up a frictionless ramp that has an incline of 25.0°.
How long is the ramp if the gravitational potential energy associated
with the stone is 2.4 × 102 J?
5. A pogo stick contains a spring with a force constant of 1.5 × 104 N/m.
Suppose the elastic potential energy stored in the spring as the pogo
stick is pushed downward is 120 J. How far is the spring compressed?
6. A 1750 kg weather satellite moves in a circular orbit with a gravita-
tional potential energy of 1.69 × 1010 J. At the satellite’s altitude above
Earth’s surface, the free-fall acceleration is only 6.44 m/s2. How high
above Earth’s surface is the satellite?
7. An automobile to be transported by ship is raised 7.0 m above the
dock. If its gravitational potential energy is 6.6 × 104 J, what is the
automobile’s mass?
8. One of the largest planes ever to fly, and the largest to fly frequently, is
the Ukrainian-built Antonov An-124 Ruslan. Its wingspan is 73.2 m
and its length is 69.2 m. The gravitational potential energy associated
with the plane at an altitude of 1.45 km is 3.36 × 109 J. What is the
airplane’s mass?
9. The force constant of the spring in a child’s toy car is 550 N/m. How
much elastic potential energy is stored in the spring if the spring is
compressed a distance of 1.2 cm?
10. With an elevation of 5334 m above sea level, the village of Aucanquilca,
Chile, is the highest inhabited town in the world. What would be the
gravitational potential energy associated with a 64.0 kg person in
Aucanquilca? Assume that the free-fall acceleration at Aucanquilca is
equal to that at sea level.
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ADDITIONAL PRACTICE
Holt Physics Problems BankCh. 5–10
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 5ECONSERVATION OF MECHANICAL ENERGY
P R O B L E MA raindrop with a mass of 0.500 g falls to Earth from a height of 1.50 km.The raindrop reaches Earth’s surface with a speed of 6.67 m/s. How muchof the raindrop’s mechanical energy is lost because of air resistance? As-sume free-fall acceleration to be the same at 1.5 km above Earth’s surfaceas it is at Earth’s surface.
S O L U T I O NGiven: m = 0.500 g
h = 1.50 km
vf = 6.67 m/s
vi = 0 m/s
g = 9.81 m/s2
Unknown: ∆ME = ?
Choose the equation(s) or situation: Use the conservation of mechanical energy
to account for energy dissipated through air resistance.
MEi + ∆ME = MEf
The zero level for gravitational potential energy is the ground. Because the rain-
drop starts at altitude h, the initial potential energy is its maximum value. Be-
cause the raindrops’ initial velocity is zero, the initial kinetic energy is zero.
MEi = PEi + KEi = PEi = mgh
When the raindrop reaches the ground, the gravitational potential energy is zero.
MEf = PEf + KEf = KEf = 12
mvf2
Substituting the last two equations into the first yields the following equation:
mgh + ∆ME = 12
mvf2
Rearrange the equation(s) to isolate the unknown(s):
∆ME = 12
mvf2 − mgh
Substitute the values into the equation(s) and solve:
∆ME = 12
(0.500 × 10−3 kg)(6.67 m/s)2
− (0.500 × 10−3 kg)(9.81 m/s2) (1.50 × 103 m)
∆ME = 1.11 × 10−2 J − 7.36 J
∆ME =
Most of the gravitational potential energy is given up through the interaction of
the raindrop with the surrounding air. Only 0.15 percent of the initial energy
remains as kinetic energy.
− 7.35 J
1. DEFINE
2. PLAN
3. CALCULATE
4. EVALULATE
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Problem 5E Ch. 5–11
NAME ______________________________________ DATE _______________ CLASS ____________________
1. What would be the kinetic energy of a 0.500 g raindrop if it fell
0.250 km without any resistance provided by air?
2. Angel’s Flight, which has been called “the world’s shortest railway,” is a
hill-climbing cable car, or funicular, that is located on Bunker Hill in
downtown Los Angeles, California. The funicular consists of two small
railway cars that are connected to each other by a cable. The cable is, in
turn, wrapped around a large pulley that is attached to a motor. As one
car rises up the hill, the other car descends to the street below.
a. The tracks of Angel’s Flight extend 96.0 m along the side of the
hill at an angle of 18.4° with respect to the horizontal. If each car
has a passenger with a mass of 70.0 kg, what is the total mechani-
cal energy associated with the two passengers when the cars are
about to leave the boarding platforms?
b. What is the total mechanical energy associated with the two
passengers when the cars arrive at their destinations?
c. Except for a brief initial and final acceleration, the cars move in
opposite directions at a constant speed of about 1.0 m/s. Suppose
the ascending car is somewhere between street level and the mid-
level. If the descending car is 20.0 m above street level, what is the
gravitational potential energy associated with the passenger in
the ascending car?
3. A toy rocket is at a height of 75.0 m and is moving upward with a
speed of 1.2 m/s when it ejects a payload with a mass of 20.0 g. The
payload has an initial upward speed relative to the rocket of 3.5 m/s.
What is the height reached by the payload when its upward velocity is
zero?
4. A 25.0 kg falling trunk strikes the ground with a speed of 12.5 m/s. As-
suming that there is no loss of energy due to air resistance, what is the
height from which the trunk falls?
5. If you were to neglect air resistance, a projectile fired straight up into
the air would land again with the speed with which it was fired. (This is
why it is dangerous to shoot a bullet directly upward.) However, some
of the energy is lost because of air resistance. Suppose a 50.0 g projec-
tile is fired upward with an initial speed of 3.00 × 102 m/s. If it lands
with a speed of 89.0 m/s, how much mechanical energy is given up
because of air resistance?
6. The air resistance that slows the projectile in problem 5 affects it both
as it rises and falls. How high would the projectile rise if there were no
air resistance? How high does it rise because of air resistance? (HINT:
Use the work–kinetic energy theorem to describe the forces on the pro-
jectile as it goes up and again as it comes down. Then use these equa-
tions to solve for the projectile’s maximum height. Use the data from
problem 5 for the final calculation.)
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ADDITIONAL PRACTICE
Holt Physics Problems BankCh. 5–12
NAME ______________________________________ DATE _______________ CLASS ____________________
7. A 50.0 kg circus performer jumps from a platform into a safety net
below. The net, which has a force constant of 3.4 × 104 N/m, is
stretched by 0.65 m. If the unstretched net is positioned 1.00 m above
the ground, what is the height of the platform? Ignore the effects of air
resistance.
8. A miniature golf course has a hole in which the fairway is 3.0 m above
the green. If you hit the ball into the middle hole in a row of three, the
ball will be directed to the green by a connecting pipe. Suppose the ball
falls down most of the length of the pipe and slides the rest of the way
without any loss of energy to friction. What is the ball’s speed as it
emerges from the pipe onto the green?
9. A 100.0 g arrow is pulled back 30.0 cm against a bowstring. If the
spring constant of the bowstring is 1250 N/m, at what speed will the
arrow leave the bow?
10. Because the wind speeds in Dayton, Ohio, are lower than at Kitty
Hawk, North Carolina, the Wright brothers improved their launch
mechanism for their 1904 flyer. Weights with a total mass of 546 kg
were dropped 5.64 m from a derrick. These weights pulled a rope that
was attached to the flyer, causing it to accelerate along a track for
5.64 m. This acceleration gave the flyer enough speed in addition to
that provided by its propellers to take off. If the flyer, which had a mass
of 273 kg, was initially at rest, what would its speed due to the falling
masses be when the masses reached the ground? (Note that your an-
swer is for a frictionless device; in reality, the speed of the flyer due to
the weights was somewhat lower.)
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Problem 5F Ch. 5–13
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 5FPOWER
P R O B L E MThe engines of the Queen Mary could deliver 174 MW to propel the mas-sive ship. How long does it take for the engines to do 7.31 1010 J of workon the ship?
S O L U T I O NGiven: P = 174 MW
W = 7.31 × 1010 J
Unknown: ∆t = ?
Use the equation for power and rearrange it to solve for time.
P = ∆W
t
∆t = W
P =
1
7
7
.3
4
1
××
1
1
0
06
10
W
J =
∆t = (4.20 × 102 s)(1 min/60 s) = 7.00 min
4.20 × 102 s
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1. The engine that moves the cables for the San Francisco cable cars de-
livers 380.3 kW of power for each line. How long does it take for 4.5 ×106 J of work (about the amount of work needed to raise a partially
loaded cable car up Nob Hill) to be done by this engine?
2. If the stairs of the Sears Tower in Chicago, Illinois, can be climbed by
an athlete with a power output of 331 W, how long does it take the ath-
lete to climb the building’s 442 m height? Assume the athlete has a
mass of 55 kg and that all of the power goes toward doing work against
gravity.
3. A runner exerts a force of 334 N against the ground while using 2100 W
of power. How long does it take the runner to run a distance of 50.0 m?
4. A ship’s diesel engine has a power output of 13.0 MW. How much
work is done by this engine in 15.0 min?
5. One horsepower (1 hp) is the unit of power based on the work that a
horse can do in one second. This is defined, in English units, as a force
of 550 lb that can move an object 1 foot in 1 s. In SI, 1 hp equals
745.7 W, or 745.7 J of work delivered in 1 s. Suppose you have a horse
that delivers 745.7 W of power, but that it does the work in only 0.55 s.
How much work has this horse done?
ADDITIONAL PRACTICE
Holt Physics Problems BankCh. 5–14
NAME ______________________________________ DATE _______________ CLASS ____________________
6. The 300-series Shinkansen train of Japan has aluminum cars, so that it
can reach high speeds more easily. Ten of the sixteen cars of a 300-
series train have their own 300.0 kW motors, one for each of their four
axles. What is the work done by one car’s four motors during 25 s?
7. When it is completed in 2002, the International Financial Center in
Taipei, Taiwan, will be the tallest building in the world. The Interna-
tional Financial Center will also have the fastest elevators in the world.
Two of the 63 elevators will travel from the ground floor to the eighty-
ninth floor in just 39 s. Suppose the power output of each elevator
motor is 158 kW. How much work will these motors do in lifting the
elevator to the eighty-ninth floor?
8. The space shuttle, which was first launched on April 12, 1981, is the
world’s first reusable space vehicle. The shuttle is placed in orbit by
three engines that do 1.4 × 1013 J of work in 8.5 min. What is the
power output of these engines?
9. Borax was mined in Death Valley, California, during the nineteenth
century. It was transported from the valley by massive wagons, each
pulled by a team of mules. Suppose the team does 2.82 × 107 J of work
on the wagon for 30.0 min? How much power is delivered, on the aver-
age, by the team? Express your answer in both watts and horsepower
(1 hp = 745.7 W).
10. James Watt did not invent the steam engine, but by adding a condenser
to an existing engine he discovered how to make steam engines more
efficient and practical. His own engine of 1778 was able to do 3.0 ×106 J of work in 5.0 min. How much power was delivered by this
engine?
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Problem 6A Ch. 6-1
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 6AMOMENTUM
P R O B L E MAn ostrich with a mass of 146 kg is running with a momentum of2480 kg•m/s to the right. What is the velocity of the ostrich?
S O L U T I O NGiven: m = 146 kg
p = 2480 kg•m/s to the right
Unknown: v = ?
Use the equation for momentum to solve for v.
p = mv
v = m
p
v = = 17.0 m/s to the right2480 kg•m/s
146 kg
1. If a blue whale has a mass of 1.46 × 105 kg and momentum of 9.73 ×105 kg•m/s to the south, what is its velocity?
2. The highest land speed for a rail-guided vehicle was set in 1982 by a
rocket sled at Holloman Air Force Base in southern New Mexico. The
sled was unmanned, but if it had a payload with a mass of 25 kg, the
magnitude of the payload’s momentum would have been 6.8 ×104 kg•m/s. What was the speed, in m/s and km/h, of the payload and
sled?
3. Thoroughbred horses are among the fastest horses in the world and are
used in famous racing events such as the Kentucky Derby. The mass ofa thoroughbred is about 5.00 × 102 kg. If a horse with this mass is gal-
loping with a momentum of 8.22 × 103 kg•m/s to the west, what is its
velocity?
4. The World Solar Challenge in 1987 was the first car race in which all
the vehicles were solar powered. The winner was the GM Sunraycer,
which had a mass of 177.4 kg, not counting the driver’s mass. Assume
that the driver had a mass of 61.5 kg, so that the total momentum of
the car and driver had a magnitude of 4.416 × 103 kg•m/s. What was
the car’s speed in m/s and km/h?
5. The current holder of the men’s world record for running 200 m is
Michael Johnson, who in 1996 ran 200.0 m in 19.32 s. Johnson’s mass
at the time of his record-breaking run was about 77 kg. What was the
magnitude of his momentum at his average speed?
ADDITIONAL PRACTICE
Holt Physics Problem BankCh. 6-2
NAME ______________________________________ DATE _______________ CLASS ____________________
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6. Although it cannot sustain its top speed for more than 8.65 s, the chee-
tah can run a distance of 274 m during that time. If a cheetah with a
mass of 50.0 kg is moving north at its top speed, what is its momentum?
7. The high-speed 300-series Shinkansen trains of Japan consist of 16 alu-
minum cars with a combined mass of 7.10 × 105 kg. The reduction in
mass from the 100-series trains enables the 300-series trains to reach a
top speed of 270 km/h. What is the magnitude of a 300-series train’s
momentum at its top speed?
8. The largest species of hummingbird is the Patagonia gigas, or the Giant
Hummingbird of the Andes. This bird has a length of 21 cm and can
fly with a speed of up to 50.0 km/h. Suppose one of these humming-
birds flies at this top speed. If the magnitude of its momentum is
0.278 kg•m/s, what is the hummingbird’s mass?
9. A hovercraft, or air-cushion vehicle, glides on a cushion of air, allow-
ing it to travel with equal ease on land or water. The first commercial
hovercraft to cross the English Channel, the V. A-3, had an average
velocity of 96 km/h to the southeast. Its average momentum was
4.8 × 104 kg•m/s to the southeast. What was the mass of the V. A-3?
10. The brightest, hottest, and most massive stars (over 10 times as massive
as the sun) are the brilliant blue stars designated as spectral class O. If a
class O star moves with a speed of 255 km/s and has a momentum of
8.62 × 1036 kg•m/s, what is the star’s mass?
Problem 6B Ch. 6-3
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 6BFORCE AND MOMENTUM
P R O B L E MA student with a mass of 55 kg rides a bicycle with a mass of 11 kg. A netforce of 125 N to the east accelerates the bicycle and student during a timeinterval of 3.0 s. What is the final velocity of the bicycle and student? As-sume the student and bicycle are initially at rest.
S O L U T I O NGiven: ms = 55 kg
mb = 11 kg
F = 125 N to the east
∆t = 3.0 s
vi = 0 m/s
Unknown: vf = ?
Use the impulse-momentum theorem to solve for vf.
F∆t = ∆p = mvf − mvi
vf = F∆t
m
+ mvi
m = ms + mb = 55 kg + 11 kg = 66 kg
vf =
vf = (125
6
N
6
)
k
(
g
3.0 s)
vf = 5.7 m/s to the east
(125 N)(3.0 s) − (66 kg)(0 m/s)
66 kg
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ADDITIONAL PRACTICE
1. A net force of 10.0 N to the right pushes a 3.0 kg book across a table. If
the force acts on the book for 5.0 s, what is the book’s final velocity?
Assume the book to be initially at rest.
2. A 60.0 g egg dropped from a window is caught by a student. If the stu-
dent exerts a net force of −1.5 N over a period of 0.25 s to bring the egg
to a stop, what is the egg’s initial speed?
3. A child riding a sled is pulled down a snowy hill by a force of 75 N. If
the child and sled have a combined mass of 55 kg, what is their speed
after 7.5 s? Assume the child and sled are initially at rest.
4. A billiard ball with a mass of 0.195 kg and a velocity of 0.850 m/s to the
right is deflected by the cushioned edge of the billiard table. The cush-
ion exerts a force of 3.50 N to the left for 0.0750 s. What is the ball’s
final velocity?
Holt Physics Problem BankCh. 6-4
NAME ______________________________________ DATE _______________ CLASS ____________________
5. A 5.00 g projectile has a velocity of 255 m/s to the right. What force is
required to stop this projectile in 1.45 s?
6. The Pacific walrus has an average mass of 1.1 × 103 kg and can swim
with a speed of about 9.7 m/s. Suppose a walrus starting from rest
takes 19 s to reach a velocity of 9.7 m/s to the east. What net force acts
upon the walrus?
7. With a mass of 3.000 × 103 kg, the Russian-made Zil-41047 is the most
massive automobile to have been manufactured on a regular basis.
Suppose one of these cars accelerates from rest to a velocity of 8.9 m/s
to the right in 5.5 s. Calculate the net force acting on the Zil-41047.
8. How much time would it take for a 0.17 kg ice hockey puck to decrease
its speed by 9.0 m/s if the coefficient of kinetic friction between the ice
and the puck is 0.050?
9. A girl pulls a 12.0 kg wagon along by exerting a force of 15.0 N on the
wagon’s handle, which makes an angle of 20.0° with the horizontal.
Friction provides a force of 11.0 N in the opposite direction. How long
does it take for the wagon, which is initially at rest, to reach a speed of
4.50 m/s?
10. The compressed-air device at Sandia Laboratories in Albuquerque,
New Mexico, accelerates small metal disks to a speed of 15.8 km/s. Sup-
pose the compressed air exerts a force of 12.0 N on a 0.20 g disk that is
initially at rest. How long will it take the disk to reach its maximum
speed?
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Problem 6C Ch. 6-5
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 6CSTOPPING DISTANCE
P R O B L E M
A high-speed train with a total mass of 9.25 105 kg travels north at aspeed of 220 km/h. Suppose it takes 16.0 s of constant acceleration for thetrain to come to rest at a station platform. Calculate the force acting onthe train during this time. What is the train’s stopping distance?
S O L U T I O N
Given: m = 9.25 × 105 kg
vi = 220 km/h to the north
vf = 0 km/h
∆t = 16.0 s
Unknown: F = ? ∆x = ?
Use the impulse-momentum theorem to solve for F. Use the kinematic equation
for ∆x in terms of initial velocity, final velocity, and time to solve for ∆x.
F∆t = ∆p
F = ∆∆
p
t =
mvf
∆−t
mvi
F =
F =
F = −3.5 × 106 N =
∆x = 12
(vi + vf)∆t
∆x = 12
(220 km/h + 0 km/h)(103 m/km)(1 h/3600 s)(16.0 s)
∆x = 490 m to the north
3.5 × 106 N to the south
−(9.25 × 105 kg)(220 km/h)(103 m/km)(1 h/3600 s)
16.0 s
(9.25 × 105 kg)(0 km/h) − (9.25 × 105 kg)(220 km/h)(103 m/km)(1 h/3600 s)
16.0 s
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ADDITIONAL PRACTICE
1. A race car has a velocity of 382 km/h to the right. If the car’s mass is
705 kg and the driver’s mass is 65 kg, what force is needed to bring the
car and driver to a stop in 12.0 s? What is the car’s stopping distance?
2. The danger that space debris poses to spacecraft can be understood in
terms of momentum. At 160 km above Earth’s surface, any object will
have a speed of about 7.82 × 103 m/s. Consider a meteoroid (a small
orbiting rock) with a mass of 42 g. Suppose this meteoroid collides
with a space shuttle and is brought to a full stop in 1.0 × 10–6 s. How
large is the force that stops the meteoroid? By how much would the
meteoroid dent the side of the shuttle?
Holt Physics Problem BankCh. 6-6
NAME ______________________________________ DATE _______________ CLASS ____________________
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3. A 63 kg astronaut drifting with 7.0 m/s to the right with respect to a
spacecraft uses a jet pack to slow down. If it takes 14.0 s to come to a
stop with respect to the spacecraft, what is the force exerted by the jet
pack? How far does the astronaut travel before stopping?
4. A polar bear with a mass of 455 kg slides for 12.2 s across the surface of
a frozen lake. If the coefficient of kinetic friction between the bear and
the ice is 0.071, what is the change in the bear’s momentum as it comes
to a stop? How far does the polar bear slide?
5. A skyrocket that has consumed all of its fuel continues to move up-
ward, slowed mostly by the force of gravity. If the rocket’s mass is
75.0 g and it takes 1.2 s for the rocket to stop, what is the change in the
rocket’s momentum? What is the rocket’s stopping distance?
6. A 4400 kg sailboat drifts in calm waters. The force of resistance that the
water exerts on the ship is 2200 N to the left. What is the change in the
ship’s momentum after 8.0 s? If the ship’s initial speed is 6.5 m/s, how
far has the ship drifted?
7. A jet of water exerts a 25.0 N force on a type of sail that is attached to a
small wagon. What is the magnitude of the change in the wagon’s mo-
mentum after 7.00 s? If the wagon’s mass is 14.0 kg and there are no
other forces acting on it, how far will the wagon travel during the
7.00 s? Assume the wagon is initially at rest.
8. How long will it take a 2.30 × 103 kg truck to go from 22.2 m/s to a
complete stop if acted on by a force of –1.26 × 104 N? What would be
its stopping distance?
9. Consider a Hornet jet landing on the flight deck of an aircraft carrier.
The jet’s mass is 1.35 × 104 kg and its initial velocity is 66.1 m/s to the
west. If the force required to bring the jet to a stop is 4.00 × 105 N to
the east, how long does it take the jet to come to rest? How far does the
jet travel during its deceleration?
10. The surface of Jupiter’s moon Europa is covered with ice. Suppose an
ice boat moving with a speed of 14.5 m/s drifts to a stop. The boat’s
mass is 1.50 × 103 kg and the coefficient of kinetic friction between the
boat’s runners and the ice is 0.065. However, free-fall acceleration on
Europa is only 1.305 m/s2. How long will it take the boat to stop? How
far does the ice boat glide before stopping?
Problem 6D Ch. 6-7
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 6DCONSERVATION OF MOMENTUM
P R O B L E M
A 20.0 kg cannonball is fired from a 2.40 × 103 kg. If the cannon recoils with a velocity of 3.5 m/s backwards, what is the velocity of thecannonball?
S O L U T I O NGiven: m1 = mass of cannonball = 20.0 kg
m2 = mass of cannon = 2.40 × 103 kg
v1,i = initial velocity of cannonball = 0 m/s
v2, i = initial velocity of cannon = 0 m/s
v2,f = final velocity of cannon = 3.5 m/s backwards = −3.5 m/s
Unknown: v1, f = final mass of cannonball = ?
Choose the equation(s) or situation: Because the momentum of the cannon-
cannonball system is conserved and therefore remains constant, the total initial
momentum of the cannon and cannonball will equal the total final momentum
of the cannon and cannonball.
m1v1,i + m2v2,i = m1v1,f + m2v2,f
Because the cannon and cannonball are initially at rest, the initial momentum for
each is zero. From momentum conservation it follows that the total final mo-
mentum is also zero.
m1v1,f + m2v2,f = 0
Rearrange the equation(s) and isolate the unknown(s):
v1,f = −m
m2
1
v2,f
Substitute the values into the equation(s) and solve:
v1,f = = 420 m/s
v1,f =
The velocity is positive, indicating the forward direction. The cannonball’s mass
is less than one-hundredth the mass of the cannon, so its speed should be over a
hundred times greater than the recoil speed of the cannon.
420 m/s forward
−(2.40 × 103 kg)(−3.5 m/s)
20.0 kg
1. DEFINE
2. PLAN
3. CALCULATE
4. EVALUATE
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ADDITIONAL PRACTICE
1. A student stumbles backward off a dock and lands in a small boat. The
student isn’t hurt, but the boat drifts away from the dock with a
velocity of 0.85 m/s to the west. If the boat and student each have a
mass of 68 kg, what is the student’s initial horizontal velocity?
Holt Physics Problem BankCh. 6-8
NAME ______________________________________ DATE _______________ CLASS ____________________
2. A coal barge with a mass of 1.36 × 104 kg drifts along a river. When it
passes under a coal hopper, it is loaded with 8.4 × 103 kg of coal. What
is the speed of the unloaded barge if the barge after loading has a speed
of 1.3 m/s?
3. A child jumps from a moving sled with a speed of 2.2 m/s and in the
direction opposite the sled’s motion. The sled continues to move in the
forward direction, but with a new speed of 5.5 m/s. If the child has a
mass of 38 kg and the sled has a mass 68 kg, what is the initial velocity
of the sled?
4. A swimmer with a mass of 58 kg and a velocity of 1.6 m/s to the north
climbs onto a 142 kg raft. The combined velocity of the swimmer and
raft is 0.32 m/s to the north. What is the raft’s velocity before the swim-
mer reaches it?
5. A 50.0 g shell fired from a 3.00 kg rifle has a speed of 400.0 m/s. With
what speed does the rifle recoil in the opposite direction?
6. Momentum conservation often assumes that the mass of an object re-
mains constant throughout a process or event. However, a change in
momentum can also occur when mass changes. Consider an automo-
bile with a full tank of gasoline traveling at a velocity of 88.0 km/h to
the east. The mass of the car when the fuel tank is full is 1292 kg. Sup-
pose that the car travels along a highway that extends eastward for
600 km. By the time the car has traveled this distance, its mass is
1255 kg. What is the car’s velocity at the end of the journey?
7. In 1976, Comet West was observed to break into four smaller parts as it
orbited near the sun. Suppose a comet with a mass of 5.0 × 1014 kg and
moving with a speed of 74.0 km/s breaks into two equal parts. One
part moves 15.0° above the original orbit with a speed of 105 km/s,
while a second fragment moves 30.0° below the original orbit. What is
the velocity of the second comet fragment?
8. A twig floating in a small pond is initially at rest. On the twig is a snail,
which begins moving along the length of the twig with a speed of
1.2 cm/s. The twig moves in the opposite direction with a speed of
0.40 cm/s. If the snail’s mass is 2.5 g, what is the mass of twig?
9. A toy that is initially at rest consists of three parts that are held together
by spring-loaded clips. At a given instant, the toy “explodes.” Two of the
pieces, which each have a mass of 25.0 g, travel with velocities of
7.0 cm/s to the south and 7.0 cm/s to the west, respectively. The third
piece has a velocity of 3.3 cm/s at an angle of 45° north of east. What is
the mass of the third piece?
10. An ice skater at rest catches a bag of sand moving to the north with a
speed of 5.4 m/s. This causes both the skater and the bag to move to
the north at a speed of 1.5 m/s. If the skater’s mass is 63 kg, what is the
mass of the bag of sand?
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Problem 6E Ch. 6-9
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 6EPERFECTLY INELASTIC COLLISIONS
P R O B L E MAn arrow is fired into a small target at rest on a frictionless surface. Thearrow’s mass is 20.0 g and the target’s mass is 2.50 kg. If the speed of thearrow and target combined is 0.67 m/s, what is the arrow’s initial speed?
S O L U T I O N
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Given: m1 = mass of arrow = 20.0 g
m2 = mass of target = 2.50 kg
v2,i = initial speed of target = 0 m/s
vf = final speed of target and arrow = 0.67 m/s
Unknown: v1,i = initial speed of arrow = ?
Use the equation for a perfectly inelastic collision and rearrange it to solve for v1,i .
m1v1,i + m2v2,i = (m1 + m2)vf
v1,i =
v1,i =
v1,i = (2.
2
5
0
2
.0
kg
×)(
1
0
0
.6−3
7
k
m
g
/s)
v1,i = 84 m/s
(20.0 × 10−3 kg + 2.50 kg)(0.67 m/s) − (2.50 kg)(0 m/s)
20.0 × 10−3 kg
(m1 + m2)vf − m2v2,imi
ADDITIONAL PRACTICE
1. A 1550 kg torpedo strikes a 770 kg target that is initially at rest. If the
combined torpedo and target move forward with a speed of 9.44 m/s,
what is the initial velocity of the torpedo? Assume that no resistance is
provided by the water.
2. An ice hockey puck with a mass of 0.17 kg collides inelastically with a
0.75 kg snowball that is sliding to the left with a speed of 0.50 m/s. The
combined puck and snowball slide along the ice with a velocity of
4.2 m/s to the right. What is the velocity of the hockey puck before the
collision?
3. A clay ball with a mass of 45 g is attached to a long string to make a
pendulum. The ball is pulled back so that the string is horizontal to the
ground, and is then released. At the bottom of the ball’s path is another
clay ball that has a mass of 75 g and is at rest. The two balls collide in-
elastically, so that they follow the path of the first ball beyond the point
of collision. What must the speed of the first ball be just before the col-
lision so that the combined balls rise to a height of 8.0 cm above the
Holt Physics Problem BankCh. 6-10
NAME ______________________________________ DATE _______________ CLASS ____________________
point of collision? How high must the first ball be raised for it to have
this speed at the bottom of its path?
4. A 5.00 × 102 kg log collides inelastically with a second log with the
same mass. These combined logs then collide with a third log with a
mass of 5.00 × 102 kg. The final speed of the three combined logs is
3.67 m/s. If the speed of the third log before collision was 3.00 m/s, and
the speed of the second log before collision was 3.50 m/s, what was the
speed of the first log before collision?
5. A railway car with a mass of 8500 kg and a velocity of 4.5 m/s to the
right collides inelastically with a railway car with a mass of 9800 kg and
a velocity of 3.9 m/s to the left. What is the final velocity of the com-
bined cars?
6. A 1400 kg automobile heading north at 45 km/h collides inelastically
with a 2500 kg truck traveling east at 33 km/h. What is the final veloc-
ity of the combined vehicles?
7. Four velcro-lined air-hockey disks collide with each other in a perfect
inelastic collision. The first disk has a mass of 50.0 g and a velocity of
0.80 m/s to the west, the second disk has a mass of 60.0 g and a velocity
of 2.50 m/s to the north, the third disk has a mass of 100.0 g and a ve-
locity of 0.20 m/s to the east, and the fourth disk has a mass of 40.0 g
and a velocity of 0.50 m/s to the south. What is the final velocity of the
disks after the collision?
8. A 25.0 kg sled carrying a 42.0 kg child is moving with a speed of
3.50 m/s when it collides with a snowman that is initially at rest. If the
speed of the snowman, sled, and child is 2.90 m/s, what is the snow-
man’s mass?
9. A remora is a type of fish that uses suckers underneath its head to at-
tach itself to other fish, notably sharks (for this reason it is also called
the “sharksucker”). Suppose a remora swimming with a velocity of
5.0 m/s to the right attaches itself to a 150.0 kg shark that is swimming
to the left with a speed of 7.00 m/s. If the remora collides inelastically
with the shark, the velocity of the two fish combined is 6.25 m/s to the
left. From this information, calculate the mass of the remora.
10. A proposed method for removing small but hazardous debris in orbit
around Earth involves a large ball composed of a type of gum or putty.
This soft ball would orbit Earth and collide inelastically with small par-
ticles of debris, sweeping them up in the process. Suppose this putty
ball moves in an orbit containing a stream of debris. The ball has a ve-
locity of 8.0 × 103 m/s to the right, while the particles of debris have a
velocity of 8.0 × 103 m/s to the left. Each particle of debris has an aver-
age mass of 2.5 g. If the putty ball sweeps up 5,000 particles before the
velocity of the ball and debris is 90.0 percent of the ball’s initial veloc-
ity, what is the ball’s mass?
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Problem 6F Ch. 6-11
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 6FKINETIC ENERGY IN PERFECTLY INELASTIC COLLISIONS
P R O B L E M
A ship with a mass of 4.50 107 kg and a velocity of 2.30 m/s to the northcollides with another ship whose mass is 2.30 107 kg. If the speed of thesecond ship is 3.40 m/s to the south, what is the change in the kinetic energy after the two ships undergo a perfectly inelastic collision?
S O L U T I O N
Given: m1 = mass of first ship = 4.50 × 107 kg
m2 = mass of second ship = 2.30 × 107 kg
v1,i = initial velocity of first ship = 2.30 m/s to the north
= +2.30 m/s
v2,i = initial velocity of second ship = 3.40 m/s to the south
= −3.40 m/s
Unknown: vf = ? ∆KE = ?
Use the equation for a perfectly inelastic collision to find vf.
m1v1,i + m2v2,i = (m1 = m2)vf
vf = m1v
m1
1
,i
++
m
m
2
2v2,i
vf =
vf =
vf =
vf = 0.38 m/s
Use the equation for kinetic energy to calculate the kinetic energy of each ship
before the collision and the final kinetic energy of the two ships combined.
Initial kinetic energy:
KEi = KE1,i + KE2,i = 12
m1v1,i2 + 1
2m2v2,i
2
KEi = 12
(4.50 × 107 kg)(2.30 m/s)2 + 12
(2.30 × 107 kg)(−3.40 m/s)2
KEi = 1.19 × 108 J + 1.33 × 108 J = 2.52 × 108 J
Final kinetic energy:
KEf = KE1,f + KE2,f = 12
(m1 + m2)vf
KEf = 12
(4.50 × 107 kg + 2.30 × 107 kg)(0.38 m/s)2
KEf = 12
(6.80 × 107 kg)(0.38 m/s)2
KEf = 4.9 × 107J
2.6 × 107 kg•m/s6.80 × 107 kg•m/s
1.04 × 108 kg•m/s − 7.82 × 107 kg•m/s
6.80 × 107 kg
(4.50 × 107 kg)(2.30 m/s) + (2.30 × 107 kg)(−3.40 m/s)
4.50 × 107 kg + 2.30 × 107 kg
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1. A ball of clay with a mass of 55 g and a speed of 1.5 m/s collides with a
55 g ball of clay that is at rest. By what percent has the kinetic energy
decreased after the inelastic collision?
2. An unconventional artist creates paintings by sloshing buckets of paint
onto large canvases. Suppose the canvas and easel on which it is placed
have a combined mass of 4.5 kg and are initially at rest when the artist
throws 1.3 kg of paint onto the canvas. The canvas, easel, and paint to-
gether slide back on the smooth floor with a speed of 0.83 m/s. What is
the change in the kinetic energy after the inelastic collision?
3. The farthest source of comets is called the Oort cloud. This is a volume
of space ranging from 1.5 × 1010 km to 7.4 × 1012 km away from the
sun in which comets are loosely held by the sun’s gravitational force.
Suppose a comet in the Oort cloud has a mass of 1.50 × 1013 kg and a
speed of 250 m/s. This comet collides inelastically with another comet
that has a mass of 6.5 × 1012 kg and a velocity of 420 m/s in the same
direction as the first comet. What is the change in the kinetic energy of
the comets after the collision?
4. Two flying fish have an inelastic collision while in mid-flight. One fish
has a mass of 0.650 kg and a velocity of 15.0 m/s to the right; the other
has a mass of 0.950 kg and a velocity of 13.5 m/s to the left. Find the
change in their kinetic energy after the collision.
5. A 75.0 kg log floats downstream with a speed of 1.80 m/s. Eight frogs
hop onto the log in a series of inelastic collisions. If each frog has a
mass of 0.30 kg and an upstream speed of 1.3 m/s, what is the change
in kinetic energy for this system?
6. What is the change in kinetic energy for the inelastic collision between
the two railway cars described in problem 5 of the previous section?
7. What is the change in kinetic energy for the inelastic collision between
the clay balls in problem 3 of the previous section?
8. What is the change in kinetic energy in the collision between the ball of
putty and space debris described in problem 10 of Section 6E?
9. What is the change in the kinetic energy for the colliding logs in prob-
lem 4 of the previous section?
10. What is the change in kinetic energy for the inelastic collision between
the four disks in problem 7 of the previous section?
Holt Physics Problem BankCh. 6-12
NAME ______________________________________ DATE _______________ CLASS ____________________
Change in kinetic energy:
∆KE = KEf − KEi = 4.9 × 107 J − 2.52 × 108 J =
By expressing ∆KE as a negative number, ∆KE indicates that the energy has left
the system to take a form other than mechanical energy.
−2.03 × 108 J
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ADDITIONAL PRACTICE
Problem 6G Ch. 6-13
NAME ______________________________________ DATE _______________ CLASS ____________________
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Holt Physics
Problem 6GELASTIC COLLISIONS
P R O B L E MIn the game of marbles, a shooter is a large marble about 2 cm in diame-ter that is used to knock smaller marbles out of the ring. Suppose ashooter with a speed of 0.80 m/s hits a 4.8 g marble that is at rest in thering. The shooter continues forward with a speed of 0.51 m/s while thesmaller marble moves forward with a speed of 1.33 m/s. What is the massof the shooter?
S O L U T I O NGiven: v1,i = initial velocity of shooter = 0.80 m/s forward
v2,i = initial velocity of marble = 0 m/s
v1,f = final velocity of shooter = 0.51 m/s forward
v2,f = final velocity of marble = 1.33 m/s forward
m2 = mass of marble = 4.8 g
Unknown: m1 = mass of shooter = ?
Choose the equation(s) or situation: Use the equation for the conservation of
momentum to determine the mass of the shooter.
m1v1,i + m2v2,i = m1v1,f + m2v2,f
Rearrange the equation(s) to isolate the unknown(s):
m1(v1,i − v1,f) = m2v2,f − m2v2,i
m1 = m2
v
v
1
2
,
,
i
f
−−
v
m
1
2
,f
v2,i
Substitute the values into the equation(s) and solve:
m1 =
m1 =
m1 =
Confirm your answer by making sure that kinetic energy is also conserved.
12
m1v,i2 + 1
2m2v2,i
2 = 12
m1v1,f2 + 1
2m2v2,f
2
12
(22 × 10−3 kg)(0.80 m/s)2 + 12
(4.8 × 10−3 kg)(0 m/s)2
= 12
(22 × 10−3 kg)(0.51 m/s)2 + 12
(4.8 × 10−3 kg)(1.33 m/s)2
7.0 × 10−3 J + 0 J = 2.9 × 10−3 J + 4.2 × 10−3 J
7.0 mJ =
The slight difference arises from rounding.
7.1 mJ
22 g
6.4 g•m/s0.29 m/s
(4.8 g)(1.33 m/s) − (4.8 g)(0 m/s)
0.80 m/s − 0.51 m/s
1. DEFINE
2. PLAN
3. CALCULATE
3. EVALUATE
1. A highly elastic rubber ball is tossed at a moveable wooden wall panel
that is initially at rest. The ball’s initial velocity is 6.00 m/s to the right.
After the elastic collision, the ball returns with a velocity of 4.90 m/s to
the left, while the panel moves 1.09 m/s to the right. If the panel’s mass
is 1.25 kg, what is the mass of the ball?
2. A 2.0 kg block of ice with a speed of 8.0 m/s makes an elastic collision
with another block of ice that is at rest. The first block of ice proceeds
in the same direction as it did initially, but with a speed of 2.0 m/s.
What is the mass of the second block? (Hint: Use the conservation of
kinetic energy to solve for the second unknown variable.)
3. A golf ball slides down a pipe from the upper level of a miniature golf
course and heads directly for the hole on the green. Unfortunately, an-
other player’s ball is directly in the way. The second ball, which is ini-
tially at rest, moves forward with a speed of 3.0 m/s, causing it to land
in the cup. The first ball comes to a complete stop after the collision.
Both balls have a mass of 45 g. What is the first ball’s speed before the
collision.
4. Suppose two ships, one with a mass of 3.0 × 107 kg and the second with
a mass of 2.5 × 107 kg, are equipped with bumpers, so that they undergo
a completely elastic collision. Before the collision, the second ship
moves north with a speed of 4.0 km/h. After the collision, the first ship
moves 3.1 km/h to the north while the second ship moves 6.9 km/h to
the south. Assume that the ships move over the ocean without friction.
Given this information, calculate the initial velocity of the first ship.
5. A basketball player throws a ball at the same time a ball from a nearby
court is thrown. The two balls collide elastically, so that the final veloc-
ity of the first ball is 4.0 m/s to the west and the final velocity of the
second ball is 3.0 m/s to the north. If the first ball’s initial velocity is
3.0 m/s to the north, what is the initial velocity of the second ball. As-
sume both balls have identical masses.
6. A red ball with a mass of 0.75 kg strikes two croquet balls, a 0.50 kg
green ball and a 0.50 kg blue ball, that are at rest next to each other on
a smooth wood floor. After the collision, the red ball has a velocity of
0.80 m/s to the east, the green ball moves 45° north of east with a speed
of 3.4 m/s, and the blue ball moves 45° south of east with a speed of
3.4 m/s. What is the red ball’s initial velocity?
7. An elevator is moving upward at a speed of 2.000 m/s. At the instant
the elevator is 20.4 m from the top of the shaft, a ball is dropped down
the shaft. The ball collides elastically with the elevator, so that it rises
up the shaft again. The elevator’s velocity immediately after the colli-
sion is 1.980 m/s upward. If the ball has a mass of 0.150 kg and the ele-
vator’s mass is 325.0 kg, what is the velocity of the ball after the colli-
sion with the elevator? How high above the elevator shaft does the
returning ball bounce?
Holt Physics Problem BankCh. 6-14
NAME ______________________________________ DATE _______________ CLASS ____________________
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ADDITIONAL PRACTICE
Problem 6G Ch. 6-15
NAME ______________________________________ DATE _______________ CLASS ____________________
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8. Suppose the initial conditions are the same as for problem 7, except
that the elevator’s initial velocity is 2.000 m/s downward and its final
velocity is 2.017 m/s downward. What is the velocity of the ball after
the collision with the elevator? How far below the top of the elevator
shaft does the returning ball bounce?
9. A steel ball with a mass of 0.50 kg is fastened to a cord that is 40.0 cm
long and is released from a height of 40.0 cm. At the bottom of its path
the ball strikes a 2.5 kg block that is initially at rest on a frictionless sur-
face. The collision is elastic. What are the final velocities of the ball and
block? (Hint: Use the conservation of kinetic energy to solve for the
second unknown variable.)
10. A bowling ball of mass 7.00 kg moves east with a speed of 2.00 m/s.
The ball collides with an identical ball at rest, which causes the first ball
to move 30.0° north of east at a speed of 1.73 m/s. What is the velocity
of the second ball after the collision?
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Problem 7A Ch. 7-1
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 7AANGULAR DISPLACEMENT
P R O B L E MA woman on vacation admires the murals on the inner wall of Coit Towerin San Francisco, California. If the woman walks 10.0 m clockwise alongthe curved wall, what will her angular displacement be? Assume the innerradius of Coit Tower is 4.20 m.
S O L U T I O NGiven: ∆s = −10.0 m
r = 4.20 m
Unknown: ∆q = ?
Use the angular displacement equation to solve for ∆q.
∆q = ∆r
s =
−4
1
.2
0
0
.0
m
m
∆q = −2.38 rad
1. At Tivoli, Italy, a circular ruin stands on the grounds of the emperor
Hadrian’s villa. A tourist walks 24.0 m counterclockwise along the edge
of the ruin. If the ruin’s radius is 3.50 m, what is the tourist’s angular
displacement?
2. In terms of volume, the largest tree is the General Sherman at Sequoia Na-
tional Park in California. Assuming the tree is perfectly circular, its radius
at the base is 5.55 m. If you were allowed to walk 31.3 m counterclockwise
along the side of the tree, what would your angular displacement be?
3. There was once a comedy skit that had a character who boasted of a stor-
age tank “the size of Rhode Island.” Suppose a circular storage tank with
this area did exist. The area of Rhode Island is 2730 km2. If you were to
drive 545 km clockwise along the perimeter of this tank, what would
your angular displacement be?
4. One of the largest stars in our galaxy is Betelgeuse, a red super giant that
has expanded as it has evolved. Betelgeuse is so large that, were it placed
in our solar system where the sun is, its surface would lie beyond Jupiter’s
orbit. Suppose a spacecraft orbits the sun at a distance equal to the radius
of Betelgeuse. The ship must travel through an arc length with a magni-
tude of 4.3 × 1011 m to have an angular displacement of magnitude
0.39 rad. From this information, determine the average radius of
Betelgeuse.
5. The only circular border in the United States is between northern
Delaware and Pennsylvania. The border has a length of about 35.0 km. If
a vehicle traveling through an arc length of this size has an angular dis-
placement with a magnitude of 1.75 rad, what is the border’s radius?
ADDITIONAL PRACTICE
Holt Physics Problem BankCh. 7-2
NAME ______________________________________ DATE _______________ CLASS ____________________
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6. A laser etches a line with an arc length of 36.6 mm clockwise along the sur-
face of a spherical grain of sand. If this distance corresponds to a clockwise
angular displacement of p6
rad, what is the sand grain’s radius?
7. Casa Rinconada is the name given to one of the largest ceremonial struc-
tures, or kivas, in the Chaco Canyon area of northwestern New Mexico.
Built nearly a thousand years ago, Casa Rinconada’s inner wall has a radius
of 10.0 m. If you were to walk counterclockwise along this wall, so that
your angular displacement was 5.7 rad, what arc length would you travel
through?
8. Of the solar system’s planets, the one whose orbit around the sun is most
circular is Venus. The mean radius of Venus’s orbit is 1.08 × 108 km. What
arc length does Venus travel through when its angular displacement is p3
rad counterclockwise?
9. After Venus, the next most circular orbit belongs to Neptune. The mean
radius of Neptune’s orbit is 4.48 × 109 km. What arc length does Neptune
travel through when its angular displacement is p3
rad counterclockwise?
10. The Pantheon in Rome is the oldest surviving domed building. The dome
is a hemisphere with a radius of 21.8 m. Suppose you were to walk inside
the Pantheon in a circular path just underneath the dome’s rim. What arc
length would you have traveled through if your angular displacement was
7.50 rad clockwise?
Problem 7B Ch. 7-3
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 7BANGULAR SPEED
P R O B L E MThe spin of a flying disk toy gives the disk stability while gliding. If one ofthese disks has an angular displacement with a magnitude of 60.0 rev in12.0 s, what is the disk’s average angular speed in rev/s and rad/s?
S O L U T I O NGiven: ∆q = 60.0 rev
∆t = 12.0 s
Unknown: wavg = ?
Use the angular speed equation to solve for waug.
wavg = ∆∆
qt
= 6
1
0
2
.0
.0
re
s
v
wavg =
wavg = (5.00 rev/s)(2π rad/rev) = 31.4 rad
5.00 rev/s
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ADDITIONAL PRACTICE
1. The propellers, or screws, of a steamship push water away from the ship,
causing the ship to move forward. Suppose a ship’s screw turns 106 rad
clockwise in 7.5 s. What is the average angular speed of the screw?
2. Devil’s Tower National Monument in northeastern Wyoming is an outcrop-
ping of volcanic rock that is 264 m tall. Suppose a golden eagle soars counter-
clockwise in a circular path around the top of the tower. If the eagle circles the
monument once in 4.56 min, what is the bird’s average angular speed in rad/s?
3. The speed of a bullet can be determined in the following way. Each end
of a long metal rod is passed through the center of a paper disk. The rod
is rapidly rotated, so that the two disks spin at the same angular speed.
The bullet, which is fired parallel to the rod, penetrates the first disk near
the rim and then travels the distance between the disks. During this time,
the disks rotate through a certain angular displacement. After the bullet
has penetrated the second disk, the angle between the two holes in the
disks can be measured. With the known distance between the disks and
the angular speed of the disks, the bullet’s speed can be calculated. Sup-
pose the bullet’s speed is 280 m/s, the disks are separated by 2.0 m, and
the counterclockwise angular displacement between the holes is
0.54 rad. What is the average angular speed of the disks in rev/s?
= 0.54 radθ
d
vω
Holt Physics Problem BankCh. 7-4
NAME ______________________________________ DATE _______________ CLASS ____________________
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4. The average angular speed of a merry-go-round is 0.75 rad/s. How long
does it take the merry-go-round to rotate through an angular displace-
ment with a magnitude of 3.3 rad?
5. Located on a 21 km2 island in the western Pacific Ocean, the Republic of
Nauru is the smallest—and in terms of its population one of the wealthi-
est—republics in the world. Suppose a tuna, which can swim at 80 km/h,
swims in a circle just off the shore of Nauru. If the tuna’s average angular
speed is 8.6 × 10–3 rad/s, how long will it take to circle the island exactly
6 times?
6. With a diameter nearly 1 m, Rafflesia arnoldii of Malaysia is considered
the world’s largest flower. Suppose a mosquito flies around the rim of a
Rafflesia with an average angular speed of 2.75 rad/s. How long will it
take the mosquito to circle the flower exactly 3 times?
7. Suppose you were to stand on Earth’s equator for 4.2 h. What would your
angular displacement be in radians? Consider Earth’s rotation to be
counterclockwise.
8. Just as the planets in the solar system move in elliptical orbits around the
sun, the solar system itself moves in an elliptical orbit around the gravita-
tional center of the Milky Way galaxy, making one revolution every 212 ×106 years. What would the magnitude of the solar system’s angular dis-
placement be over its approximate lifetime of 4.50 × 109 years? Give your
answer in radians.
9. Venus is the only planet in the solar system to rotate in a clockwise (retro-
grade) direction with respect to its north pole. It also rotates slower than
any other planet: once every 243 days. Express this average angular speed
in terms of rad/day, then calculate the angular displacement of Venus
during one Earth year (365.25 Earth days) and one Venus year (224.7
Earth days). (Note that one Venus year is shorter than one Venus day.)
10. Imagine the following experiment: you are standing at the top rim
looking down into a amusement park centrifuge ride in which a motion-
picture camera has been mounted to the wall. The camera and ride are
started, and at a point where the camera is on the side of the ride oppo-
site you, you toss a marble into the ride. Although from your viewpoint
the marble moves in a straight line, the image the camera records shows
the marble to be deflected through a certain angular displacement (actu-
ally the camera’s angular displacement). If the diameter of the ride is
6.0 m, the marble has a horizontal speed of 5.0 m/s, and the average an-
gular speed of the ride is 6.00 rad/s, what is the magnitude of the cam-
era’s actual (or marble’s apparent) angular displacement?
Problem 7C Ch. 7-5
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 7CANGULAR ACCELERATION
P R O B L E MFor a certain high-speed elevator, the drive wheel on the motor under-goes an average angular acceleration of 52.0 rad/s2. If the wheel is ini-tially at rest, what is its angular speed after 4.00 s?
S O L U T I O N
Given: αavg = 52.0 rad/s2
∆t = 4.00 s
w1 = 0 rad/s
Unknown: w2 = ?Use the angular acceleration equation and rearrange it to solve for w2.
αavg = ∆∆wt =
w2
∆−t
w1
w2 = w1 + αaug ∆t = 0 rad/s + (52.0 rad/s2)(4.00 s)
w2 = 208 rad/s
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ADDITIONAL PRACTICE
1. A turntable has an average angular acceleration of 1.5 rad/s2 and an ini-
tial angular speed of 3.0 rad/s. What is the turntable’s angular speed 4.0 s
later?
2. The take-up reel of a cassette tape must pull the tape across the magnetic
head at a constant speed. To do this, the reel’s angular speed must vary
with the increasing amount of tape on the reel. An empty reel has an an-
gular speed of 9.5 rad/s and an average angular acceleration of –5.4 ×10–3 rad/s2. What is the reel’s angular speed after 22 min?
3. The gasoline engine for a power lawnmower has an average angular ac-
celeration of 32 rad/s2 when first started. What is the engine’s angular
speed after 1.5 s, assuming an initial angular speed of 0 rad/s?
4. A tire on an accelerating car goes from rest to an angular speed of
76 rad/s. How long does it take the tire to reach this angular speed if the
tire’s average angular acceleration is 9.5 rad/s2?
5. A rolling hula hoop undergoes an average angular acceleration of
3.91 rad/s2. How long will it take for the hula hoop to increase its angular
speed from 2.50 rad/s to 7.70 rad/s?
6. As it turns around, a steamship has an angular speed of 5.14 × 10–2 rad/s
at the beginning of the turn. Its speed a short time later is 3.09 × 10–2 rad/s.
How much time does it take the ship to change speed if the average angu-
lar acceleration of the ship is –1.75 × 10–3 rad/s2?
Holt Physics Problem BankCh. 7-6
NAME ______________________________________ DATE _______________ CLASS ____________________
7. The speed of a steam engine is automatically controlled by a device called
a governor. A governor consists of two metal balls attached by rods to the
top of a rotating shaft. The balls are also attached by rods to a lever that
controls the flow of steam to the engine. As the engine’s speed increases,
the shaft’s angular speed also increases, the balls spin outward, and the
control lever is raised. This reduces the amount of steam that enters the
engine, causing it to slow down. The reverse process causes the engine to
speed up again. If a governor increases its angular speed from 4.0 rad/s to
5.0 rad/s in 7.5 s, what is its average angular acceleration?
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8. A toy consists of a windup car and a circular metal base with a shallow
groove that guides the car in a circular path. At first the car has an angu-
lar speed of 7.14 rad/s, but after 9.00 s its angular speed is 2.38 rad/s.
What is the car’s average angular acceleration?
9. A driver turns a corner when she realizes that oil has been spilled on the
pavement. By letting the car’s angular speed slow from 2.07 rad/s to
1.30 rad/s over a period of 2.2 s, the driver is able to make the turn with-
out skidding. What is the average angular acceleration of the car and
driver?
10. Because of the gravitational forces between Earth and the moon, the an-
gular speed of Earth is decreasing at about 1.7 × 10–3 s each century. This
means that 70.0 million years ago, Earth made one complete rotation in
just 23.66 h, whereas now it takes 24.00 h. What is Earth’s average angular
acceleration over the last 70.0 × 106 years?
controllever
for steamflow
rotationarrow
Problem 7D Ch. 7-7
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 7DANGULAR KINEMATICS
P R O B L E MThe screws of an ocean liner undergo a constant angular acceleration of0.656 rad/s2. How long does it take for the screws to reach a final angularspeed of 10.5 rad/s, assuming that they are initially at rest?
S O L U T I O N
Given: α = 0.656 rad/s2
wf = 10.5 rad/s
wi = 0 rad/s
Unknown: ∆t = ?
Use the angular kinematic equation for wf in terms of wi, α, and ∆t, and re-
arrange it to solve ∆t.
wf = wi + α∆t
∆t = wf
α− wi =
10.5
0.
r
6
a
5
d
6
/s
ra
−d
0
/s
r2ad/s
∆t = 16.0 s
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ADDITIONAL PRACTICE
1. A model airplane is kept in a circular path by means of a control line.
The airplane starts at rest and reaches an angular speed of 3.33 rad/s after
a constant angular acceleration of 0.183 rad/s2. How long does it take the
airplane to reach its final angular speed?
2. A millstone is a large stone wheel once used to grind grain for flour. Sup-
pose a stone in a windmill is moved from rest with a constant angular
acceleration of 0.13 rad/s2. How long does it take the millstone to move
through an angular displacement of 1.6 rad?
3. A ceiling fan’s angular speed increases from 5.2 rad/s to 20.9 rad/s. Dur-
ing this constant angular acceleration, the fan moves through an angular
displacement of 216 rad. How long does it take the fan to reach its final
angular speed?
4. A runner on a circular track increases his angular speed from
0.111 rad/s to 0.178 rad/s. If the runner’s angular acceleration is 1.1 ×10–2 rad/s2, what is his angular displacement?
5. Even rarer than phonograph turntables for long-playing (LP) records are
those turntables used in the first half of the twentieth century. Unlike the
LP turntables, which rotated at 33.3 rev/min, these older turntables
turned at 78.0 rev/min. Suppose a turntable with this angular speed is
shut off, so that it comes to a stop 30.0 s later. If the turntable has a con-
stant angular acceleration of –0.272 rad/s2, what is its angular displace-
ment? Give your answer in both radians and number of revolutions.
Holt Physics Problem BankCh. 7-8
NAME ______________________________________ DATE _______________ CLASS ____________________
6. A rotary saw with an angular speed of 298 rad/s cuts through a piece of
wood. The saw undergoes a constant angular acceleration of –44.0 rad/s2
while it goes through an angular displacement of 276 rad. What is the
saw’s final angular speed?
7. The lumberjack sport of logrolling, or birling, involves two competitors
running along the circumference of a floating log. Each competitor tries
to alter the log’s speed in such a way that the other competitor will fall
into the water. Suppose a lumberjack is practicing logrolling skills alone
on a log. Initially at rest, the log undergoes a constant angular accelera-
tion, so that after 13.0 s the log has turned through 10.0 rev. What is the
final angular speed of the log in rad/s?
8. An automobile engine’s angular speed is increased from 1200 rev/min to
3600 rev/min in 12 s. What is the engine’s angular acceleration in rad/s2?
9. The drive wheel on a vacuum pump has an angular displacement of
158 rad as it accelerates from rest to 70.0 rad/s. What is the wheel’s angu-
lar acceleration?
10. A pet hamster runs inside its exercise wheel. The wheel has an initial an-
gular speed of 3.29 rad/s. Over the next 2.50 s, the hamster increases its
speed and the wheel undergoes a constant angular acceleration. If the an-
gular displacement of the wheel during this time is 12.3 rad, what is the
wheel’s angular acceleration?
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Problem 7E Ch. 7-9
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 7ETANGENTIAL SPEED
P R O B L E MA pog disk rolls on its edge with an angular speed of 44.5 rad/s. If the ra-dius of the pog is 1.75 cm, what is the tangential speed of the pog’s rim?
S O L U T I O NGiven: w = 44.5 rad/s
r = 1.75 cm
Unknown: vt = ?
Use the tangential speed equation to solve for vt.
vt = rw = (1.75 cm)(44.5 rad/s)
vt = 77.9 cm/s
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ADDITIONAL PRACTICE
1. The Maelstrom is a whirlpool formed in a narrow strait among the Lo-
foten Islands, off the coast of northern Norway. It occurs during changes
in the tides, when currents of water move in opposite directions through
the strait. The angular speed of the waters in the Maelstrom is about 2.1 ×10–3 rad/s, though this varies with location. Suppose a boat is at a distance
of 1.5 km from the center of the Maelstrom and the angular speed of the
boat at that point is 2.07 × 10–3 rad/s. What is the boat’s tangential speed?
2. A professionally pitched baseball has an angular speed of 188.5 rad/s.
The radius of a baseball is 3.73 cm. What is the tangential speed of a
point on the baseball’s surface?
3. A wind turbine is a type of windmill that generates electricity. In order
for the turbine to be reasonably efficient, the blades must be very long.
Suppose one of the many turbines located near Palm Springs, California,
has blades that are 15.2 m long and rotate at an angular speed of
6.28 rad/s. What is the tangential speed at the tip of one of these blades?
4. A particular unicycle has a wheel with a radius of 0.30 m. If the unicycle is
ridden with a linear speed of 4.5 m/s, what is the wheel’s angular speed?
5. The main rotor of a certain helicopter consists of blades that extend
2.00 m from the center of the drive shaft. If the tangential speed at the
tips of these blades is 94.2 m/s, what is the rotor’s angular speed?
6. The “Barrel of Fun” is found in “fun houses” at certain amusement parks.
It consists of a cylindrical tunnel that rotates, so that anyone trying to
walk the length of the tunnel is pulled off to one side. Suppose the inner
wall of a “Barrel of Fun” has a tangential speed of 0.63 m/s. If the barrel
has a radius of 1.5 m, what is the barrel’s angular speed?
Holt Physics Problem BankCh. 7-10
NAME ______________________________________ DATE _______________ CLASS ____________________
7. The windmills of the Netherlands are not the only ones in Europe, nor
are they the oldest, but they are among the most famous. These mills
have been used for grinding grain and for draining water to reclaim land
from the sea. Suppose the sails of one of these mills turn with an angular
speed of 3.14 × 10–2 rad/s. If the tangential speed at the tips of the sails is
0.45 m/s, what is the radius of each sail?
8. A grindstone for sharpening knives and axes turns with an angular speed
of 10.0 rad/s. If the tangential speed of the grindstone’s rim is 4.60 m/s,
what is the radius of a grindstone?
9. A toy consists of a track made from two parallel metal wires that are sepa-
rated by a distance of 3 cm. A wheel with a magnetic shaft is placed be-
tween the wires. The wheel spins when the track is tilted, and it remains
on the wires at the end of the track because of the magnetized shaft. In
this way, the angular speed of the wheel can be increased simply by
tilting the track back and forth when the wheel is at the ends of the track.
Suppose the wheel has an angular speed of 11 rad/s and a tangential
speed of 4.0 cm/s. What is the radius of the wheel’s shaft?
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10. An ice skater moves in a circle on a frozen pond. The skater has a tangen-
tial speed of 1.5 m/s and an angular speed of 0.33 rad/s. What is the ra-
dius of the circle in which the skater moves?
Problem 7F Ch. 7-11
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 7FTANGENTIAL ACCELERATION
P R O B L E M
One of the fastest angular speeds for a laboratory centrifuge is 2.5 106 rev/min. Suppose the angular acceleration for this centrifuge is 5800 rad/s2. If the radius of the centrifuge is 0.15 m, what is the tangen-tial acceleration of a point on the rim of the centrifuge?
S O L U T I O N
Given: α = 5800 rad/s2
r = 0.15 m
Unknown: at = ?
Use the tangential acceleration equation to solve for at.
at = rα = (0.15 m)(5800 rad/s2)
at = 870 m/s2
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1. A compact disc has a radius of 6.0 cm. Suppose this disc undergoes an
angular acceleration of 35.2 rad/s2. What is the tangential acceleration of
a dust speck located on the rim of the disc?
2. The drive shaft of an automobile engine undergoes an angular accelera-
tion of 105 rad/s2. If the shaft’s radius is 1.75 cm, what is the tangential
acceleration along the side of the shaft?
3. The drive wheels of the river steamship Grand Republic each had a radius
of 5.87 m. Suppose the angular acceleration of each wheel was 1.40 ×10–2 rad/s2. What would the tangential acceleration at the rim of either
wheel have been?
4. Though it was built by an engineer named Basset rather than by Ferris,
one of the largest and oldest Ferris wheels in the world is located in Vi-
enna at the Prater amusement park. Suppose this wheel accelerates from
rest to its final angular speed of 1.23 × 10–2 rad/s in 10.0 s. The corre-
sponding tangential acceleration along the outer rim of the wheel is
7.50 × 10–2 m/s2. What is the wheel’s radius?
5. An airplane propeller undergoes an angular acceleration of 42 rad/s2. If
the tangential acceleration of the propeller’s tip is 64 m/s2, what is the
propeller’s radius?
6. A large carousel undergoes an angular acceleration of 6.25 × 10–2 rad/s2
before it reaches its final angular speed. During this time a rider on the
edge of the carousel experiences a tangential acceleration of 0.75 m/s2.
What is the carousel’s radius?
ADDITIONAL PRACTICE
Holt Physics Problem BankCh. 7-12
NAME ______________________________________ DATE _______________ CLASS ____________________
7. An experimental bus in Switzerland used the kinetic energy of a massive
spinning flywheel as a means of propulsion. Eventually the flywheel
would slow down, and because there was no engine in the bus, it would
have to be driven periodically to a power station. At this station, the fly-
wheel’s angular speed would be increased by an electric motor. Suppose
that while the flywheel was being brought up to speed, a point on its rim
underwent a tangential acceleration of 0.157 m/s2. If the radius of the
flywheel was 0.90 m, what was its angular acceleration?
8. The largest tire ever made had a radius of 1.75 m. Suppose this tire un-
derwent a linear acceleration 0.83 m/s2. What would the angular acceler-
ation of the tire have been?
9. The old kinds of bicycles with the enormous front wheels and high seats
are called pennyfarthings. This is because the ratio of the big wheel’s area
to the small wheel’s area is similar to the ratio of the areas of an old En-
glish penny to an old English farthing. The radius of the big wheel on a
pennyfarthing is 0.50 m. Suppose a rider accelerates from rest to a linear
speed of 5.0 m/s in 8.5 s. What would the angular acceleration of the big
wheel be?
10.The small wheel of a pennyfarthing has a radius of 16 cm. Suppose the
tangential acceleration along the small wheel’s rim is the same as for the
big wheel in problem 9. What would the angular acceleration of the small
wheel be?
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Problem 7G Ch. 7-13
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 7GCENTRIPETAL ACCELERATION
P R O B L E MThe most violent part of a hurricane is at the edge of the hurricane’s eye.This region, called the eyewall, can have winds with speeds of more than300 km/h. Suppose winds in a hurricane’s eyewall have a tangential speedof 82.2 m/s. If the eyewall is 25 km from the center of the hurricane, whatis the magnitude of the centripetal acceleration of particles in the eye-wall?
S O L U T I O NGiven: vt = 82.2 m/s
r = 25 km
Unknown: ac = ?
Use the centripetal acceleration equation for ac in terms of vt and r.
ac = v
rt2
= (
2
8
5
2.
×2
1
m
0
/3s
m
)2
ac = 0.27 m/s2
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ADDITIONAL PRACTICE
1. Because of the conditions that give rise to them, tornadoes do not have
the widespread destructive effects of hurricanes. Nevertheless, the winds
encountered in some tornadoes are even greater than those at the eyewall
of a hurricane. Suppose a small pebble is swept up in a tornado. The peb-
ble is 3.81 m from the center of the tornado and has a tangential speed
equal to that of the surrounding wind: 124 m/s. What is the magnitude
of the centripetal acceleration of the pebble?
2. A type of pinball machine uses a spinning rubber disk to deflect the ball.
The disk has a radius of 6.50 cm and an angular speed of 30.0 rad/s. Sup-
pose this machine is laid flat so that a ball on the edge of the disk isn’t
pulled to the side by gravity. How large is the ball’s centripetal accelera-
tion when the ball is at the edge of the disk?
3. A customer sits in a revolving restaurant 11 m from the center. If the cus-
tomer’s tangential speed is 1.92 × 10–2 m/s, how large a centripetal accel-
eration does the customer experience?
4. NASA uses large centrifuges to study the effects of large forces on astro-
nauts prior to their going into space. A subject in the 20-G centrifuge,
which has a radius of 8.9 m, can have a centripetal acceleration as large as
20.0g, where g equals 9.81 m/s2. What is the tangential speed of the subject?
5. A rotating furnace has been developed recently to give a rough parabolic
curve to molten glass. This makes the manufacture of very large telescope
mirrors easier and more economical than ever before. The largest mirror
Holt Physics Problem BankCh. 7-14
NAME ______________________________________ DATE _______________ CLASS ____________________
made in this furnace to date has a radius of 4.2 m. While in the furnace,
the centripetal acceleration of the molten glass for this mirror was
2.13 m/s2. What was the tangential speed at the edge of the molten glass?
6. For several decades the idea of an orbiting space colony has been dis-
cussed. The colony would consist of a large hollow cylinder that rotates
at a constant angular speed. Colonists would live on the inner wall of the
cylinder, where centripetal acceleration would simulate free-fall accelera-
tion at Earth’s surface. If the structure has an inner radius of 150 m, what
would the tangential speed of a colonist standing on the cylinder’s inner
wall be?
7. The Indianapolis Motor Speedway has four banked curves, each of which
forms a quarter of a circle. Suppose a race car speeds along one of these
curves with a constant tangential speed of 75.0 m/s. Neglecting the
effects due to the banking of the curve, the centripetal acceleration on
the car is 22.0 m/s2. What is the radius of the curve?
8. A turntable spins with an angular speed of 3.5 rad/s. A quarter placed on
the turntable has a centripetal acceleration of 2.0 m/s2. How far is the
quarter from the center of the turntable?
9. A model electric train moves along a circular track. The train has a tan-
gential speed of 0.35 m/s and has a centripetal acceleration of 0.29 m/s2.
What is the radius of the track?
10.A roller-coaster has a loop-the-loop in which the centripetal acceleration
on the cars and passengers just equals 9.81 m/s2. If the tangential speed
of the roller-coaster cars is 15.7 m/s, what is the radius of the loop-the-
loop?
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Problem 7H Ch. 7-15
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 7HFORCE THAT MAINTAINS CIRCULAR MOTION
P R O B L E MA model airplane with a mass of 3.2 kg moves in a circular path with a ra-dius of 12 m. If the airplane’s speed is 45 m/s, how large is the force thatthe control line exerts on the plane to keep it moving in a circle?
S O L U T I O NGiven: m = 3.2 kg
r = 12 m
vt = 45 m/s
Unknown: Fc = ?
Use the equation for force that maintains circular motion that is expressed in
terms of vt.
Fc = m v
rt2
= 3.2 kg (45
12
m
m
/s)2
Fc = 540 N
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1. A 40.0 kg child takes a ride on a Ferris wheel that rotates with an angular
speed of 0.50 rad/s. Find the magnitude of the force that maintains the
circular motion of the child if the distance from the center of the Ferris
wheel to the child is 18.0 m.
2. A toy model of an amusement park ride has a central shaft that rotates
while carts attached to the top of the shaft by threads “fly” outward. The
force that keeps the carts in a circular path is provided by the tension in
the thread. When the carts are 0.25 m from the center of the shaft, the
largest tangential speed that the carts can have without the threads
breaking is 5.6 m/s. If the mass of a cart is 0.20 kg, how large is the maxi-
mum force that maintains circular motion?
3. An automobile with a tangential speed of 48.0 km/h follows a circular
road that has a radius of 35.0 m. The pavement is wet and oily, so the co-
efficient of kinetic friction between the car’s tires and the pavement is
only 0.500. How large is the force needed to maintain the car’s circular
motion? How large is the available frictional force? Is the available fric-
tional force large enough to maintain the automobile’s circular motion?
Assume the automobile has a mass of 1250 kg.
4. Another automobile with a mass of 1250 kg follows a curved road with a
radius of 35 m. This time, however, the road is banked, so that it tilts
toward the center of the curve with an angle of 9.5°. If the coefficient of
kinetic friction is 0.500, how large is the centripetal force from friction
and gravity? If this force is equal to the force needed to maintain the auto-
mobile’s circular motion, what is the auto’s maximum tangential speed?
ADDITIONAL PRACTICE
Holt Physics Problem BankCh. 7-16
NAME ______________________________________ DATE _______________ CLASS ____________________
5. A small asteroid with a mass of 2.05 × 108 kg is pulled into a circular
orbit around Earth. The distance from the asteroid to Earth’s center is
7378 km. If the gravitational force needed to keep the asteroid in orbit
has a magnitude of 3.00 × 109 N, what is the asteroid’s tangential speed?
6. The first of the great amusement parks on Coney Island, in Brooklyn,
New York, was Steeplechase Park. In 1905, one of the rides in the fun
house at Steeplechase Park was the “Human Roulette Wheel.” This ride
consisted of a large wooden wheel, nearly 6 m in diameter, on which sev-
eral people climbed. The wheel would then spin, causing all but the pas-
sengers closest to the center to slide off. Suppose a passenger with a mass
of 55 kg was able to stay on the wheel, which rotated with an angular
speed of 2.0 rad/s. If the magnitude of the frictional force that held the
passenger on the wheel was 135 N, how far was the passenger from the
center of the wheel?
7. The comets with the longest periods between appearances, as well as
comets that appear only once, come from a region of the solar system
called the Oort cloud. In the Oort cloud, comets have slow tangential
speeds as they orbit the distant sun. Suppose one of these comets has a
mass of 7.55 × 1013 kg and moves with a tangential speed of 0.173 km/s
relative to the sun. If the magnitude of the gravitational force that keeps
the comet in orbit is 505 N, how far is the comet from the sun?
8. The governor for a steam engine rotates with an angular speed of
36.7 rad/s. The balls of the governor are each 0.10 m from the rotation
axis. A force with a magnitude of 670 N is provided by a metal rod, so
that each ball is kept in a circular path. Determine the mass of each ball.
9. To encourage donations of loose change, a zoo has placed an interesting
type of coin well at its visitor’s center. The well is about 1 m tall and is
shaped like the flared bell of a trumpet, with the widest part at the top
and the hole perpendicular to the ground. A coin placed in a chute and
knocked into the well does not simply drop in, but rolls on its edge
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coinwell
coin
rotationvelocity
Problem 7H Ch. 7-17
NAME ______________________________________ DATE _______________ CLASS ____________________
around the rim of the well, gradually moving lower down it. If the coin is
placed correctly, it can still roll around the well’s wall even when the wall
is nearly vertical. Consider a half dollar rolling around the top rim of this
coin well. The radius of the top of the well is 35.0 cm and the coin rolls
around its edge with a tangential speed of 2.21 m/s. If the well’s inner wall
exerts a force of 0.158 N on the rim of the coin, what is the coin’s mass?
10.Since antiquity people have used the sling to increase the speed of a rock
and send it swiftly in a specific direction. While the rock is being spun
overhead, the force that keeps the rock moving in a circle is provided by
the tensile strength of the sling material. Leather has a fairly high tensile
strength, so that a strip of leather with a cross-sectional area of 0.25 cm2
can withstand a pulling force of 800 × 102 N. Assume that, for a certain
sling, 8.00 × 102 N is the largest force that can keep a rock in a circular
path. If the rock in the sling is 0.40 m from the center of rotation and has
a tangential speed of 6.0 m/s, what is the largest mass the rock can have?
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Holt Physics Problem BankCh. 7-18
NAME ______________________________________ DATE _______________ CLASS ____________________
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Holt Physics
Problem 7IGRAVITATIONAL FORCE
P R O B L E MOne of the more mysterious objects in the solar system is a large asteroidor comet called Chiron. Chiron’s orbit lies between the orbits of Jupiterand Uranus, and crosses the orbit of Saturn. Assuming a mass for Chironof 4.5 × 1019 kg, what is the gravitational force between the sun and Chi-ron when the distance between the two is 2.0 × 1012 m? The sun’s mass is1.99 × 1030 kg.
S O L U T I O N
Given: m1 = mass of Chiron = 4.5 × 1019 kg
m2 = mass of the sun = 1.99 × 1030 kg
r = distance between the sun and Chiron = 2.0 × 1012 m
G = 6.673 × 10−11 N
k
•
g
m2
2
Unknown: Fg = ?
Use the equation for Newton’s Universal Law of Gravitation to solve for Fg.
Fg = G m1
r 2m2 = 6.673 × 10−11
N
k
•
g
m2
2
Fg = 1.5 × 1015 N
(4.5 × 1019 kg)(1.99 × 1030 kg)
(2.0 × 1012 m)2
ADDITIONAL PRACTICE
1. The largest fish is the whale shark, which can have a mass of 2.04 × 104 kg.
The largest mammal, and indeed the largest animal ever to have lived on
Earth, is the blue whale, which can have a mass of 1.81 × 105 kg. If the
distance between these two creatures is 1.5 m, how large is the gravita-
tional force between them?
2. Although it contributes only 0.02 percent to Earth’s total mass, the water
in Earth’s oceans is still quite massive. Suppose the water of the oceans
could somehow be drained, kept in liquid form, and moved as far from
Earth as the moon is. How large would the gravitational force between
the water and Earth be? Assume the mass of the ocean’s water to be
1.4 × 1021 kg, the mass of Earth to be the same with water (5.98 × 1024 kg),
and the Earth-moon distance to be 3.84 × 108 m.
3. A gravimeter is a highly sensitive device that measures changes in the
gravitational force in a given area. These measurements reveal variations
in the density of underground rock. This information can be used to in-
dicate whether resources like oil are possibly present. Suppose a certain
type of gravimeter has a test mass of 0.500 kg inside it. How large is the
gravitational force between this mass and a mountain that is 10.0 km
away and that has a mass of 2.50 × 1012 kg?
Problem 7I Ch. 7-19
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4. The magnitude of the gravitational force between two Sumo wrestlers
just before they collide (r = 0.025 m) is 2.77 × 10–3 N. If one wrestler has
a mass of 157 kg, what is the mass of the other wrestler?
5. Jupiter’s largest moon, Ganymede, is also the eighth largest known object
in the solar system. The magnitude of the gravitational force between
Ganymede and Jupiter is 1.636 × 1022 N. Given that Jupiter’s mass is
1.90 × 1027 kg and the distance between Jupiter and Ganymede is
1.071 × 106 km, calculate Ganymede’s mass.
6. The largest known asteroid in the solar system is Ceres, which is located
in an orbit between Mars and Jupiter. Although the existence of Ceres
has been known for two hundred years, its mass is still not well deter-
mined. Suppose the magnitude of the gravitational force between the
sun and Ceres equals 1.17 × 1018 N. Given that the sun’s mass is 1.99 ×1030 kg and the distance between the sun and Ceres is 4.12 × 1011 m,
what is Ceres’ mass?
7. The nearest galaxy with a size similar to that of the Milky Way is the so-
called Andromeda galaxy. By all appearances, this galaxy is very similar to
our own. Suppose both the Milky Way and Andromeda galaxies have
identical masses equal to 500 billion solar masses, or 9.95 × 1041 kg. If
the gravitational force between the two galaxies has a magnitude of
1.83 × 1029 N, what is the distance separating the galaxies?
8. At the sun’s surface, which is called the photosphere, the gravitational
force between the sun and a 1.00 kg mass of hot gas has a magnitude of
274 N. Given that the sun’s mass equals 1.99 × 1030 kg and assuming that
the sun is spherical, what is the sun’s mean radius?
9. At the surface, or photosphere, of the red super giant star Betelgeuse, the
gravitational force between the star and a 1.00 kg mass of hot gas is only
2.19 × 10–3 N. This is because the mean radius of Betelgeuse is so large.
Given that the mass of Betelgeuse is 20 times that of the sun, or
3.98 × 1031 kg, what is the mean radius of Betelgeuse?
10.Comets from the Oort cloud do not frequently enter the solar system
close to the sun. This occurs only when they are perturbed by other
masses, such as other nearby comets, or even the closest stars. Suppose
two comets in the Oort cloud interact with a gravitational force of just
125 N. If the masses of the comets are 4.5 × 1013 kg and 1.2 × 1014 kg,
what is the distance separating them?
Problem 8A Ch. 8–1
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Holt Physics
Problem 8AROTATIONAL EQUILIBRIUM AND DYNAMICS
P R O B L E MWhile driving an automobile, the driver makes a left turn. To performthis maneuver, the driver exerts a torque with a magnitude of 3.5 N•m onthe rim of the steering wheel. If the radius of the wheel is 0.15 m, what isthe magnitude of the force applied by the driver?
S O L U T I O NGiven: t = 3.5 N•m
d = 0.15 m
Unknown: F = ?
Choose the equation(s) or situation: Apply the definition of torque. Note that
the force is applied to the rim of a wheel, so that the angle between the force and
the lever arm (radius) is always 90.0°.
t = Fd(sin q) = Fd
Rearrange the equation(s) to isolate the unknown(s):
F = d
t
Substitute the values into the equation(s) and solve:
F =
F =
Note that the magnitude of torque is not a good indicator of the force that pro-
duces the torque. A small torque can be the product of a large force acting on a
small lever arm, or a small force acting on a large lever arm.
23 N
3.5 N•m0.15 m
1. DEFINE
2. PLAN
3. CALCULATE
4. EVALUATE
ADDITIONAL PRACTICE
1. A lever is used to lift a boulder. The fulcrum is placed 1.60 m away
from the end at which you exert a downward force, producing a torque
with a magnitude of 4.00 × 102 N•m. If the angle between the force
and the lever is 80.0°, what is the magnitude of the applied force? As-
sume that the lever is massless.
2. Suppose the applied force in problem 1 produces a counterclockwise
torque. If the net torque exerted on the lever in problem 1 is 14.0 N•m
counterclockwise, what is the weight of the boulder? Assume that the
lever arm between the boulder’s center of mass and the fulcrum is
0.200 m and that the angle between the boulder’s weight and the lever
arm is 80.0°.
Holt Physics Problem BankCh. 8–2
3. Small windmills have been used for over a century to pump water on
farms and ranches in the United States. The rotors of these mills con-
sist of 18 metal blades called “sails.” Even a small wind can provide
enough torque for drawing water from underground wells. If the
length of a sail is 2.44 m and the torque exerted by the wind is
50.0 N•m counterclockwise, what is the magnitude of the wind’s force?
Assume that this force is exerted perpendicular to the blade and at the
blade’s tip.
4. A force is applied to a door at an angle of 60.0° and 0.40 m from the
hinge. What force produces a torque with a magnitude of 1.4 N•m?
How large is the maximum torque this force can exert?
5. The force exerted by the driving rods of a steam locomotive has a mag-
nitude of 2.27 × 105 N. Each rod is connected to one of the driving
wheels at a point halfway between the center and the rim of the wheel.
Suppose the driving wheel has a radius of 0.660 m. How large is the
maximum torque exerted on the driving wheels by the driving rods?
6. The world’s narrowest street, which is located in a small Italian village,
is only 43 cm wide. Suppose a fish with a mass of 1.6 kg is hung from a
string attached to a stick. The stick, slightly longer than the street is
wide, is placed horizontally across the narrow street with each end rest-
ing on a windowsill. The fish hangs a horizontal distance of 15 cm
from the windowsill on the right. If the axis of rotation for the stick is
taken to be the end farthest from the fish, what is the magnitude of the
torque produced by the fish? Assume the stick has negligible mass.
7. A golfer produces a torque with a magnitude of 0.46 N•m on a golf
club. If the club exerts a force with a magnitude of 0.53 N on a station-
ary golf ball, what is the length of the club?
8. In 1902, fresh water was provided to San Francisco, California, by two
large Dutch-style windmills on the western edge of the city. Though
not in use, both mills are still standing. Suppose a worker is restoring
one of these windmills when the ladder shifts to the side. The worker
grabs the end of one of the rotor vanes and hangs onto it until fellow
workers come to the rescue. The worker hangs at an angle of 65.0° to
the vane, exerting a counterclockwise torque of 8.25 × 103 N•m. If the
worker weighs 587 N, what is the length of the windmill vane?
9. A Foucault pendulum consists of a large balanced mass hanging on the
end of a long wire. The pendulum swings freely, so that the rotation of
Earth causes the pendulum’s surroundings to slowly shift position with
respect to the pendulum’s motion. The pendulum’s apparent path dur-
ing one rotation of Earth is an indication of both Earth’s rotation and
the location of the pendulum on Earth’s surface. At the point where a
28-kg pendulum has the greatest potential energy, the angle between
the pendulum’s weight and the wire is 89°. What is the wire’s length if
the magnitude of the torque exerted by the pendulum’s weight at this
position is 1.84 × 104 N•m?
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Problem 8A Ch. 8–3
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10. At the moment before a diver jumps from a diving board, a force of
1.200 × 103 N is exerted on the diver at an angle of 90.0° to the board.
This force produces a torque in the clockwise direction. At the same
time, the diver’s weight produces a torque in the counterclockwise di-
rection. The diver’s mass is 60.0 kg, and the angle between the diver’s
weight and the board is 87.7°. If the net torque acting on the diver is
2985 N•m clockwise, what is the length of the diving board?
Holt Physics Problem BankCh. 8–4
Holt Physics
Problem 8BROTATIONAL EQUILIBRIUM
P R O B L E MThe narrowest navigable strait lies between the mainland of Greece andthe island of Euboea. Imagine that a uniform bridge spans the gap andthat a car with a mass of 8.2 102 kg is on the bridge. The car is 10.0 mfrom the Euboea side of the strait. The upward force exerted by the landsupporting the bridge on the Greece side is 2.40 104 N, while the up-ward force exerted by the land on the Euboea side is 2.80 104 N. What isthe length of the bridge?
S O L U T I O N
Given: mc = mass of car = 8.2 × 102 kg
dc = distance of car from Euboea = 10.0 m
F1 = upward force on Greece side of bridge = 2.40 × 104 N
F2 = upward force on Euboea side of bridge = 2.80 × 104 N
g = 9.81 m/s2
Unknown: l = length of bridge = ?
Diagram:
Greece Euboea
F1
mbg mcg
F2
dc
1. DEFINE
2. PLAN
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Choose the equation(s) or situation:
Apply the first condition of equilibrium: To find the length of the bridge, the
weight of the bridge must first be determined. All forces are in the vertical (y)
direction.
F1 + F2 − mc g − mbg = 0
Choose a point for calculating net torque: Either end of the bridge may be cho-
sen. Choosing the Euboea side, note that the torque produced by F2 is zero.
Apply the second condition of equilibrium: The torques produced by the
bridge’s and car’s weights are counterclockwise and therefore positive. The nor-
mal force F1 exerts a clockwise, and therefore negative, torque. The lever arm for
F1 is l, while the lever arm for the weight of the bridge is l /2.
tnet = mbg2
l + mcgdc − F1l = 0
Problem 8B Ch. 8–5
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Rearrange the equation(s) to isolate the unknown(s):
mbg = F1 + F2 − mcg
m
2bg − F1l + mcgdc = 0
l =
Substitute the values into the equation(s) and solve:
mbg = 2.40 × 104 N + 2.80 × 104 N − (8.2 × 102 kg)(9.81 m/s2)
mbg = 5.20 × 104 N − 8.0 × 103 N
mbg = 4.40 × 104 N
l =
l =
l =
l =
While the weights of the bridge and car remain constant, the normal forces vary
as the car moves across the bridge (F1 increases as F2 decreases by an equal
amount). The increase in F1, and the resulting increase in the clockwise torque, is
balanced by the increase in the counterclockwise torque produced by the car. The
clockwise torque increases because of an increase in force, while the counter-
clockwise torque increases because of the lengthened lever arm, dc.
4.0 × 101 m
(8.2 × 102 kg)(9.81 m/s2)(10.0 m)
0.20 × 104 N
(8.2 × 102 kg)(9.81 m/s2)(10.0 m)
2.40 × 104 N − (2.20 × 104 N
(8.2 × 102 kg)(9.81 m/s2)(10.0 m)
2.40 × 104 N − 4.40 ×2
104 N
mcgdcF1 −
m
2bg
3. CALCULATE
4. EVALUATE
ADDITIONAL PRACTICE
1. The largest measured ostrich egg—and therefore the largest bird egg—
had a mass of 2.3 kg. By contrast, an adult dwarf rabbit that was born
in France was reported to have a mass of only 0.40 kg. Suppose you
balance this egg and this rabbit on a twin-pan balance, shifting the ful-
crum of the balance to the point along the beam at which the torques
exerted by the two masses cancel. If the beam’s total length is 1.00 m,
and the pans holding the masses are placed at each end of the beam,
where should the fulcrum be placed? Assume that the beam and pans
have negligible mass.
2. A meter stick with a mass of 139 g is found to balance at the 49.7 cm
mark when placed on a fulcrum. Suppose a 50.0 gram mass is attached
at the 10.0 cm mark. To which mark on the meter stick must the ful-
crum be moved for balance?
3. The Firth of Forth Bridge, which spans the Forth River near Edin-
burgh, Scotland, is one of the oldest bridges with a span greater than 2
km. Built in the late nineteenth century, it is also one of the strongest,
Holt Physics Problem BankCh. 8–6
safest, and most expensive bridges ever built. The Forth Bridge (as it is
often called) is a cantilever-type bridge, which means that the most of
the bridge spans are balanced over massive supports and anchored at
one end. Short spans are placed at the free ends of the cantilevers, thus
completing the bridge. Consider a single cantilever shown in the figure
below. The anchored arm of the bridge, A, is shorter and more massive
than the cantilever arm, C. For the Forth Bridge, A has a mass of
roughly 4.64 × 107 kg and a length of 3.00 × 102 m. If the connecting
span between cantilevers exerts a downward force of 3.22 × 107 N on
the end of the cantilever arm, and the support exerts an upward force
of 7.55 × 108 N, what is the length of the cantilever arm?
4. A special ladder used by firefighters is mounted on a hydraulic pump
that permits the ladder to be rotated, tilted, and extended from a set of
controls on the fire engine. When a 70.0 kg firefighter is at the end of
the fully-extended ladder, and the ladder makes an angle of 10.0° with
the vertical, a torque of 7.08 × 103 N•m must be exerted by the ma-
chinery at the base of the ladder in order to keep the ladder from rotat-
ing. If the upward force exerted on the ladder at its base is 3.14 × 103 N,
what is the length of the ladder?
5. A cable is attached 32.0 m from the base of a flagpole that is about to
be raised. The raising of the pole is temporarily halted when the pole is
at an angle of 60.0° with respect to the ground. If the cable exerts a ver-
tical force of 1.233 × 104 N downward and a horizontal force of 1.233 ×104 N to the left, what is the length of the flagpole?
6. Archimedes is supposed to have said, “Give me a place to stand on and
I will move the Earth.” While in principle the claim is valid, the prob-
lems of testing it are vast. Consider the following, highly unrealistic
conditions. Suppose you had a mass equal to that of the Earth—5.98 ×1024 kg— and that it rests on one end of a lever that is balanced on a
fulcrum. The fulcrum is placed on a vast, flat surface in a gravitational
field equal to that at Earth’s surface. Let Earth be placed 1.00 m from
the fulcrum, while the distance from the fulcrum to the applied force is
3.8 × 1016 m, or the approximate distance from the sun to the next
closest star. How large must the applied force be to balance the Earth’s
weight? How large is the upward force exerted by the fulcrum on the
lever? Assume the lever to be massless.
support(fulcrum)
cantileverarm (C)
anchorarm (A)
anchoredend
FcsmAg mcg
Fs
dA dC
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Problem 8B Ch. 8–7
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7. A storefront sign is hung from the end of a 2.00 m uniform massless
rod over the doorway of the business. Additional support for the sign is
provided by a guy wire that is anchored to the wall of the building and
is attached to the end of the rod. The angle between the rod and the
guy wire is 30.0°. If the guy wire provides a torque of 1.47 × 103 N•m
in the counterclockwise direction, what is the weight of the sign? What
is the magnitude of the force exerted by the wire on the sign? What is
the magnitude of the force exerted by the wall on the rod?
8. A garage door with a weight of 7.10 × 102 N is designed to slide side-
ways by being mounted on an overhead rail. Two small wheels on ei-
ther side at the top edge of the door roll along the rail. Suppose the
wheels have become rusted and no longer roll, and that the coefficient
of static friction is 0.75. The wheels are separated by a distance of
1.22 m. A force is applied on one side of the door at a distance of
1.00 m below the rail. How large can this force be without causing the
wheels to rise from the track or the door to slide sideways?
9. A pane of plate glass is tilted upward from the ground until the pane is
at an angle of 70.0° with respect to the ground. Suppose the mass of
the pane is 307 kg and its height is 2.44 m. If the force required to hold
the pane in place is applied perpendicular to the pane and 1.22 m from
the pane’s base, what is the magnitude of the lifting force? How large is
the force exerted on the pane by the ground?
10. Many vending machines have a label on them warning of the dangers
of trying to tip the machine forward. Fortunately, such machines are so
heavy that an average person cannot exert a large enough force to tilt
the machines. Suppose a hinge is placed along the front part of the ma-
chine’s base. With this an additional safety feature, the machine is an-
chored, so that in the event it tips over it rotates around the hinge
rather than tilting forward and sliding backwards. This increases the
force with which it falls forward. Now suppose that the machine is
tilted forward, so that the center of mass has a horizontal position of
0.250 m to the left of the hinge. A horizontal force is applied 1.50 m
above the hinge to prevent the machine from tilting further. If the ma-
chine weighs 1.96 × 103 N, what is the magnitude of the applied force?
What is the magnitude of the force exerted on the machine at the
hinge?
d1
F
Fg
d2
Holt Physics Problem BankCh. 8–8
Holt Physics
Problem 8CNEWTON’S SECOND LAW FOR ROTATION
P R O B L E MA torque of –11 N •m is applied perpendicular to the radius of a potter’swheel. If the angular acceleration on this wheel equals –2.6 rad/s2, what isthe moment of inertia of the wheel?
S O L U T I O NGiven: t = −11 N•m
α = −2.6 rad/s2
Unknown: I = ?
Use the equation for Newton’s second law for rotating objects, and rearrange the
equation to solve for moment of inertia.
t = Ia
I = at
= = 4.2 kg•m2−11 N•m−2.6 rad/s2
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1. A wheel is made to “stand up” by using its diameter as the axis of rota-
tion and by spinning it rapidly. Suppose a torque of 2.98 N•m acts on
this wheel, so that its angular speed increases from 0 rad/s to 55 rad/s
in just 0.75 s. What is the wheel’s moment of inertia? If the wheel’s ra-
dius is 20.0 cm, what is the wheel’s mass, assuming that the wheel is a
thin hoop?
2. A torque of 1.7 N•m must be exerted on a croquet mallet in order for
the mallet to have an angular acceleration of 5.5 rad/s2. What is the
moment of inertia of the mallet?
3. A torque of 0.750 N•m is applied to a plastic flying disk, giving the disk
an angular acceleration of 499 rad/s2. What is the disk’s moment of
inertia?
4. Francis Johnson of Minnesota made a ball of string with a mass of
7.91 × 103 kg and a radius of 1.83 m. Suppose this ball rolls down an
incline with an angular acceleration of 6.13 rad/s2. What is the torque
acting on the ball?
5. The tidal force exerted on Earth by the moon causes Earth’s rotation
speed to slowly decrease. This angular acceleration is equal to –6.53 ×10–22 rad/s2. If the moment of inertia for Earth is described by the
equation 0.331MR2, where M is 5.98 × 1024 kg and R is 6.37 × 106 m,
what is the torque that causes the angular acceleration?
ADDITIONAL PRACTICE
Problem 8C Ch. 8–9
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6. The largest centrifuge equipped for human testing at NASA is capable
of simulating free-fall acceleration up to 20 times that at Earth’s sur-
face. To reach the speeds necessary for a centripetal acceleration as
great as 12g, the centrifuge undergoes a constant angular acceleration
of 1.05 rad/s2. If the moment of inertia of the centrifuge is 8.14 × 104 kg•m2, what is the torque acting on the centrifuge?
7. A clutch consists of two disks whose flat, circular surfaces can be
moved together or apart. In this way, the rotational energy of one disk
can be transferred or removed from the other disk. Suppose the rotat-
ing disk applies a torque of 108 N•m to the stationary disk, which has a
moment of inertia of 5.40 kg•m2. What is the second disk’s angular ac-
celeration?
8. A gyroscope consists of a spinning wheel and shaft that is mounted in a
frame. The stability of the wheel resulting from its large angular mo-
mentum made gyroscopes useful for guiding rockets in the 1940s and
1950s. Suppose a small gyroscope has a moment of inertia of 3.85 ×10–5 kg•m2. If a torque of 1.01 N•m is applied to the rotor of this gy-
roscope, what is the rotor’s angular acceleration?
9. A toy catapult propels a 0.15 kg stone into the air. Assume that the
length of the catapult arm is 0.35 m and that its mass (and therefore its
moment of inertia) is negligible. What is the stone’s angular accelera-
tion if the net torque on the stone is 1.5 N •m? What is the stone’s tan-
gential speed if the angular acceleration lasts for 0.26 s?
10. A 15 kg grindstone with a radius of 0.25 m is initially rotating at an an-
gular speed of 9.5 rad/s. A lawn mower blade pressed against the grind-
stone’s outer edge produces a constant torque of −0.80 N•m. Assum-
ing that the grindstone is a uniform solid disk, what is the angular
acceleration of the grindstone? How long does it take for the grind-
stone to stop?
Holt Physics Problem BankCh. 8–10
Holt Physics
Problem 8DCONSERVATION OF ANGULAR MOMENTUM
P R O B L E MAn experiment aboard a spacecraft consists of a cylindrical bucket, inwhich a disk of ice is placed. The bucket and ice are placed on a frictionlessturntable that rotates with an angular speed of 8.000 rad/s. The tempera-ture inside the bucket is then increased until the ice melts, at which pointthe liquid water forms a ring of water along the wall of the bucket. The an-gular speed of the turntable with the bucket and water is now 7.990 rad/s.If the moment of inertia of the turntable, bucket, and ice is 1.620 kg •m2,what is the moment of inertia of the turntable, bucket, and water?
S O L U T I O NGiven: w1 = 8.000 rad/s
w2 = 7.990 rad/s
I1 = 1.620 kg•m2
Unknown: I2 = ?
Choose The equation(s) or situation: Because there are no external torques, the
angular momentum of the turntable, bucket, and water/ice is conserved.
L1 = L2
I1w1 = I2w2
Rearrange the equation(s) to isolate the unknown(s):
I2 = I
w1w
2
1
Substitute the values into the equation(s) and solve:
I2 =
I2 =
The liquid water redistributes itself because of the bucket’s motion. The ring of
water has a moment of inertia that is larger than a disk of ice with the same mass.
Because the total moment of inertia increases, the angular speed must decrease.
1.622 kg•m2
(1.620 kg•m2)(8.000 rad/s)
7.990 rad/s
1. DEFINE
2. PLAN
3. CALCULATE
4. EVALUATE
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ADDITIONAL PRACTICE
1. A bucket with an inner radius of 0.120 m is rotated on a frictionless
turntable. Inside the bucket are 25 osmium 22.0-g ball bearings. The
turntable, bucket, and bearings rotate with an angular speed of
50.00 rad/s. Suppose that the bucket has small holes underneath and
that these holes can be uncovered by remote control. At a certain point
the holes are opened and the ball bearings fly out. If the angular speed
of the bucket and turntable is now 50.24 rad/s, what is the combined
moment of inertia of the bucket and turntable?
Problem 8D Ch. 8–11
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2. A type of satellite has a cylindrical shape when launched. In orbit, how-
ever, long arms are extended outward, so that the satellite ultimately
takes the form of a long rod. The satellite rotates about an axis that is
perpendicular to its length and passes through its center. Suppose that
before the arms are extended, the satellite’s mass is 755 kg, the distance
between the axis to the end of the satellite is 1.75 m, and its angular
speed of rotation is 1.25 rad/s. After the arms have been extended, the
angular speed of the satellite is 1.70 × 10–2 rad/s. What is the satellite’s
final moment of inertia?
3. The “Human Roulette Wheel” at Steeplechase Park on Coney Island
was a horizontal spinning wheel with a radius of 3.00 m. Passengers
would try to climb toward the center, where the centripetal accelera-
tion was smaller and the passenger would be more able to stay on the
wheel. Suppose the wheel is set to spinning at an angular speed of
2.00 rad/s and then is allowed to spin freely without any external
torque (including friction) applied to it. Four students, each 90° away
from the next, sit on the rim of the wheel, then crawl at the same speed
toward the center of the wheel. When they reach the wheel’s center,
where they are 0.20 m from the axis of rotation, the angular speed of
the wheel is 2.35 rad/s. If the students each have a mass of 55.0 kg,
what is the moment of inertia of the wheel?
4. A tetherball revolves around a pole with an angular speed of 2.1 rad/s
when the rope, which has a length of 1.2 m, is fully extended. What is
the angular speed of tetherball after the rope has wrapped around the
pole several times, so that the ball is only 0.50 m from the axis of
rotation?
5. Most asteroids in the solar system move in orbits outside that of Earth.
However, there are asteroids whose orbital paths cross Earth’s orbit,
causing concern over the possibility of collisions. Some of these aster-
oids move primarily outside Earth’s orbit, crossing it only when they
are closest to the sun. Others move within Earth’s orbit, crossing it only
when they are farthest from the sun. One of the largest of these second
types of asteroids has a tangential speed of 43.5 km/s when it is 7.00 ×107 km from the sun. What is its speed when it crosses Earth’s orbit at a
distance of 1.49 × 108 km from the sun?
6. A student standing at the center of a large turntable holds a spinning
bicycle wheel by a short axle that passes through its center. The bicycle
wheel is held so that it is over the student’s head, so that its axis of rota-
tion is the same as that for the turntable. The angular speed of the bicy-
cle wheel is 57.7 rad/s, while the turntable and student are at rest. The
bicycle wheel has a radius of 0.350 m and a mass of 3.81 kg, which is
concentrated in the wheel’s rim. The combined moment of inertia of
the turntable, student, and wheel is 2.09 kg•m2. If the student suddenly
stops the spinning wheel, what is the angular speed of the turntable,
student, and wheel?
Holt Physics Problem BankCh. 8–12
7. Suppose a satellite identical to the one in problem 2 is launched. The
only difference between this second satellite and the first is that the ini-
tial angular speed is 1.50 rad/s, and the final angular speed is 2.04 ×10–2 rad/s. Use the values for mass and initial arm length from prob-
lem 2 to calculate the final length of the satellite’s arms after they are
completely extended.
8. The planet Pluto is, for most of its orbit, the farthest known planet in
the solar system. However, because its orbit is much more elliptical
than those of the other planets, Pluto is closer to the sun during a
twenty year period of its orbit. It is in the middle of this period of time
that Pluto is closest to the sun. Pluto’s tangential speed is 3.68 km/s
when it is farthest from the sun, at a distance of 7.35 × 109 km. Pluto’s
tangential speed is 6.14 km/s when it is closest to the sun. How close
does Pluto come to the sun?
9. A spacecraft for exploring Mars enters an elliptical orbit around the
planet. The craft travels with a tangential speed of 3403 m/s when it is
3593 km from Mars’s center of mass, or about 2.00 × 102 km from
Mars’s surface. How far is the craft from Mars’s center of mass when its
speed is 3603 m/s? How far is the spacecraft from Mars’s surface?
10. Two skaters, each with a mass of 55.0 kg, approach each other along
parallel paths separated by 5.00 m. They have equal and opposite
speeds of 5.00 m/s. The first skater carries a very light 5.00 m pole, the
end of which the second skater grabs when passing by the first skater.
Now the skaters pull themselves closer to the center of the pole. How
far apart are the skaters if they each have a tangential speed of
15.0 m/s?
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Problem 8E Ch. 8–13
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Holt Physics
Problem 8ECONSERVATION OF MECHANICAL ENERGY
P R O B L E MIn the sport of trapshooting, a small clay disk called a “clay pigeon” islaunched into the air by a device called a “trap.” There are many differentdesigns and levels of complexity to a trap, but a simple device consists of arack in which the clay pigeons are placed. The rack is hinged at one end,and is rapidly pulled forward by a large spring. Suppose this spring isused in a different device that provides a clay pigeon with kinetic energyto roll along a flat surface. The spring has a force constant of 1.05 104
N/m and is pulled 4.0 10–2 m from its equilibrium position. Assumingthat the clay pigeon is a uniform disk with a mass of 95.0 g and a radius of5.60 cm, what is its rotational kinetic energy upon launching?
S O L U T I O N
Given: k = 1.05 × 104 N/m
x = 4.0 × 10−2 m
m = 95.0 g
r = 5.60 cm
Unknown: KErot = ?
Choose the equation(s) or situation: Apply the principle of conservation of me-
chanical energy.
MEi = MEf
Initially, the system possesses only elastic potential energy.
MEi = PEelastic = 12
kx2
When the clay pigeon initially moves away from the spring, all of the potential
energy has been converted to translational and rotational kinetic energy.
MEf = KEtrans + KErot = 12
mvf2 + 1
2Iwf
2
Noting that vf = rwf, and for a uniform solid disk, I = 12
mr2, the equation for MEf
takes the following form.
MEf = 12
mr2wf2 + 1
2Iwf
2 = Iwf2 + 1
2Iwf
2
MEf = 32
Iwf2
The translational kinetic energy can now be expressed in terms of MEf.
KEtrans = Iwf2 = 2
3MEf
From conservation of mechanical energy, MEf can be equated to PEelastic.
MEf = MEi = 12
kx2
KEtrans = 23
MEf = 23
MEi = 23
(12
kx2)
Rearrange the equation(s) to isolate the unknown(s):
12
kx2 = 23
(12
kx2) + KErot
KErot = (12
− 13
)kx2 = 16
kx2
1. DEFINE
2. PLAN
Holt Physics Problem BankCh. 8–14
Substitute the values into the equation(s) and solve:
KErot = 16
(1.05 × 104 N/m)(4.00 × 10−2m)2 =
From the general treatment of the conservation of mechanical energy, it may be
noted that the rotational kinetic energy for a rolling disk is always one-third of
the total kinetic energy. Therefore, it is total potential energy.
2.8 J
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3. CALCULATE
4. EVALUATE
ADDITIONAL PRACTICE
1. Using the information given in the Sample problem, determine the
translational kinetic energy of the clay pigeon.
2. A child’s ball flies a few centimeters above the ground at a speed of
2.2 m/s. Suppose it lands on the ground without bouncing and rolls
forward. What is the rotational kinetic energy of the ball? Treat the ball
as a hollow sphere with a mass of 55 g.
3. An empty barrel falls off a wagon and rolls down a steep hill. If the hill
is 46.0 m high at the point where the barrel starts rolling, what is the
barrel’s translational kinetic energy when it reaches the bottom of the
hill? Assume the barrel is a thin hoop with a mass of 25.0 kg.
4. Suppose that the same arrangement described in the Sample problem
is used to set a metal hoop into motion. Assume a spring constant of
1.05 × 104 N/m and that the spring is stretched 4.0 cm. If the final an-
gular speed of the hoop is 43.5 rad/s, what is its moment of inertia?
5. A bowling ball has an initial kinetic energy of 45 J when it begins to roll
down a bowling lane. What is the moment of inertia of the ball if its an-
gular speed is 27 rad/s? What is the mass of the ball if its radius is 0.11 m?
6. A returning bowling ball rolls in a track between lanes until it reaches
the front of the lane. The ball then rolls up a sloped track; it rolls very
slowly to a circular track near the bowler. The height of the sloped
track is about 0.60 m. What linear speed must the returning bowling
ball have for it to come to a stop at the top of the sloped track?
7. A plunger surrounded by a spring is used in pinball machines to
launch the ball into the field of play. Suppose the force constant for this
spring is 150 N/m, that the spring is compressed 6.0 cm, and that the
ball has a mass of 67 g. Assuming that the pinball machine is level,
what is the linear speed of the ball after the spring has returned to its
equilibrium position?
8. Pinball machines are slightly tilted, so that gravity provides a constant
force on the ball. Suppose the ball has just been launched with the
speed calculated in problem 7. If the ball has a mass of 67 g and the
machine is tilted 3.5° above the horizontal, how far up the incline can
the ball go? Considering that most pinball machines are about 1.5 m
long, would the ball reach the back of the machine and still have ki-
netic energy?
Problem 8E Ch. 8–15
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9. A uniform solid sphere rolls along a horizontal surface at 4.6 m/s. It
then encounters an inclined ramp. What maximum height will the
sphere reach as it rolls up the ramp?
10. A fun slide is constructed so that riders must climb a tower to reach the
top of the slide, at which point they slide down on protective mats.
Suppose a hollow ball with a radius of 15 cm is rolled down the slide.
Assuming that there is no slippage or air resistance, the ball will have
the remarkable angular speed of 102 rad/s when it reaches the end of
the slide. What is the height of the slide?
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Problem 9A Ch. 9–1
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 9ABUOYANT FORCE
P R O B L E MThe average density of Saturn is just 0.685 times that of water. Suppose acube with sides 1.00 m in length could be cut out of Saturn and deliveredto Earth. The cube is allowed to float in fresh water, which has a densityof 1.00 103 kg/m3. What would be the buoyant force acting on the cube?How much of the cube’s volume would be submerged?
S O L U T I O NGiven: rs = (0.685)rw
rw = 1.00 × 103 kg/m3
Vs = (1.00 m)3 = 1.00 m3
g = 9.81 m/s2
Unknown: FB = ?
Vw = ?
Choose the equation(s) or situation: For floating objects, the buoyant force
equals the floating object’s weight.
FB = msg = rsVsg
The volume of the object that is submerged can be calculated by recalling that the
weight of the displaced water equals the buoyant force. The volume of the dis-
placed water equals the submerged volume of the object.
FB = mwg = rwVwg
Rearrange the equation(s) to isolate the unknown(s):
Vw = rF
w
B
g
Substitute the values into the equation(s) and solve:
FB = (0.685)(1.00 × 103 kg/m3)(1.00 m3)(9.81 m/s2) =
Vw = =
The fraction of a floating object’s volume that is submerged in a fluid is equal to
the ratio of the object’s density to the fluid’s density.
0.685 m36.72 × 103 N(1.00 × 103 kg/m3)(9.81 m/s2)
6.72 × 103 N
1. Now suppose the cube from Saturn described in the sample problem is
submerged in gasoline, which has a density of 675 kg/m3. What would
be the buoyant force acting on the cube?
2. One of the rarest non-radioactive elements is rhenium, which is part of
the same chemical group as manganese. The pure metal also has the
third highest melting point of any solid element and the fourth greatest
density of any element. Imagine a cube of rhenium, which has a den-
ADDITIONAL PRACTICE
1. DEFINE
2. PLAN
3. CALCULATE
4. EVALUATE
Holt Physics Problem BankCh. 9–2
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sity of 2.053 × 104 kg/m3, that is 10.0 cm long on each side. Suppose
this cube is partially submerged in fresh water. If the apparent weight
of the rhenium cube is 192 N, what is the buoyant force acting on
the cube?
3. The buoyant force on a floating object is equal to its weight as long as
the average density of the object is smaller than the density of the fluid
in which it is floating. Thus, by taking steel, which is nearly 8 times as
dense as water, and shaping it into a large hollow shell, the average den-
sity of the shell can be made less than that of the water and the shell
will float. This is the principle that allows steel ships to float. However,
the side of the ships shell, or hull, must be high enough so that the hull
does not fill with water, causing its density to increase and the buoyant
force supporting it to decrease. Consider a steel hull with a mass of
1.47 × 106 kg. What is the buoyant force exerted by the water on the
hull? Suppose the hull has a flat base that is 2.50 × 103 m2 in area.
Given that the density of sea water is 1.025 × 103 kg/m3, how deep does
the hull sink into the water?
4. Suppose the hull in problem 3 is made of gold instead of steel. Assum-
ing that the density of the steel in the hull of problem 3 was 7.86 ×103 kg/m3 and given the density of gold as 1.93 × 104 kg/m3, what
would the buoyant force be on the gold hull? If the flat base of the gold
hull has an area of 2.50 × 103 m2, how deep does the gold hull sink into
the sea water?
5. The element osmium has the greatest density of any element. Suppose
a cube of osmium with a volume of 166 cm3 is submerged in fresh
water. The apparent weight of the cube is 35.0 N. Given that fresh
water has a density of 1.00 × 103 kg/m3, what is the density of osmium?
6. Ebony is one of the densest of all woods. Suppose a block of ebony
with a volume of 2.5 × 10–3 m3 is submerged in fresh water. If the ap-
parent weight of the block is 7.4 N, what is the density of ebony? As-
sume the water has a density of 1.0 × 103 kg/m3.
7. Gallium is one of only three non-radioactive elements that melts be-
tween 0°C and 40°C, and the only one for which the solid form has a
lower density than the liquid form. Suppose that a cube of solid gal-
lium with a volume of 7.62 cm3, mass of 45.0 g floats in a container of
liquid gallium. Assuming that the measurement is made before any of
the solid gallium melts, the volume of liquid gallium displaced by the
solid is 7.38 × 10–6 m3. Use this information to calculate the densities
of solid and liquid gallium.
8. The composition Density 21.5 is so-called because it was written for a
musician playing a platinum flute (the density of platinum being
21.5 g/cm3). Suppose this flute is submerged in fresh water, which has a
density of 1.00 g/cm3. If the apparent weight of the submerged flute is
40.2 N, what is the flute’s mass?
Problem 9A Ch. 9–3
NAME ______________________________________ DATE _______________ CLASS ____________________
9. With a density of only 534 kg/m3, lithium has the lowest density of any
metallic element. Suppose a block of lithium floats on a surface of
gasoline, which has a density of 675 kg/m3. If 5.93 × 10–4 m3 of gaso-
line is displaced by the lithium, what is the mass and volume of the
lithium block?
10. A boat floating in a large tank of fresh water contains a passenger and a
heavy object of uniform density. When the boat, passenger, and object
are placed in the water, the surface of the water rises 5.00 cm because of
the volume of water displaced by the buoyant force. The object is then
dropped into the water, where it sinks to the bottom of the tank. The
surface of the water drops from 5.00 cm above its original level to
4.30 cm above its original level. (The increased displacement of the
water due to the submerged object equals the mass of the object, not its
volume. Because the object is denser than the water, the overall volume
of displaced water is less than when the object was floating with the
passenger in the boat.) Use the equation for buoyant force on floating
objects to calculate the mass of the submerged object, given that the
area of the displaced water is 3.4 m2 and that the density of the water is
1.0 × 103 kg/m3.
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Holt Physics Problem WorkbookCh. 9–4
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 9BPRESSURE
P R O B L E MThe surface of the planet Jupiter is believed to consist of liquid hydrogen.Above this surface lies a thick atmosphere that exerts a pressure of 1.00 107 Pa on Jupiter’s surface. If the total force exerted by this atmosphere is6.41 1023 N, what is the area of Jupiter’s surface?
S O L U T I O N
Given: P = 1.00 × 107 Pa
F = 6.41 × 1023 N
Unknown: A = ?
Use the equation for pressure, and rearrange it to solve for the area.
P = A
F A =
P
F
A = 6
1
.
.
4
0
1
0
××
1
1
0
0
2
7
3
P
N
a = 6.41 × 1016 m2
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ADDITIONAL PRACTICE
1. In a certain gasoline engine, the ignited gasoline and air mixture in a
single cylinder exert a pressure of 1.50 × 106 Pa on the upper surface of
the piston. If the force exerted by the gas on the piston is 1.22 × 104 N,
how large is the area on the piston’s upper surface?
2. The two Echo satellites, which were launched in 1960 and 1964, were
early examples of communications satellites. The Echo satellites con-
sisted of aluminum-coated Mylar balloons that reflected radio signals
from one point on Earth to another location on Earth’s surface. Suppose
the gas pressure inside Echo 1 was the same as the atmospheric pressure
at Earth’s surface. If the total force exerted on the inner surface of Echo 1
was 2.86 × 108 N, what was the surface area of the satellite? Given that
the surface area of a sphere is 4pr2, what was the radius of the satellite?
3. A ball strikes the pavement with a force of 5.0 N. If the pressure exerted
on the pavement is 9.6 × 103 Pa, what is the area of contact between the
ball and the pavement?
4. A hydraulic lift consists of a piston, on which is placed an automobile
with a mass of 1.40 × 103 kg. The other piston consists of a block of ice
with a uniform thickness of 0.076 m. Given that the density of ice is
917 kg/m3, what is the area of the piston holding the automobile? As-
sume that the weight of the automobile is uniformly distributed over
the piston’s surface. Assume the piston to be massless.
Problem 9B Ch. 9–5
NAME ______________________________________ DATE _______________ CLASS ____________________
5. A door with an airtight fit around its edge has an area of 1.54 m2. Sup-
pose the pressure on the inside of the door is 1.00 percent greater than
the pressure on the other side of the door. If the outside pressure is
1.013 × 105 Pa, how large a force must you exert on the door to pull it
open? How large a mass can just be lifted by this force?
6. A neutron star is a remnant of a massive star that has undergone an
explosion called supernova. The neutron star’s mass, which mainly con-
sists of neutrons, can be 1.5 times that of the sun. Yet the neutron star’s
radius is only about 6.0 km. Suppose the neutron star is spherical with a
radius of 6.0 km, its surface consists of closely packed neutrons, and the
pressure beneath the surface layer of neutrons is 1.2 × 1016 Pa. If the
surface area of a sphere is 4pr2, what force is exerted by the neutrons?
7. An engineer devises an elevator for tall buildings that does not rely on
cables suspending a car. The elevator consists of a piston that is raised
by a non-compressible fluid, which is pushed upward by another,
larger piston. Suppose the elevator piston is designed to rise 448 m
above street level and that the maximum load that can be lifted is
4.45 × 104 N. What force must be exerted on the larger piston if its
maximum displacement is 8.00 m? Assume the pistons to be massless.
8. Atmospheric pressure is often given in units of millimeters of mercury.
This refers to the height of a mercury column above the mercury’s sur-
face at the base of a barometer. The force exerted by the atmosphere on
the surface of the mercury in the reservoir equals the weight of the mer-
cury in the column. If the mercury column extends 760 mm above the
mercury’s surface in the reservoir, what is the atmosphere’s pressure
over the mercury? Use 13.6 × 103 kg/m3 for the density of mercury.
9 Osmium, the densest of all known elements, is rarely found in Earth’s
crust. This is partly because, being more dense than other elements,
most osmium present on Earth has drifted toward Earth’s center. Sup-
pose Earth’s supply of osmium was mined and formed into a single
massive cylinder. If the mass of this cylinder were 2.4 × 1013 kg and the
area of the cylinder were 3.14 km2, the cylinder would be 340 m high.
If this cylinder could be placed on a hydraulic lift with a piston area of
3.14 km2, what pressure would be necessary to support the osmium
cylinder? Assume the piston to be massless.
10. When a can of soda is shaken vigorously, carbon dioxide is temporarily
separated from the water. This causes the pressure inside the can to in-
crease. If the soda is opened before the carbon dioxide can be redis-
solved in the water, the gas—and soda—will rapidly escape through
the opening. Suppose the force that the carbon dioxide gas exerts on
the inside of the sealed can is 4.4 × 103 N. If the surface area of the can
is 2.9 × 10–2 m2, what is the absolute pressure exerted by the carbon
dioxide? What is the gauge pressure of the gas, assuming that the pres-
sure of the air outside the can is 1.0 × 105 Pa?
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Holt Physics Problem WorkbookCh. 9–6
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 9CPRESSURE AS A FUNCTION OF DEPTH
P R O B L E MSaturn’s largest moon, Titan, is actually larger than the planet Mercury. Itis also one of the few bodies in the solar system that has a dense atmos-phere. Data from the Voyager 1 probe indicates that the atmosphericpressure on Titan’s surface is about 1.60 105 Pa, even though free-fallacceleration at Titan’s surface is only 1.36 m/s2. Assuming that Titan’s at-mosphere has a uniform density of 2.32 kg/m3, at what height above thesurface is Titan’s atmospheric pressure equal to 1.01 105 Pa?
S O L U T I O N
Given: Po = 1.60 × 105 Pa
gT = 1.36 m/s2
r = 2.32 kg/m3
P = 1.01 × 105 Pa
Unknown: h = ?
Use the equation for fluid pressure as a function of depth, and rearrange it to
solve for height, which yields a negative value for h.
P = Po + rgh
h = P
r−
g
Po
h =
h =
h = −1.9 × 104 m
h = 19 km above Titan’s surface
−5.9 × 104 Pa(2.32 kg/m3)(1.36 m/s2)
1.01 × 105 Pa − 1.60 × 105 Pa
(2.32 kg/m3)(1.36 m/s2)
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ADDITIONAL PRACTICE
1. The barheaded goose breeds in Tibet and spends the winter in India.
This migration requires the goose to fly at extremely high altitudes, so
that it can fly over the Himalayan mountains. The average atmospheric
pressure at the goose’s minimum altitude is 6.9 × 104 Pa. Because air is
not an incompressible fluid, its density is not uniform at all altitudes;
therefore assume that the air has a much lower average density of
0.55 kg/m3. From this data, calculate the altitude at which the bar-
headed goose flies. Use the value of 1.01 × 105 Pa for the atmospheric
pressure at sea level.
Problem 9C Ch. 9–7
NAME ______________________________________ DATE _______________ CLASS ____________________
2. Throughout Yucatán, Mexico, there are sinkholes called cenotes. Filled
with water and often connected to underground caves, these cenotes
form some of the world’s deepest pools. Rumored to be the deepest of
these is the Cenote Azul, which has sheer walls and a true depth that
has never been measured. At the cenote’s greatest known depth, the
pressure is 1.47 × 106 Pa greater than the pressure at the water’s sur-
face. Using a density of 1.00 × 103 kg/m3 for the density of the water,
determine the maximum known depth of the Cenote Azul.
3. The surface of Lake Tahoe, California, lies 1899 m above sea level. At
this elevation, the average atmospheric pressure is about 9.0 × 104 Pa.
At the deepest part of the lake, the pressure is 5.00 × 106 Pa. If the den-
sity of the water in Lake Tahoe is 1.0 × 103 kg/m3, what is the maxi-
mum depth of the lake?
4. At the greatest depths that scuba divers can work (and then only
briefly), the pressure is 4.03 × 105 Pa. Noting that the density of sea
water is 1.025 × 103 kg/m3, what is the maximum depth for scuba
divers?
5. The world’s first underwater hotel was built in 1986 at Key Largo,
Florida. The hotel is located 9.1 below the surface of a mangrove
swamp. If the water had a density of 1.0 × 103 kg/m3, what would be
the gauge pressure across a window in the hotel at a depth of 9.1 m?
6. The lowest elevation in the western hemisphere is 86 m below sea level,
in Death Valley, California. Assuming the density of the air to be
1.29 kg/m3, what is the atmospheric pressure at this location?
7. Cinnabar is the mineral from which mercury is extracted. Consider a
pool of mercury in a cinnabar mine. Given that mercury’s density is
13.6 × 103 kg/m3 and that the atmospheric pressure above the pool is
1.01 × 105 Pa, what is the pressure 1.50 m below the mercury’s surface?
8. The planet Venus has one of the densest atmospheres in the solar sys-
tem. At the surface of Venus, the atmosphere’s pressure is nearly 9.10 ×106 Pa. Even at an elevation of 1.00 km, the atmospheric pressure is
8.60 × 106 Pa. Given that free-fall acceleration on the surface of Venus
is 8.87 m/s2, what is the density of Venus’ atmosphere?
9. Imagine a giant vat containing liquid bromine. The pressure at the sur-
face of the bromine is 1.01 × 105 Pa, but at a depth of 3.99 m below the
liquid’s surface the pressure is 2.23 × 105 Pa. What is the density of liq-
uid bromine?
10. Now imagine the same vat as in problem 9, but containing the organic
compound diethyl ether. If the pressure at a depth of 3.99 m below the
ether’s surface is 1.29 × 105 Pa, what is the density of the diethyl ether?
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Holt Physics Problem WorkbookCh. 9–8
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 9DBERNOULLI’S EQUATION
P R O B L E MA feature of dams is a spillway, which provides for excess water from up-stream flooding to bypass the dam. At Hoover Dam on the border of Ari-zona and Nevada, the Arizona spillway is a tunnel with a 15 m diameterthat opens to Lake Mead near the top of the dam. Water levels in the lakehave not been high enough for the spillway to be used since it was firsttested, but if they were to reach that height they would be channeledthrough the spillway to a location downriver from the dam. Assumingthat the bottom of the spillway is on the same level as the bottom of thedam, if water enters the spillway with a speed of nearly 0 m/s, it will leavethe spillway with a speed of 65.83 m/s. Given that the pressure differencebetween the top and the bottom of the dam is 1.2 103 Pa and that freshwater has a density of 1.00 103 kg/m3, what is the height of Hoover dam?
S O L U T I O NGiven: v1 = 0 m/s
v2 = 65.83 m/s
∆P = P2 − P1 = 1.2 × 103 Pa
r = 1.00 × 103 kg/m3
g = 9.81 m/s2
Unknown: ∆h = h1 − h2 = ?
Choose the equation(s) or situation: Because this problem involves fluid flow, it
requires the application of Bernoulli’s equation.
P1 + 12
rvi2 + rgh1 = P2 + 1
2rv2
2 + rgh2
Rearrange the equation(s) to isolate the unknown(s):
rg(h1 − h2) = P2 − P1 + 12
r(v22 − v1
2)
∆h = h1 − h2 =
∆h = ∆r
P
g +
v22
2
−g
v12
∆h = +
∆h = 0.12 m + 221 m
∆h =
The difference in atmospheric pressures on the flowing water does not contribute
significantly to the water’s motion. In this case, Bernoulli’s equation simplifies to
rgh = 12
rv22
This is simply the principle of mechanical energy conservation applied to fluid of
undetermined volume. The units on either side of the equation are J/m3.
221 m
(65.83 m/s)2 − (0 m/s)2
(2)(9.81 m/s2)
1.2 × 103 Pa(1.00 × 103 kg/m3)(9.81 m/s2)
P2 − P1 + 12
(v22 − v1
2)
g
1. DEFINE
2. PLAN
3. CALCULATE
4. EVALUATE
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Problem 9D Ch. 9–9
NAME ______________________________________ DATE _______________ CLASS ____________________
1. A pressure difference of 1.5 × 103 Pa draws water from a well. The
water emerges from the pump with a speed of 17.0 m/s. Given that the
density of the water is 1.00 × 103 kg/m3, what is the depth of the well?
2. A major tourist attraction in the town of Collinsville, Illinois, is the
“world’s largest catsup bottle.” The bottle is actually a water tower
shaped like the bottle of a once-locally produced catsup. The 100 000-
gallon tank is mounted on a tower that is about 69 m above the
ground. Imagine a water tank that is as tall as the catsup bottle, but is
cylindrical, open to the air, and filled to the top. A hole at the base of
the tank causes a horizontal stream of water to land 120 m horizontally
from the hole and 69 m below the base of the tank. If the difference in
atmospheric pressure is nearly the same at the top and base of the tank,
how tall is the tank?
3. A hydraulic ram is a device that uses the pressure created by a large
amount of water falling a small distance to raise a portion of that water
to a much greater height. It is suitable in valleys where springs feed a
stream below, and where farms needing irrigation are at the top of the
valley. Suppose a spring feeding a hydraulic ram pours water with a
speed of 2.00 m/s into a pipe, which descends a certain distance to the
base of the ram. At the ram’s base, the water has a speed of 7.93 m/s.
How far has the water descended? Assume that the pressure at the two
levels is the same.
4. Now water is raised up the hill by the hydraulic ram to a water storage
tank. The water leaves the ram with a speed of 24.45 m/s and arrives
with a speed of just 0.55 m/s. If there is no significant change in pres-
sure between the two levels, what is the height to which the water is
raised?
5. Rainwater flows through a drainage conduit with an initial speed of
2.50 m/s. How fast does it emerge from the other end of the conduit,
which is 3.00 m below the upper opening? Assume that the pressure is
the same at both ends of the conduit.
6. Sea water flows through a level pipe in a tidal power station. The initial
speed of the sea water is 0.90 m/s. If the water’s pressure increases by
311 Pa as it passes through another section of pipe, what is the water’s
speed in the second section? The density of sea water is 1.025 × 103 kg/m3.
7. A device called a Pitot tube uses the change in height a liquid’s height
to measure the speed a fluid passing through the tube. Suppose a Pitot
tube is attached to the wing of an airplane, which flies at a low altitude
above the ground. The air flowing through the Pitot tube has a density
of 1.3 kg/m3. The part of the tube used for measurement contains mer-
cury, which has a density of 1.36 × 104 kg/m3. If the displaced mercury
column is 3.5 cm high, what is the speed of the air flowing across the
airplane’s wing? (HINT: Solve Bernoulli’s equation for the change in
pressure of each fluid; then substitute one equation into the other.)
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Holt Physics Problem BankCh. 9–10
NAME ______________________________________ DATE _______________ CLASS ____________________
8. Oil transported along a level pipeline has a density of 950 kg/m3. The
oil moves with a speed of 10.00 m/s at one point along the pipeline, but
after traveling 1 km the oil has slowed to a speed of 9.90 m/s. What is
the difference in pressure between the two points in the pipeline?
9. Water leaves a fire hydrant pipe, which has an area of 7.8 × 10–3 m2,
with a speed of 2.0 m/s. The firehose attached to the hydrant has a noz-
zle with an area of 3.1 × 10–4 m2. If the nozzle is 10.5 m higher than the
hydrant, what is the gauge pressure of the water? Use 1.00 × 103 kg/m3
as the density of water. (HINT: Use the continuity equation to express
the Bernoulli equation in terms of the known fluid speed.)
10. A Venturi meter, like the Pitot tube, is a device that measures the
change in pressure in a moving fluid. The pressure change occurs when
the fluid passes through sections of the tube with different cross-
sectional areas. Suppose fresh water flows through a level Venturi
tube. At one section of the tube, where the cross-sectional area is
9.2 × 10–2 m2, the speed of the water is 8.3 m/s. At a second section
of the tube, the cross-sectional area of the tube is 4.6 × 10–2 m2. If
density of the water is 1.0 × 103 kg/m3, what is the pressure difference
between the two sections of the Venturi tube? (HINT: Use the continu-
ity equation to express the Bernoulli equation in terms of the known
fluid speed.)
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Problem 9E Ch. 9–11
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 9ETHE IDEAL GAS LAW
P R O B L E M
An ideal gas kept under constant pressure has a volume of 1.5 m3 whenthe gas is at a temperature of 380.0 K, which is the maximum tempera-ture of the two-week lunar day. The moon’s lowest temperature, whichoccurs during the two weeks of night, is 100.0 K. If the ideal gas is cooledto this temperature, what is its new volume?
S O L U T I O N
Given: V1 = 1.5 m3
T1 = 380.0 K
T2 = 100.0 K
Unknown: V2 = ?
Choose the equation(s) or situation: To solve for the volume under the new
conditions, apply the form of the ideal gas law that relates initial and final states.
P
T1V
1
1 = P
T2V
2
2
Because pressure remains constant, the equation simplifies to
V
I1
1 = V
I2
2
Rearrange the equation(s) to isolate the unknown(s):
V2 = V
T1T
1
2
Substitute the values into the equation(s) and solve:
V2 = (1.5 m
38
3)
0
(
.0
10
K
0.0 K) =
The estimated answer, using T1 ≈ 4.0 K, is
V2 ≈ 1.5
4
m3
≈ 0.4 m3,
which corresponds closely to the result.
0.39 m3
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1. An ideal gas kept under constant pressure is heated from the lowest
recorded temperature on Earth, 184 K, to the highest recorded temper-
ature, 331 K. If the initial volume of the gas is 3.70 m3, what will its
volume be after heating?
ADDITIONAL PRACTICE
1. DEFINE
2. PLAN
3. CALCULATE
4. EVALUATE
Holt Physics Problem BankCh. 9–12
NAME ______________________________________ DATE _______________ CLASS ____________________
2. An ideal gas held at constant temperature has a volume of 0.455 m3
and a pressure of 9.1 × 106 Pa, which is equal to the atmospheric pres-
sure on the surface of the planet Venus. Suppose this pressure is low-
ered to 5.0 × 104 Pa, which is equal to the atmospheric pressure on the
surface of Mars. What will be the volume of the gas after this reduction
of pressure?
3. A balloon filled with helium has a volume of 65.4 m3. The balloon is
insulated so that the temperature of the helium remains constant. The
balloon is launched from the Dead Sea, the lowest point on Earth’s sur-
face, where the pressure is 1.03 × 105 Pa. What will be the volume of
the balloon when it is at an elevation equal to the summit of Mt. Ever-
est, where the atmospheric pressure is 5.84 × 104 Pa?
4. The lowest level of Earth’s atmosphere is the troposphere. At Earth’s
surface in mid-latitude regions, the average temperature and pressure
are 295 K and 1.01 × 105 Pa, respectively. At the top of the troposphere,
at an altitude of about 10 km, the average temperature is about 225 K
and the pressure is 5.10 × 104 Pa. What would be the volume of 5.55 ×1022 particles of an ideal gas at the top and bottom of the troposphere?
5. An ideal gas at a pressure of 7.5 × 104 Pa (the approximate pressure at
the top of Saturn’s atmosphere) has a temperature of 250 K. Suppose
the pressure of the gas is increased to 2.0 × 106 Pa (the estimated pres-
sure on Saturn’s surface). If the volume of the gas is kept constant,
what will its new temperature be?
6. A volume of 1.00 m3 of helium is placed in a balloon. The temperature
of the gas is 295 K. If the gas is kept at constant pressure, what temper-
ature must the helium have for the volume of the balloon to expand to
65.4 m3?
7 Although the conditions are extreme, the ionized gas within the sun’s
core behaves like an ideal gas. The core contains about 2.1 × 1057 gas
particles. The pressure of the gas in the sun’s core is 2.1 × 1016 Pa, while
the volume of the core is 2.1 × 1025 m3. What is the temperature at the
sun’s core?
8. A tank of helium is kept at a constant volume with a pressure of 2.50 ×105 Pa and a temperature of 295 K. Suppose the warehouse where the
tank is stored catches fire, so that the tank and the helium reach a tem-
perature of 506 K. What is the pressure of the gas at this temperature?
9. Hydrogen is the most common element in the universe. Large clouds
of hydrogen are often found in regions between the stars. Some of
these clouds have high temperatures and pressures as stars form within
them. Others have lower temperatures near 1.0 × 102 K. Suppose the
volume of a hydrogen cloud at this temperature is 3.3 × 1043 m3, and
that the density of the hydrogen is 10.0 atoms/cm3. What would the
pressure of the hydrogen gas in the cloud be?
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Problem 9E Ch. 9–13
NAME ______________________________________ DATE _______________ CLASS ____________________
10. The tallest mountain on Earth is Mauna Kea, which is the taller of the
two large volcanoes on the island of Hawaii. The island rises 5995 m
from the ocean floor, and then Mauna Kea rises another 4205 m above
sea level for a total height of 10 200 m (or nearly 2 km higher than
Mt. Everest). Suppose the oceans were drained and all of Mauna Kea
were exposed to air. At the base of the mountain, the air pressure
would be 1.42 × 105 Pa, assuming that air pressure at what was previ-
ously sea level is 1.01 × 105 Pa. If a balloon containing 1.00 m3 of he-
lium is launched from the base of the mountain, its volume will in-
crease to 1.83 m3 by the time it reaches the summit of the mountain. If
the balloon is insulated to keep the helium at constant temperature,
what is the pressure at the mountain’s summit?
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Problem 10A Ch. 10–1
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NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 10ATEMPERATURE
P R O B L E MSummertime temperatures on Pluto’s surface average 45 K—the onlyplanet cold enough to keep methane solid. What is this temperature onthe Celsius and Fahrenheit scales?
S O L U T I O NGiven: T = 45 K
Unknown: TC = ? TF = ?
Choose the equation(s) or situation:
Use the Celsius-Kelvin equation and the Celsius-Fahrenheit equation.
TC = (T − 273)°C = (45 – 273)°C =
TF = 9T
5C + 32 =
9
5 (−228)°F + 32°F = (–4.10 + 102)°F + 32°F
TF = −378°F
−228°C
ADDITIONAL PRACTICE
1. When Mercury was farthest from the Sun in 1974 and 1975, the U.S.
probe Mariner 10 flew by Mercury three times to collect data. Surface
temperatures ranged from 463 K during the day to 93 K at night. Ex-
press this temperature range in degrees Fahrenheit and in degrees
Celsius.
2. On July 10, 1913, the temperature reached 330.0 K Death Valley, Cali-
fornia—the hottest temperature ever reached in the United States. Cal-
culate this temperature in degrees Fahrenheit and in degrees Celsius.
3. On January 21, 1918, Granville, North Dakota had a surprising change
in temperature. Within 12 hours, the temperature changed from 237 K
to 283 K. What is this change in temperature in the Celsius and
Fahrenheit scales?
4. In only 15 minutes, the temperature in Fort Assiniboine, Montana,
went from −5°F to +37°F on January 19, 1892. Calculate this change in
temperature in kelvins.
5. Hypothermia is a condition in which the body gives up to much energy
by heat to its colder surroundings. If a person’s body temperature
drops to 90.0°F, they can lose consciousness, and if their temperature
falls to 78°F, they can die. What is this second temperature in kelvins?
Holt Physics Problem BankCh. 10–2
NAME ______________________________________ DATE _______________ CLASS ____________________
6. Just as the human body cannot survive if its temperature falls to too
low a temperature, it also cannot survive if its temperature is too high.
In a condition called hyperthermia, energy is transferred to the body
from its surroundings, causing the body’s temperature to increase. The
condition known as heat stroke is a severe form of hyperthermia. Nor-
mally, the human body cannot survive for long at a temperature of
about 42°C, although a recent survivor of heat stroke had a high tem-
perature of nearly 47°C. Express both of these temperatures in kelvins.
7. Air that slowly falls from high altitudes can result in cold fronts that
sweep across many states. At 30,000 feet above Earth’s surface, the air
temperature can be around –67°F. Find this temperature in kelvins.
8. The Hawaiian lavas at Kilauea Crater have the highest temperatures
measured on Earth’s surface—over 2192°F! Express this temperature in
degrees Celsius.
9. Much of the hot water in Reykjavík, Iceland comes from wells bored
into the hot springs of Reykir. The water temperature from the wells is
188.6°F. Express this temperature in degrees Celsius.
10. The present temperature of the background radiation in the universe is
2.7 K. What is this temperature in degrees Celsius?
Problem 10B Ch. 10–3
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 10BCONSERVATION OF ENERGY
P R O B L E MA bright fireball zoomed across four states on October 9, 1992. Supposethe meteorite had a speed of 35.0 km/s just before it smashed through thetrunk of a Chevrolet Malibu in Peekskill, New York. At the time the car’sowner discovered the 11.99 kg meteorite on the ground beneath the car,the meteorite was still warm. Suppose that, at the time of impact, the in-ternal energy of the meteorite had increased by an amount equal to 0.001times the kinetic energy before impact, while the rest of the kinetic en-ergy was transferred to the ground, car and atmosphere. If 5348 J of en-ergy were required to increase the meteorite’s temperature by a degreeCelsius, what was the meteorite’s temperature upon impact?
S O L U T I O NGiven: vi = meteorite speed = 35.0 km/s
m = meteorite mass = 11.99 kg
k = energy needed to raise meteorite’s temperature by 1°C
= 5348 J
∆Umeteorite = (0.00100)KEi
Unknown: ∆T = ?
Choose the equation(s) or situation: Use the equation for energy conservation,
including the term for the change in internal energy.
∆PE + ∆KE + ∆U = 0
∆U stands for the change in the internal energy of the meteorite, car, and
ground. The change in potential energy from just before to after impact and
the final kinetic energy are both zero.
∆PE + ∆KE + ∆U = 0 + KEf – KEi + ∆U = – KEi + ∆U = 0
∆U = KEi = 12
mvi2
The change in the meteorite’s internal energy is 0.00100 times the total
change in internal energy.
∆Umeteorite = (0.00100)∆U = (0.00100)KEi = (0.0010
2
0)mvi2
To find the change in the meteorite’s temperature, the change in the mete-
orite’s internal energy must be divided by the conversion constant, k.
∆T = ∆Um
keteorite =
(0.001
2
0
k
0)mvi2
Substitute the values into the equation(s) and solve:
∆T =
∆T =
The answer can be estimated by taking the ratio of m to k as 10
2
00. This
∆T ≈ (35)2 °C = 1200° C, which is close to the calculated value.
1370°C
(0.00100)(11.99 kg)(35.0 × 103 m/s)2
(2)(5348 J/°C)C
opyr
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©by
Hol
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nd W
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on.A
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hts
rese
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.
1. DEFINE
2. PLAN
3. CALCULATE
4. EVALUATE
NAME ______________________________________ DATE _______________ CLASS ____________________
1. A football game begins by flipping a penny to decide which team will get
the ball first. The 5.25 g penny has a speed of 3.27 m/s just before it hits
the ground. If half of the increase in the internal energy of the ground
and penny is transferred to the penny’s internal energy, by how much
does the penny’s temperature increase after impact? Assume that the
penny’s temperature increases by 1.00° C as 2.03 J are added to its mass.
2. A squirrel drops an acorn from the top of a tree that is 9.5 m tall. When
the acorn lands, 85 percent of the increase in the internal energy of the
ground and acorn is transferred to the acorn’s internal energy. If it
takes 1200 J/kg to increase the acorn’s temperture by 1.0°C, by how
much does the acorn’s temperature increase upon landing?
3. On a hot summer’s day, a child’s triple-dip ice cream cone falls to the
sidewalk. The gravitational potential energy associated with the cone is
initially 6.2 J. After impact, the internal energy of the ice cream cone
increases by an amount equal to 10.0 percent of the increase in the in-
ternal energy of the ground and ice cream cone. What is the increase in
the ice cream cone’s temperature if after impact it takes 180 J to in-
crease the ice cream’s temperature by 1.0°C?
4. Although the peregrine falcon can fly faster in a dive, the fastest-flying
bird in general is the Asian spine-tailed swift, which can fly at 47.5 m/s.
Suppose this bird drops a 7.32-g twig she is carrying to build her nest
while she is flying 151 m above the ground. If after impact 10.0 percent
of the total internal energy of the twig and ground is transferred to the
twig’s internal energy and it takes 8.5 J to raise the twig’s temperature
by 1.0°C, by how much does the twig’s temperture increase when it
strikes the ground?
5. Balls Pyramid, a rock pinnacle off the eastern coast of Australia, is
561.7 m tall. Suppose erosion causes a stone to fall from the summit.
When the stone lands, its kinetic energy is transferred to the internal
energy of the ground and the stone, which increases by 105 J. What is
the mass of the stone?
6. Bones found from a Tyrannosaurus suggest that the dinosaur was 5.5 m
high and up to 14.3 m long. Suppose a Tyrannosaurus dropped its prey
while eating. When it lands, all of the prey’s kinetic energy, 2.77 × 103 J,
is transferred to the internal energy of the ground and prey. What was
the mass of the prey?
7. A person riding a bicycle on level ground reaches a velocity of 13.4 m/s
before stopping. While braking, the internal energy of the brakes,
wheels, and road increase by 5836 J. What is the mass of the cyclist? Cop
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Holt Physics Problem BankCh. 10–4
ADDITIONAL PRACTICE
NAME ______________________________________ DATE _______________ CLASS ____________________
8. You can start a campfire by rubbing two dry sticks together. To create
sparks, rest the larger stick on the ground. Press the point of the
smaller stick into the larger stick so that the sticks are at right angles.
You then rotate the smaller stick between your hands. Suppose you
begin to see smoke when you rotate the stick so that you transfer a ki-
netic energy of 2.15 × 104 J. What is the internal energy at the point
where the sticks meet?
9. The fastest roller skater skated over 402 m, reaching a final velocity of
20.5 m/s. Suppose this roller skater has a mass of 61.4 kg. What is the
total change in the internal energy of the skates, wheels, and ground as
the skater skates past the finish line?
10. Suppose on a cold winter day, you forgot to bring your gloves. To warm
your hands, you rub them together vigorously with a kinetic energy of
7320 J. What is the change in the internal energy of the skin on your
hands if 30.0% of the total mechanical energy was transferred to the air
by heat?
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Problem 10B Ch. 10–5
Holt Physics Problem BankCh. 10–6
NAME ______________________________________ DATE _______________ CLASS ____________________
P R O B L E M
S O L U T I O N1. DEFINE
2. PLAN
3. CALCULATE
4. EVALUATE
1. DEFINE
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Holt Physics
Problem 10CCALORIMETRY
In 1906, a 636.73-g diamond was found at the Premier Mine in SouthAfrica—making it the world’s largest uncut diamond. After being cut, thediamond pieces were dropped into an insulated water bath containing1.00 kg of water. The water temperature increased by 1.30C, and the dia-mond’s temperature decreased by 15.54°C. What is the specific heat ca-pacity of diamond? The specific heat capacity of water is 4186 J/kg•C.
S O L U T I O NGiven: md = 636.73 g mw = 1.00 kg ∆Td = 15.54°C
∆Tw = 1.30°C cp,w = 4186 J/kg•°C
Unknown: cp,d = ?
Choose the equation(s) or situation: Use the equation for specific heat capacity to
equate the energy removed from the diamond to the energy absorbed by the water.
Energy removed from diamond = Energy absorbed by water
cp,dmd∆Td = cp,wmw∆Tw
Rearrange the equation(s) to isolate the unknown(s):
cp,d =
Substitute values into the equation(s) and solve:
cp,d = =
The answer can be estimated using rounded values for cp,w (4200 J/kg•°C), md 23
kg, and
the ratio of ∆Tw to ∆Td 1
1
2. The resulting value for cp,d is then 530 J/kg•°C, which is close
to the calculated value.
5.50 × 102 J/kg•°C(4186 J/kg•°C)(1.00 kg)(1.30°C)
(0.63673 kg)(15.54°C)
cp,wmw ∆Twmd∆Td
ADDITIONAL PRACTICE
1. Mixing equal parts of hydrogen peroxide and water to use as a mouth-
wash disinfects the mouth and whitens teeth. Suppose you mix 15 g of
each into a plastic foam cup. The water’s temperature changes 1.0°C
and the hydrogen peroxide changes 1.6°C. Disregarding energy transfer
as heat to the solution’s surroundings, what is the specific heat capacity
of hydrogen peroxide?
NAME ______________________________________ DATE _______________ CLASS ____________________
2. Vinegar, which contains acetic acid, can be used as an effective and
environmentally-friendly household cleanser. Suppose you mix 0.340 kg
of vinegar at 21.0°C with 1.00 kg hot water at 90.0°C in a plastic bucket.
The solution of vinegar and water reaches a final equilibrium tempera-
ture of 73.7°C. Disregarding energy transfer as heat to the surrounding
air and bucket, what is the specific heat capacity of vinegar?
3. After eating a hearty stew you cooked over a campfire with your
0.250-kg aluminum-alloy pot, you place the pot in a plastic bucket
containing 1.00 kg of water. The water’s temperature increases 1.00°C
and the temperature of the pot decreases 17.5°C. Disregarding energy
transfer as heat to the surrounding air and bucket, what is the specific
heat capacity of the pot?
4. Suppose you bake cornbread in a 3.0-kg cast iron skillet. After remov-
ing the cornbread from the oven, you place the hot skillet in a sink
filled with 5.0 kg of dishwater. The water’s temperature increases
2.25°C, and the temperature of the skillet decreases 29.6°C. Disregard-
ing energy transfer as heat to the surrounding air and sink, what is the
specific heat capacity of the skillet?
5. The water in a swimming pool transfers 1.09 × 1010 J of energy as heat
to the cool night air. If the temperature of the water, which has a spe-
cific heat of 4186 J/kg•°C, decreases by 5.0°C, what is the mass of the
water in the pool?
6. Bismuth’s specific heat is 121 J/kg•°C, the lowest of any non-
radioactive metal. What is the mass of a bismuth sample if 25 J raises
its temperature 5.0°C?
7. The temperature of air in a foundry increases when molten metals cool
and solidify. Suppose 45 × 106 J of energy is added to the surrounding
air by the solidifying metal. The air’s temperature increases by 55 °C,
and the air has a specific heat capacity of 1.0 × 103 J/kg•°C. What is the
mass of the heated air?
8. A 0.190 kg piece of copper is heated and fashioned into a bracelet. The
amount of energy transferred as heat to the copper is 6.62 × 104 J. If
the specific heat of copper is 387 J/kg•°C, what is the change in the
temperature of the copper?
9. A 0.225 kg sample of tin, which has a specific heat of 2.2 × 103 J/kg•°C,
is cooled in water. The amount of energy transferred to the water is 3.9
× 104 J. What is the change in the tin’s temperature?
10. Tantalum is an element used, among other things, in making aircraft
parts. Suppose the properties of a tantalum part are being tested at
high temperatures. Tantalum has a specific heat of about 140 J/kg•°C.
The part, which has a mass of 0.23 kg, is cooled by being placed in
water. If 3.0 × 104 J of energy is transferred to the water, what is the
change in the part’s temperature?
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Problem 10C Ch. 10–7
Holt Physics Problem BankCh. 10–8
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 10D
P R O B L E MCarbon tetrachloride, a toxic compound used in dry cleaning, evaporatesat 76.7°C. It is therefore important to use it in well-ventilated areas. If ittakes 4.85 × 105 J to evaporate 2.50 kg of liquid carbon tetrachloride,what is the compound’s latent heat of vaporization?
S O L U T I O N
Given: Q = 4.85 × 105 J
m = 2.50 kg
Unknown: Lv = ?
Use the equation for latent heat, rearranging it to solve for Lv.
Q = mLv
Lv = m
Q =
4.8
2
5
.5
×0
1
k
0
g
5 J = 1.94 × 105 J/kg
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ADDITIONAL PRACTICE
1. Prior to the Bronze age, which began around 2200 B. C., the metal used
occasionally for tools was copper. Copper is not as hard or sturdy as
bronze, an alloy that combines copper with tin; but it is easy to locate
in mineral deposits by its color and the color of copper compounds. In
order to melt copper, it must be raised to a temperature of 1083°C. At
this temperture, 1.10 106 J of energy must be absorbed by 5.33 kg of
copper in order for the copper to completely melt. What is the latent
heat of fusion of copper?
2. Bronze, which is made by combining copper and tin in different pro-
portions, offers advantages that copper or tin alone lack. Not only is
bronze harder than either copper or tin, but bronze has a lower melting
point than copper, so that bronze is somewhat easier to manufacture. A
bronze alloy containing 15 percent tin and 85 percent copper has a
melting point approximately 950°C. At this temperature, 9.6 × 105 J of
energy must be transferred as heat to 5.33 kg sample of bronze in order
for the bronze to completely melt. What is the latent heat of fusion of
this particular bronze alloy?
NAME ______________________________________ DATE _______________ CLASS ____________________
3. At atmospheric pressure, solid carbon dioxide, or dry ice, undergoes a
process of evaporation called sublimation. In this process, at the tem-
perature of phase change, the solid dry ice goes directly into the gas
phase without becoming liquid. For carbon dioxide, the temperature of
sublimation is –78.44°C. At this temperture, 3.72 × 105 J of energy
must be transferred as heat to a 0.650 kg block of dry ice in order for
the carbon dioxide to completely sublime. What is the latent heat of
sublimation of carbon dioxide at atmospheric pressure?
4. A blacksmith has been hammering a piece of red-hot iron into the
shape of a horseshoe. To cool and temper the iron, he places the still
hot horseshoe into a barrel of water. This causes some of the water to
evaporate. Suppose the liquid water is initially at 100°C. The horseshoe
gives up 8.5 × 104 J of energy as heat to the water. Given that the latent
heat of vaporization for water a 100°C is 2.26 × 106 J/kg, what is the
mass of the water that evaporates?
5. A quantity of ethyl alcohol at a temperature of 78°C (or ethanol) ab-
sorbs 2.11 × 106 J from the surrounding air, causing it to completely
evaporate. If the latent heat of vaporization of ethyl alcohol at 78°C is
8.45 × 105 J/kg, what is the mass of the ethanol that is vaporized?
6. At a foundry, 250 kg of molten iron is poured into a mold. The iron so-
lidifies at a temperature near 1500°C and then further cools as a solid.
During this process, 1.380 × 108 J of energy is given up as heat by the
iron. What is the overall change in the temperature of the iron? The
specific heat capacity of solid iron at high temperatures is 605 J/kg•°C,
and the latent heat of fusion for iron is 2.47 × 105 J/kg.
7. A sample of lead with a mass of 1.45 kg is heated until it reaches its
melting point at 330°C. The lead is then heated further until it has en-
tirely melted. During this process the lead absorbs a total of 4.46 × 104 J
as heat. What is the temperature at which the lead is initially heated?
The specific heat capacity of solid lead over the range of temperatures
in question is 120 J/kg•°C, and the latent heat of fusion for lead is 2.45
× 104 J/kg.
8. Water vapor with a mass of 0.75 g and a temperature of 100.0°C con-
denses to form a raindrop with the same mass. By the time the rain-
drop reaches the ground, it has given up 2.0 × 103 J as heat. What is the
raindrop’s final temperature? Assume that the entire process takes place
at atmospheric pressure, that the specific heat capacity for liquid water
is 4200 J/kg•°C, and that the latent heat of vaporization for water is
2.26 × 106 J/kg.
9. Gallium melts at 29.78°C, a temperature that can be achieved by the at-
mosphere at Earth’s surface. In fact, gallium can melt in your hand.
Suppose a person holds 0.35 kg of gallium at its melting poiint. How-
much energy must be transferred as heat to the gallium in order for it
to completely melt? Gallium’s latent heat of fusion is 8.02 × 104 J/kg.
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Problem 10D Ch. 10–9
NAME ______________________________________ DATE _______________ CLASS ____________________
10. A 55.0-g sample of mercury has an initial temperature of 20.0°C. It is
heated to 357°C, at which it changes phase from a liquid to a gas. How
much energy must be transferred as heat to the mercury for it to com-
pletely evaporate? The average specific heat capacity for mercury over
this temperature range is 130 J/kg•°C, and the latent heat of vaporiza-
tion for mercury is 2.95 × 105 J/kg.
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Holt Physics Problem BankCh. 10–10
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Problem 11A Ch. 11–1
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 11AWORK DONE ON OR BY A GAS
P R O B L E MTo measure blood pressure, an inflatable rubber cuff is wrapped aroundthe upper arm and inflated by squeezing a rubber bulb connected to it by
a tube. Suppose the cuff is inflated from 0 m3 to 8.45 104 m3. If the airin the cuff does 16.9 J of work on an arm, what is the pressure of the air inthe cuff in excess of atmospheric pressure?
S O L U T I O NGiven: ∆V = 8.45 × 10–4 m3 W = 16.9 J
Unknown: P = ?
Choose the equation(s) or situation:
Use the equation that defines work in terms of changing volume, and rearrange
the equation to solve for pressure.
W = P∆V
Rearrange the equation(s) and isolate the unknown(s):
P = ∆W
V =
8.45
1
×6
1
.9
0−J4 m3 = 2.00 × 104 Pa
1. Three people inflate an empty hot-air balloon to a volume of 2190 m3.
If the air inside the balloon does 3.29 × 106 J of work on the balloon,
how much in excess of atmospheric pressure must the pressure of the
air be inside the balloon?
2. The air in a research balloon does 1.06 × 106 J of work to raise the bal-
loon to an altitude of 35 km. The balloon inflates from 2.80 × 103 m3
to 8.50 × 105 m3. How much in excess of atmospheric pressure must
the pressure of the air be inside the balloon?
3. Steam moves into the cylinder of a steam engine at a constant pressure
and does 1.3 J of work on a piston. The volume increases by 5.4 × 10–4 m3.
How much in excess of atmospheric pressure is the pressure of the
steam?
4. Suppose 472.5 J of work was required to fill an empty fire extinguisher
with compressed nitrogen gas. If the fire extinguisher was filled at a
constant pressure that was 25.0 kPa in excess of atmospheric pressure,
what was the change in the volume of the nitrogen gas?
5. Suppose you must change a flat inner tube in a bicycle tire. It requires
393 J of work to inflate the empty inner tube at a pressure of 655 kPa in
excess of atmospheric pressure. What is the change in the inner tube’s
volume?
ADDITIONAL PRACTICE
Holt Physics Problem BankCh. 11–2
NAME ______________________________________ DATE _______________ CLASS ____________________
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6. A child blows on a wand dipped in a soapy solution to form bubbles.
Suppose 0.20 J of work is required to inflate a soap bubble. If the bub-
ble was filled at a constant pressure that was 39 Pa in excess of atmos-
pheric pressure, what is the change in the bubble’s volume?
7. After the combustion of fuel in a car’s engine, a piston in one of the
cylinders moves outward. The work done by the gas on the piston is 3.2
× 102 J. If the gas maintains a constant pressure of 4.0 × 105 Pa, what is
the change in the volume of the gas?
8. An empty scuba tank is filled with air at a constant pressure of 2.07 ×107 Pa in excess of atmospheric pressure. The gas expands to fill the
tank, which has a volume of 0.227 m3. How much work is done on the
gas?
9. You may have noticed your ears pop if you yawn at a high altitude.
Suppose the volume of air in your ear increases from 3.375 × 10–6 m3
to 5.694 × 10–6 m3 as the plane you’re in flies to a high altitude. Before
you yawn, the pressure in your middle ear is 101.33 kPa—the same as
it was on the ground. What is the work done by the air on the
eardrum? (Assume that you still haven’t yawned.)
10. When an athlete inhales during strenuous exercise, the volume of air in
the lungs increases from 2.0 × 10–3 m3 to 5.0 × 10–3 m3. During this in-
halation, the pressure of air in the lungs reaches 0.90 kPa in excess of
atmospheric pressure. How much work is done by the air on the lungs?
Problem 11B Ch. 11–3
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 11BTHE FIRST LAW OF THERMODYNAMICS
P R O B L E MThe internal energy of a gas decreases by 256 J. If the process is adiabatic,how much work is done on or by the gas?
S O L U T I O NGiven: ∆U = –256 J Q = 0 J
Unknown: W = ?
Choose the equation(s) or situation: Use the equation for the first law of ther-
modynamics.
∆U = Q − W
In an adiabatic process, no energy is transferred as heat, so Q = 0 J
W = Q − ∆U = 0 J − (–256 J) = 256 J
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ADDITIONAL PRACTICE
1. A 2.0 kg metal pipe in the center of a compost pile supports a load of
15 kg. During the day the pipe’s temperature increases from 28°C to
52°C, causing the rod to thermally expand and raise the load 2.3 mm.
What is the work done in this process?
2. A system’s initial internal energy is 39 J. Then 114 J of energy is trans-
ferred to the system as heat. If the final internal energy is 163 J, how
much work is done on or by the system?
3. A system’s initial internal energy is 8093 J. Then 6932 J of energy is
transferred to the system as heat. If the final internal energy is 2.092 ×104 J, how much work is done on or by the system?
4. A steam engine’s boiler converts water to steam by transferring 4.50 ×108 J of energy as heat to the water. If steam escaping through a safety
valve does 3.21 × 108 J of work expanding against the outside atmos-
phere, what is the net change in the internal energy of the water-steam
system?
5. A pressure cooker converts water to steam by transferring 632 kJ of en-
ergy as heat to the water. If steam escaping through a safety valve does
102 kJ of work expanding against the outside atmosphere, what is the
net change in the internal energy of the water-steam system?
6. A gas expands when 867 J of energy is added to it by heat. The expand-
ing gas does 623 J of work on its surroundings. What is the overall
change in the internal energy of the gas?
Holt Physics Problem BankCh. 11–4
NAME ______________________________________ DATE _______________ CLASS ____________________
7. The expanding steam from a geyser does 192 kJ of work, and the inter-
nal energy of the system increases by 786 kJ. How much energy is
transferred to the system as heat?
8. At a nuclear power plant, heat from radioactive rods in the reactor
causes water to vaporize into steam. The expanding steam does 602 kJ
of work, and the internal energy of the system increases by 1.09 × 105 J.
How much energy is transferred to the system as heat?
9. The internal energy of the air in a closed car increases by 873 J. How
much energy is transferred as heat into the car?
10. The internal energy of the gas in a closed greenhouse increases by
986 J. How much work is done on or by the gas? How much energy is
transferred as heat into the greenhouse?
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Problem 11C Ch. 11–5
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 11CHEAT-ENGINE EFFICIENCY
P R O B L E MIf a gasoline engine has an efficiency of 33 percent and receives 660 J of en-ergy as heat from combustion during each cycle, how much work is doneby the engine?
S O L U T I O NGiven: eff = 0.33 Qh = 660 J
Unknown: Wnet = ?
Choose the equation(s) or situation: Rearrange the equation for efficiency.
Wnet = eff Qh = (0.33) (660 J) = 220 J
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ADDITIONAL PRACTICE
1. A steam engine of a locomotive has an efficiency of 17 percent and re-
ceives 5.5 × 109 J of energy by heat from its boiler. How much work is
done by the engine?
2. A coal-burning power plant has an efficiency of 35 percent. If the
power plant uses 7.37 × 108 J of energy as heat, how much work is
done by the power plant?
3. A geothermal power plant has an efficiency of 15 percent. If the power
plant takes in 9.36 × 108 J of energy as heat, how much work is done by
the power plant?
4. If a gasoline engine has an efficiency of 29 percent and receives 693 J of
energy by heat during each cycle, how much work is done by the engine?
5. Suppose an engine has an efficiency of 11 percent and performs 1150 J
of work each cycle. How much energy is taken in as heat?
6. Suppose a steam engine has an efficiency of 19 percent and performs
998 J of work each cycle. How much energy is received by the steam
engine as heat?
7. A certain propane engine performs 544 J of work in each cycle with an
efficiency of 22.25 percent. How much energy is received from the en-
gine to the exhaust and cooling system as heat?
8. Find the efficiency of a gasoline engine that, during one cycle, receives
365 J of energy from combustion and loses 223 J as heat to the exhaust.
NAME ______________________________________ DATE _______________ CLASS ____________________
9. Find the efficiency of an engine that, during one cycle, receives 571 J of
energy and loses 463 J of energy as heat.
10. A test model for an experimental engine that uses a new clean-burning
fuel does 128 J of work in one cycle and receives 581 J of energy as heat
from combustion. What is the engine’s efficiency?
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Holt Physics Problem BankCh. 11–6
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Problem 12A Ch. 12–1
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 12AHOOKE’S LAW
P R O B L E MSome bathroom scales work by stepping on a spring. Suppose a personsteps on a scale, compressing the spring 1.5 cm. If the spring constant is650 N/m, what is the spring force acting on the scale when the personsteps off?
S O L U T I O NGiven: k = 650 N/m x = 1.5 cm × 1.5 10–2 m
Unknown: F = ?
Choose the equation(s) or situation:
Use the equation for Hooke’s law to determine the spring’s restoring force.
Felastic = −kx = −(650 N/m) (1.5 10–2 m) = −9.8 N
1. A shopper places some fruit in a spring scale at a supermarket. If the
spring has a spring constant of 420 N/m and is compressed from its
equilibrium position by 4.3 cm, what is the spring force on the scale at
the moment it is released?
2. A fuzzy ball attached to an elastic cord is suspended from a ceiling to
be a toy for a cat. As the cat plays, the toy is pulled 15 cm and released.
If the toy has a spring constant of 65 N/m, what is the spring force act-
ing on the toy at the moment it is released?
3. You see a pair of joke glasses at a toy store. Each lens is connected to a
loosely coiled spring which, in turn, is connected to a plastic “eyeball.”
One spring is pulled 12 cm from its equilibrium position and released.
If the spring constant is 49 N/m, what is the magnitude of the spring
force acting on the toy at the moment it is released?
4. A lock tight curly hair is pulled a distance of 5.0 cm from its equilib-
rium position and released. If the lock of hair has a spring constant of
26 N/m, what is the magnitude of the spring force acting on the lock of
hair at the moment it is released?
5. When a person weighing 669 N sits in a hanging chair, a giant spring
suspending the load expands 6.5 cm. What is the spring constant?
6. A 550 N jumper attached to a bungee cord dives off a precipice. The
bungee cord stretches 15 m beyond its equilibrium point before it
bounces back. What is the spring constant?
ADDITIONAL PRACTICE
Holt Physics Problems WorkbookCh. 12–2
NAME ______________________________________ DATE _______________ CLASS ____________________
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7. As a 620 N mountain biker rides across rough terrain, the spring in the
seat compresses a distance of 7.2 cm. What is the spring constant?
8. A child exerts a force of 12 N to shoot a rubber band across the room.
If the rubber band has a spring constant of 180 N/m, what is the rub-
ber band’s displacement?
9. An archer applies a force of 52 N on a bowstring to shoot an arrow. If
the bow string has a spring constant of 490 N/m, what is the bow
string’s displacement?
10. A mass of 3.0 kg is attached to a spring scale with a spring constant of
36 N/m. What is the spring’s displacement?
Problem 12B Ch. 12–3
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 12BSIMPLE HARMONIC MOTION OF A SIMPLE PENDULUM
P R O B L E MA simple pendulum with a length of 1.00 m would have a period of 13.3 son Saturn’s icy moon, Dione. Find the acceleration of gravity on Dione.
S O L U T I O N
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ADDITIONAL PRACTICE
1. A simple pendulum with a length of 3.0 × 10–1 m would have a period
of 1.16 s on Venus. Calculate the acceleration of gravity on Venus.
2. On Mars, a simple pendulum with a length of 65.0 cm would have a
period of 2.62 s. Calculate the acceleration of gravity on Mars.
3. On Mercury, a simple pendulum with a length of 1.14 m would have a
period of 3.55 s. What is the acceleration of gravity on Mercury?
4. A simple pendulum with a length of 50.0 cm would have a period of
2.99 s on Pluto. Calculate the acceleration of gravity on Pluto.
5. Find the length of a pendulum that oscillates with a frequency of 1.0 Hz.
6. Find the length of a pendulum that oscillates with a frequency of 0.50 Hz.
7. Find the length of a pendulum that oscillates with a frequency of 2.5 Hz.
8. Calculate the period and frequency of a 6.200 m long pendulum in
Oslo, Norway, where g = 9.819 m/s2.
9. Calculate the period and frequency of a 2.500 m long pendulum in
Quito, Ecuador, where g = 9.780 m/s2.
10. Calculate the period and frequency of a 3.120 m long pendulum in
Cairo, Egypt, where g = 9.793 m/s2.
Given: L = 1.00 m T = 13.3 s
Unknown: g = ?
Choose the equation(s) or situation: Use the equation for the period of a
simple pendulum and rearrange it to solve for g.
T = 2pT2 =
4pg
2L
g = 4
T
p2
2L
= 4p
(
2
13
(1
.3
.0
s
0
)2m)
= 0.223 m/s2
Lg
Holt Physics
Problem 12CSIMPLE HARMONIC MOTION OF A MASS SPRING SYSTEM
P R O B L E MThe antennae of male mosquitoes have many hairs that receive sound signals from female mosquitoes. Female mosquitoes emit a frequency ofabout 230 Hz. Suppose a mass is attached to a spring with a spring con-stant of 1.14 104 N/m. How large is the mass if its oscillation frequency isthe same as a mosquito’s?
S O L U T I O N
Given: f = 230 Hz k = 1.14 × 104 N/m
Unknown: m = ?
Choose the equation(s) or situation:
Use the equation for the period of a mass-spring system to solve for m:
T = 2p = 1
f
f
12 =
4pk
2m
m = 4p
k2f 2 =
(
1
4
.1
p42)
×(2
1
3
0
0
4 N
s–/1m
)2 = 5.46 × 10−3 kg = 5.46 g
mk
Holt Physics Problems WorkbookCh. 12–4
NAME ______________________________________ DATE _______________ CLASS ____________________
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ADDITIONAL PRACTICE
1. Honeybee scouts inform other honeybees where to find food by flap-
ping their wings and “waggle-dancing.” During part of the dance, a
scout bee’s wings flap with a maximum frequency of 3.00 102 Hz. Sup-
pose a mass is attached to a spring with a spring constant of 8.65 ×104 N/m. How large is the mass if its oscillation frequency is the same
as the wings of a waggle-dancing bee?
2. On Halloween, you see an “alien” that has one antenna made of a glit-
tery foam ball connected to a spring. The springs oscillate with a pe-
riod of 0.079 s, and have a spring constant of 63 N/m. Find the mass of
the ball.
3. A farmer rides over a bumpy field on his tractor. The tractor seat is sup-
ported by a spring with a spring constant of 2.03 103 N/m. As the
farmer drives over a bump, the seat oscillates at a frequency of 0.79 Hz.
For the first few seconds, the vibration approximates simple harmonic
motion. Find the farmer’s mass if the tractor seat acts like a spring scale.
4. A 32 N sack of potatoes vibrates with a period of 0.42 s placed on a
spring scale. What is the spring constant?
Problem 12C Ch. 12–5
NAME ______________________________________ DATE _______________ CLASS ____________________
5. A 66 N pumpkin vibrates with a period of 2.9 s when attached to the
end of a spring scale. What is the spring constant?
6. As the wind moves the bough of a tree, it oscillates up and down. Dur-
ing the first few seconds, it approximates simple harmonic motion. If
the bough has a weight of 87 N and oscillates with a period of 0.64 s,
what is the spring constant of the bough?
7. A certain trampoline acts like a single spring with a spring constant of
364 N/m. If a 24 kg child jumps on the trampoline, what would be the
period of oscillation?
8. Two children jump on their parent’s bed (when the parents are not
looking). The combined mass of both kids is 55 kg. The mattress is
supported by 36 springs, each with a spring constant of 458 N/m.
If the children jump at the same time, what would be the period
of oscillation?
9. An 8.2 kg infant is placed in a jumper that is made of a seat that is sus-
pended from a door frame by a spring. If the spring has a spring con-
stant of 221 N/m. Calculate the period of oscillation.
10. Your friend’s key chain is coiled like a spring. Three keys, each with a
mass of 24 g, are on the chain. When your friend removes the keys
from a pocket, the keys bob up and down. If the key chain has a spring
constant of 99 N/m, what is the frequency of oscillation?
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Holt Physics Problems WorkbookCh. 12–6
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 12DWAVE SPEED
P R O B L E MEarthquakes generate shock waves that travel through Earth’s interior toother parts of the world. The fastest of these waves are longitudinalwaves, like sound waves, and are called primary waves, or just p-waves. Ap-wave has very low frequencies, typically around 0.050 Hz. If the wave-length of a p-wave is 160 km, what is its speed?
S O L U T I O N
Given: l = 160 km = 1.6 × 105 m f = 0.050 Hz
Unknown: v = ?
Choose the equation(s) or situation: Use the equation relating speed, wave-
length, and frequency for a wave.
v = fl = (1.6 × 105 m) (0.050 s–1) = 8.0 × 103 m/s
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1. Earthquakes also produce transverse waves that move more slowly than
the p-waves. These waves are called secondary waves, or s-waves. If the
wavelength of an s-wave is 2.3 × 104 m, and its frequency is 0.065 Hz,
what is its speed?
2. A dolphin can hear sounds with frequencies up to 280 kHz. What is the
speed of sound in water if a wave with this frequency has a wavelength
of 0.51 cm?
3. Waves in a lake are 6.0 m apart and pass a person on a raft every 2.0 s.
What is the speed of the waves?
4. Sonar is a device that uses reflected sound waves to measure underwa-
ter depths. If a sonar signal has a frequency of 288 Hz, and the wave-
length is 5.00 m, what is the speed of the sonar signal in water?
5. A buoy on the ocean bobs up and down. The waves have a wavelength
of 2.5 m, and a frequency of 1.6 Hz. What is the speed of the waves?
6. A dog whistle is designed to produce a sound with a frequency
beyond that which can be heard by humans (between 20 000 Hz and
27 000 Hz). If a particular whistle produces a sound with a frequency
of 2.5 × 104 Hz, what is the sound’s wavelength? Assume the speed of
sound in air to be 331 m/s.
7. The lowest pitch that the average human can hear has a frequency of
20.0 Hz. If sound with this frequency travels through air with a speed
of 331 m/s, what is its wavelength?
ADDITIONAL PRACTICE
Problem 12D Ch. 12–7
NAME ______________________________________ DATE _______________ CLASS ____________________
8. A ship anchored at sea is rocked by waves whose crests are 14 m apart.
The waves travel at 7.0 m/s. How often do the wave crests reach the
ship?
9. One of the largest organ pipes is in the Auditorium Organ in the At-
lantic City Convention Hall, New Jersey. The pipe is 38.6 ft long and
produces a sound with a wavelength of about 10.6 m. If the speed of
sound in air is 331 m/s, what is the frequency of this sound?
10. A drum is struck, producing a wave with a wavelength of 110 cm and a
speed of 2.42 × 104 m/s. What is the frequency of the wave?
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Problem 13A Ch. 13–1
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NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 13AINTENSITY OF SOUND WAVES
P R O B L E MYour friend whispers a secret to you with a power output of 2.05 10–10 W.If the whisper has a sound intensity 4.1 10–10 W/m2, how far are youfrom your friend?
S O L U T I O NGiven: Intensity = 4.1 × 10–10 W/m2 P = 2.05 × 10–10 W
Unknown: r = ?
Choose the equation(s) or situation:
Use the equation for the intensity of a spherical wave.
Intensity = 4p
P
r2
r = 4p(Int
P
ensity) = = 0.20 m
2.05 × 10–10 W(4p)(4.1 × 10–10 W/m2)
1. Your friend tells you about what happened last weekend with a power
output of 5.88 × 10–5 W and a sound intensity of 3.9 × 10–6 W/m2. How
far are you from your friend?
2. The power output of heavy street traffic is 1.57 × 10–3 W. At what dis-
tance is the sound intensity of the traffic 5.20 × 10–3 W/m2?
3. A subway train in New York City produces sound with a power output of
4.80 W and an intensity of 7.2 × 10–2 W/m2. How far are you from the
subway train?
4. A loud clap of thunder has a power output of 151 kW and a sound inten-
sity of 0.025 W/m2. How far are you from the thunder’s source?
5. What is the intensity of the sound waves produced by the jet engine of a
plane taking off at a distance of 32 m when the power radiated as sound
from the engine is 402 W? Assume that the sound waves are spherical.
6. Calculate the intensity of the sound waves from a car stereo at a distance
of 0.50 m when the sound has power output of 3.5 W.
7. At a maximum level of loudness, the power output of portable radio
headphones radiated as sound is 2.76 × 10–2 W. What is the intensity of
these sound waves to a jogger whose ear is 5.0 cm from the headphone’s
speaker?
ADDITIONAL PRACTICE
NAME ______________________________________ DATE _______________ CLASS ____________________
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Holt Physics Problems WorkbookCh. 13–2
8. If the intensity of a mosquito’s buzzing is 9.3 × 10–8 W/m2 at a distance
of 0.21 m, how much sound power does that mosquito generate?
9. How much power is radiated as sound from a vacuum cleaner whose in-
tensity is 4.5 × 10–4 W/m2 at a distance of 1.5 m?
10. To perforate an eardrum, an intensity of 1.0 × 104 W/m2 at a distance of
1.0 m is required. Calculate how much sound power must be generated.
Problem 13B Ch. 13–3
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 13BHARMONICS
P R O B L E MA piano wire vibrates with a fundamental frequency of 440 Hz when thespeed of sound on the wire is 550 m/s. What is the length of this wire?
S O L U T I O NGiven: v = 550 m/s n = 1 f1 = 440 Hz
Unknown: L = ?
Choose the equation(s) or situation: The fundamental frequency can be found
by using the equation for standing waves on a vibrating string:
fn = 2
n
L
v, n = 1, 2, 3, …
Rearrange the equation(s) to isolate the unknown(s): Rearrange the equation
above to solve for the length of the wire.
L = 2
n
f
v
n =
(
(
1
2
)
)
(
(
5
4
5
4
0
0
m
H
/
z
s
)
) = 0.625 m
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ADDITIONAL PRACTICE
1. A saxophone plays a tune in the key of B-flat. The saxophone has a sec-
ond harmonic frequency of 466.2 Hz when the speed of sound in air is
331 m/s. What is the length of the pipe that makes up the saxophone?
Recall that a saxophone should be treated as a pipe closed at one end.
2. A clarinetist plays a clarinet on a cold day. At one point she produces the
sound of middle F sharp, which has a frequency of 370 Hz, by playing
the third harmonic of low B. If the speed of sound in the air is 331 m/s,
what is the length of the clarinet? Recall that a clarinet resembles a pipe
closed at one end.
3. A penny whistle plays a tune in the key of G with a fundamental fre-
quency of 392.0 Hz. The speed of sound in air is 331 m/s. What is the
length of the penny whistle? Treat the penny whistle as a pipe closed
at one end.
4. An organ pipe that is open at both ends has a fundamental frequency of
370.0 Hz when the speed of sound in air is 331 m/s. What is the length of
this pipe?
5. What is the fundamental frequency of a viola string that is 35.0 cm long
when the speed of waves on this string is 346 m/s?
6. What is the fundamental frequency of a mandolin string that is 42.0 cm
long when the speed of waves on this string is 329 m/s?
NAME ______________________________________ DATE _______________ CLASS ____________________
7. What is the fundamental frequency of a cello string that is 0.85 m long
when the speed of waves on this string is 499 m/s?
8. A pipe that is open at both ends has a fundamental frequency of 277.2 Hz.
If the pipe is 0.75 m long, what is the speed of the waves in the pipe?
9. A pipe that is closed on one end has a seventh harmonic frequency of
466.2 Hz. If the pipe is 1.53 m long, what is the speed of the waves in
the pipe?
10. A pipe that is open at both ends has a fundamental frequency of 125 Hz.
If the pipe is 1.32 m long, what is the speed of the waves in the pipe?
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Holt Physics Problems WorkbookCh. 13–4
Problem 14A Ch. 14–1
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NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 14AELECTROMAGNETIC WAVES
P R O B L E MHow fast does light with a frequency of 5.4999 1014 Hz and a wave-length of 545.00 nm travel?
S O L U T I O NGiven: λ = 5.450 0 × 10–7 m f = 5.4999 1014 Hz
Unknown: c = ?
Choose the equation(s) or situation:
Use the wave speed equation for electromagnetic waves.
c = f λ = (5.4999 1014 s–1) (5.4500 × 10–7 m) = 2.9974 × 108 m/s
1. How fast does a radio wave travel that has a frequency of 7.6270 × 108
Hz and a wavelength of 39.296 cm? Does this radio wave travel
through Earth’s atmosphere or in space? Light travels in a vacuum at
2.997 924 58 × 108 m/s and in air at 2.997 09 × 108 m/s.
2. How fast does microwave radiation that has a frequency of 1.173 06 ×1011 Hz and a wavelength of 2.555 6 mm travel? Does this microwave
travel through Earth’s atmosphere or in space? Light travels in a vac-
uum at 2.997 924 58 × 108 m/s and in air at 2.997 09 × 108 m/s.
3. Scientists at Lucent Bell Labs use high-resolution microscopes to make
images of tiny organisms that provide a lot of information. By using
3.2 nm x-rays on human tissue, images can be made showing micro-
tubules in the nuclei of cells. What is the frequency of these x-rays?
4. In order to see objects, the wavelength of the light must be smaller than
the object. The Bohr radius of a hydrogen atom is 5.291 770 × 10–11 m.
a. What is the lowest frequency that can be used to locate a hydro-
gen atom?
b. The visible part of the spectrum ranges from 400 nm to 700 nm.
Why aren’t individual atoms visible?
5. Your skin tans when melanin within the skin oxidizes. Your skin sun-
burns when it receives more ultraviolet radiation than the protection
provided by the melanin. Generally, ultraviolet (UV) radiation has
been divided into two classes: UVA and UVB. You are more likely to be
sunburned if you are exposed to radiation in the UVB range (280 nm–
320 nm) than in the UVA range (320 nm–400 nm). To what range of
frequencies do these wavelength ranges correspond?
ADDITIONAL PRACTICE
NAME ______________________________________ DATE _______________ CLASS ____________________
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Holt Physics Problem BankCh. 14–2
6. Suppose you must decide whether a pre-Columbian mask is genuine
before buying it for a museum. It looks genuine, but to ensure its au-
thenticity, you shine X rays of wavelength 1.67 × 10–10 m on it to see if
a certain element is present. What is the frequency of this radiation?
7. Suppose you use ultraviolet light of frequency 9.5 × 1014 Hz to deter-
mine whether a mineral is fluorescent. To what wavelength does this
correspond?
8. Meteorologists use Doppler radar to watch the movement of storms.
If a weather station uses electromagnetic waves with a frequency of
2.85 × 109 Hz, what is the length of the wave?
9. PCS cellular phones have antennas that use radio frequencies from
1800–2000 MHz. What range of wavelengths corresponds to these
frequencies?
10. Suppose the microwaves in your microwave oven have a frequency of
2.5 × 1010 Hz.
a. What is the wavelength of these microwaves?
b. The holes in the door of a microwave oven have a radius of
1.2 mm. Why don’t microwaves pass through these holes?
c. Visible light has a wavelength that ranges from 400 nm to
700 nm. Would visible light be able to pass through the holes?
Why or why not?
Problem 14B Ch. 14–3
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 14BCONCAVE MIRRORS
P R O B L E MSuppose you place a 5.0 cm tall spring in front of a concave mirror. Themirror has a focal length of 24 cm. The spring forms an image that ap-pears to be at the same position as the spring, but the image is inverted.Where did you place the spring? How tall is the spring’s image?
S O L U T I O NGiven: h = 5.0 cm f = 24 cm q = p
Unknown: h′ = ? p = ?
Choose the equation (s) or situation: Use the mirror equation to find the posi-
tion of the spring, and the equation for magnification to find the height of the
image.
1
f =
p
1 +
1
q =
p
1 +
p
1 =
p
2
M = h
h′= −
p
q
Rearrange the equation(s) to isolate the unknown(s):
p = 2f = (2) (24 cm) =
q = p = 48 cm
h′ = − q
p
h = −
(48 cm
48
)(
c
5
m
.0 cm) = −5.0 cm
48 cm
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ADDITIONAL PRACTICE
1. You can appear to shake hands with yourself using a concave mirror.
Suppose you have a mirror with a focal length of 32.0 cm.
a. Where would you place your right hand so that you appear to be
shaking hands with yourself?
b. When your hand is in the shaking position, it is 7.5 cm tall. What
is the height of the image?
2. As you eat your soup, you look in the concave part of the spoon and
see your face. The focal length of the spoon is 9.5 cm, and the image of
your eye appears to be 15.5 cm from the mirror.
a. How far in front of the mirror is your eye?
b. If your eye is 3.0 cm tall, how tall is the image?
NAME ______________________________________ DATE _______________ CLASS ____________________
3. Suppose you bend a sheet of aluminized Mylar™ to form a reflective
surface that resembles a concave mirror when the axis is vertical. The
bent reflective sheet has a focal length of 17 cm.
a. Where must you stand so that the image of your eye appears at
23 cm?
b. If your eye is 2.7 cm tall, how tall will the image be?
4. A car’s headlamp is made of a light bulb in front of a concave spherical
mirror. If the bulb is 5.0 cm in front of the mirror, what is the radius
and focal length of the mirror?
5. Suppose you are 19 cm in front of the bell of your friend’s trumpet and
you see your image at 14 cm. Treating the trumpet’s bell as a concave
mirror, what would be its focal length and radius of curvature?
6. You look in to a metallic mixing bowl which resembles a spherical con-
cave mirror. When you are 35 cm in front of the bowl, you see an image
at 42 cm. What is the focal length and radius of curvature of the bowl?
7. You place an electric heater 3.00 m in front of a concave spherical mir-
ror that has a focal length of 30.0 cm.
a. Where would your hand feel warm?
b. If the heater is 15 cm tall, how tall is the image?
8. A new line of makeup offers concave spherical mirrors in their pressed
powder compacts. The focal length for one of these mirrors is 17.5 cm.
a. If someone used this mirror at a distance of 15.0 cm, where
would the image appear?
b. What is the magnification of the image?
9. A concave spherical mirror on an actor’s dressing table has a focal
length of 60.0 cm. Suppose the actor sits 35.0 cm in front of the mirror.
a. Where does the image appear?
b. What is the magnification of the image?
10. You place a book in front of a concave spherical mirror. The mirror has
a focal length of 23.0 cm.
a. Where does the image form when the book is 3.00 cm in front of
the mirror? What is the magnification of the image? Will you be
able to read the writing on the image?
b. Where does the image form when the book is 33.0 cm in front of
the mirror? What is the magnification of the image? Will you be
able to read the writing on the image?
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Holt Physics Problem BankCh. 14–4
Problem 14C Ch. 14–5
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 14CCONVEX MIRRORS
P R O B L E MYou have just received a silver key ring as a gift. The ring is connected to aspherical silver ball that acts like a convex spherical mirror. When youhold the ball 21 cm from your eye, your image forms 7.0 cm behind themirror. What is the magnification of the image? What is the mirror’sfocal length and radius of curvature?
S O L U T I O NGiven: p = 521 cm q = –7.0 cm
Unknown: M = ? f = ? R = ?
Choose the equation (s) or situation: Use the mirror equation to find the focal
length and radius of curvature, and the equation for magnification to find the
height of the image.
1
f =
p
1 +
1
q =
21
1
cm +
−70.
1
0 cm
f = [0.048 cm−1 − 0.143 cm−1]−1 = −10.5 cm
R = 2f = 2 (−11 cm) =
M = −p
q = −
(−(2
7
1
.0
c
c
m
m
)
) = +0.33
−22 cm
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ADDITIONAL PRACTICE
1. Your car has a side-view mirror with a convex spherical mirror on the
passenger’s side. When you pass a car, you see the car’s image in the mir-
ror. The image appears to be 9.0 cm tall, but the car is really 1.5 m tall.
a. What is the magnification of the mirror?
b. If the car is 3 m from the mirror, what is the focal length of the
mirror?
c. What is the mirror’s radius of curvature?
2. Sitting beside a Christmas tree, you notice your face is reflected in a
hanging spherical ornament. The image appears 5.2 cm behind the or-
nament when you are 17 cm in front of the ornament.
a. What is the ornament’s focal length and radius of curvature?
b. If your eye is 3.2 cm tall, how tall is the image?
3. You see an image of your hand as you reach for a polished brass door-
knob. The doorknob has a focal length of 6.3 cm. How far from the
doorknob is your hand when the image appears at 5.1 cm behind the
doorknob? What is the magnification of the image?
Holt Physics Problem BankCh. 14–6
NAME ______________________________________ DATE _______________ CLASS ____________________
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4. As you turn the knob of a faucet to draw bath water, you see your re-
flection in the water spout. The focal length of the spout is −33 cm.
How far away from the spout are you if your image appears to be
16.1 cm behind the spout? What is the magnification of the image?
5. You see your reflection in your friend’s mirrored sunglasses. If each
lens has a focal length of −12 cm, and your image appears 9.0 cm be-
hind the sunglasses, how far from your friend are you standing? What
is the magnification of the image?
6. To supervise customers, many stores install spherical convex mirrors in
strategic locations. Suppose one store has a spherical convex mirror
with a magnification of 0.11. Suppose you are 1.75 m tall.
a. How tall is the image?
b. How far in front of the mirror are you when the image appears
42 cm behind the mirror?
7. A stainless-steel ladle, used to serve soup, is like a spherical convex mir-
ror. If the focal length of the ladle is 27cm and you are 43 cm in front
of the ladle, where does the image appear? What is the magnification of
the image?
8. Just after you dry a spoon, you look into the convex part of the spoon.
If the spoon has a focal length of −8.2 cm and you are 18 cm in front of
the spoon, where does the image appear? What is the magnification of
the image?
9. The base of an art deco lamp is made of a convex spherical mirror with
a focal length of −39 cm.
a. Where does the image appear when you are 16 cm from the base?
b. If your nose is 6 cm long, how long does the image appear?
10. The button on many electric hand dryers is a convex mirror. You see
the image of your hand as you reach to press the button. If the magnifi-
cation of the image is 0.24 and your hand is 12 cm away from the but-
ton, where does the image appear?
Problem 15A Ch. 15–1
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NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 15ASNELL’S LAW
P R O B L E MA ray of light traveling in air strikes the surface of a polished agate slab (n 1.544) on display in your friend’s home. If the ray in the agate makesan angle of 29.0° with the normal, what is the angle of incidence?
S O L U T I O NGiven: nr = 1.544 qr = 29.0° ni = 1.00
Unknown: qi = ?
Choose the equation(s) or situation:
Use the equation for Snell’s law.
ni (sin qi) = nr (sin qr)
qi = sin–1nr (s
n
in
i
qr) = sin−1 (1.544)
1
(
.
s
0
i
0
n 29.0°) = 48.5°
1. An old Greek superstition was that amethyst would protect those who
wore it from drunkenness—which is why they called it améthystos,
meaning “not drunken.” Until the discovery of the large Brazilian and
Uruguayan deposits at the end of the nineteenth century, deep-colored
amethyst was highly prized. Suppose a ray of light traveling in air
strikes the surface of an amethyst crystal (n = 1.553). If the ray in the
amethyst makes an angle of 35° with the normal, what is the angle of
incidence?
2. If you were to set a calcite crystal on top of this sentence, you would see
a double image of each word. This phenomenon is called “double re-
fraction.” Suppose a ray of light traveling in air strikes the surface of a
calcite crystal (n = 1.486) used to demonstrate this phenomenon. If the
ray in the calcite makes an angle of 41° with the normal, what is the
angle of incidence?
3. The Chinese have skillfully carved figurines made of a translucent
greenish material called serpentine. A ray of light traveling in air strikes
the flat surface of a serpentine figurine (n = 1.555). If the ray in the ser-
pentine makes an angle of 33° with the normal, what is the angle of
incidence?
4. When light in air enters an opal mounted on a ring, it travels at a speed
of 2.07 × 108 m/s. What is opal’s index of refraction?
5. When light enters a pearl in a necklace, it travels at a speed of 1.97 × 108
m/s. What is the pearl’s index of refraction?
ADDITIONAL PRACTICE
NAME ______________________________________ DATE _______________ CLASS ____________________
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Holt Physics Problem BankCh. 15–2
6. When light enters albite, also called “moonstone”, it has a luminous
albedo—like a full moon. When light in air enters albite, it travels at a
velocity of 1.95 × 108 m/s. What is albite’s index of refraction?
7. Nephrite jade was once used virtually everywhere by Neolithic man for
polished stone weapons. Nephrite jade was also important in ancient
oriental art. Suppose light passes from air at an angle of incidence of
59.2° into a thin ornate handle of a nephrite jade vase (n = 1.61) on
display at a museum. Determine the angle of refraction in the jade.
8. Malachite is characterized by wavy light- and dark-green bands and
has double refraction (n =1.91 and n = 1.66). Suppose a ray of light
traveling in air strikes a malachite carving of an Aztec calendar at an
angle of 35.2° with the normal. What are the two angles of refraction?
9. Amber is a fossil resin of trees that lived tens of millions of years ago.
Sometimes insects were trapped by the resin and fossilized inside. Sup-
pose a ray of light traveling in air strikes a 2 mm thick clear amber pen-
dant (n = 1.54) at an angle of 17° with the normal. Find the angles of
refraction at each surface.
10. The Museo degli Argenti, in Florence, displays a plate that was carved
out of rock crystal in the sixteenth century. Suppose a ray of light trav-
eling in air strikes this plate (n = 1.544) at an angle of 22° with the nor-
mal. Trace the light ray through the plate, and find the angles of refrac-
tion at each surface.
Problem 15B Ch. 15–3
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 15BLENSES
P R O B L E MAn object is placed 49 cm in front of a converging lens. The image forms23 cm behind the lens and is 8.0 cm tall. Determine the focal length of thelens and the height of the object.
S O L U T I O NGiven: h′ = 8.0 cm p = 49 cm q = 23 cm
Unknown: f = ? h = ?
Choose the equation(s) or situation: Use the thin-lens equation to find the
focal length, and the equation for magnification to find the height of the object.
1
f =
p
1 +
1
q =
49
1
cm +
23
1
cm = 0.0204 cm−1 + 0.0435 cm−1
f =
M = − p
q =
h
h
′
Rearrange the equation(s) to isolate the unknown(s):
h = – p
q
h′ = –
(49 cm
23
)(
c
8
m
.0 cm) = 17 cm
16 cm
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ADDITIONAL PRACTICE
1. An object is placed 13 cm in front of a converging lens. The image
forms 19 cm behind the lens and is 3.0 cm tall. Determine the focal
length of the lens and the height of the object.
2. An object that is 15 cm tall is placed 44 cm in front of a diverging lens.
A virtual image appears 14 cm in front of the lens. Determine the focal
length of the lens and the height of the image.
3. A magnifying glass has a diverging lens with a 13.0 cm focal length. At
what distance from a toothpick should you hold this lens to form an
image with a magnification of +5.00?
4. An object with a height of 18 cm is placed in front of a converging lens.
The image has a height of –9.0 cm.
a. What is the magnification of the lens?
b. If the focal length of the lens is 6.0 cm, how far in front of the
lens is the object?
c. Where does the image appear?
NAME ______________________________________ DATE _______________ CLASS ____________________
5. A lighthouse places a 1000-watt bulb 4 m in front of a converging lens.
The focal length of the lens is 4 m. What is the image distance and the
magnification?
6. A searchlight is constructed by placing a 500-watt bulb 0.5 m in front
of a converging lens. The focal length of the lens is 0.5 m. What is the
image distance and the magnification?
7. A microscope slide is placed in front of a converging lens with a focal
length of 3.6 cm. The lens forms a real image of the slide 15.2 cm be-
hind the lens. How far is the lens from the slide?
8. Where must an object be placed to form an image 12 cm in front of a
diverging lens with a focal length of 44 cm?
9. In the projection booth of a movie theatre, film is placed in front of a
converging lens with a focal length of 9.0 cm. The lens forms a magni-
fied real image on a screen 18 m behind the lens. How far is the lens
from the film?
10. An object is placed in front of a converging lens with a focal length of
5.5 m. A virtual image appears 5.5 cm in front of the lens. How far is
the object from the lens?
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Holt Physics Problem BankCh. 15–4
Problem 15C Ch. 15–5
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 15CCRITICAL ANGLE
P R O B L E MThe critical angle for light traveling from a red spinel gemstone into air is35.8. What is the index of refraction for red spinel?
S O L U T I O NGiven: qc = 35.8° nr = 1.00
Unknown: ni = ?
Choose the equation(s) or situation: Use the equation for critical angle.
sin qc = n
nr
i
Rearrange the equation(s) to isolate the unknown(s):
ni = sin
nr
qc =
sin
1.
3
0
5
0
.8° = 1.71
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ADDITIONAL PRACTICE
1. The critical angle for light traveling from a green tourmaline gemstone
into air is 37.8°. What is tourmaline’s index of refraction?
2. The critical angle for light traveling from an aquamarine gemstone into
air is 39.18°. What is the index of refraction for aquamarine?
3. The critical angle for light traveling from almandine garnet into air
ranges from 35.3° − 33.1°. Calculate the range of almandine garnet’s
index of refraction.
4. Light moves from olivine (n =1.670) into onyx. If the critical angle for
olivine is 62.85°, what is the index of refraction for onyx?
5. Light moves from spessartite garnet (n = 1.80) —also called spessar-
tine—into obsidian. If the critical angle for spessartine is 57.0°, what is
the index of refraction for obsidian?
6. Light moves from a clear andalusite (n =1.64) crystal into ivory. If the
critical angle for andalusite is 69.9°, what is the index of refraction for
ivory?
7. Find the critical angle for light traveling from ruby (n = 1.766) into air.
8. Find the critical angle for light traveling from sapphire (n = 1.774) into
air.
9. Find the critical angle for light traveling from blue topaz (n = 1.61)
into air.
10. Find the critical angle for light traveling from emerald (n = 1.576) into air.
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Problem 16A Ch. 16-1
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 16AINTERFERENCE
P R O B L E MYou can make a hologram by splitting a laser beam into two beams thatconstructively interfere with each other. Each beam acts as a slit in a double-slit experiment—one beam is sent directly to a photographicfilm, and the other beam reflects off of an object before reaching the film. The two beams constructively interfere with each other on the film,making fringes. Suppose a hologram is made by using red light from a helium-neon laser ( = 632.8 nm). The first bright fringe is imprinted onthe film at an angle of 0.426 from the central maximum. What is the distance between the beams?
S O L U T I O N
Given: l = 632.8 nm = 6.328 × 10–7 m m = 1 q = 0.426°
Unknown: d = ?
Choose the equation(s) or situation:Use the equation for constructive interference, given on page 601.
d = s
m
in
lq
= 2 (6
s
.
i
3
n
28
(0
×.4
1
2
0
6
–
°
7
)
m) = 1.70 × 10–4 m = 0.170 mm
ADDITIONAL PRACTICE
1. A grating with 14 450 lines/cm is illuminated by light with a wavelength
of 625.0 nm. What is the highest order number that can be observed
with this grating? (Hint: Remember that sin q can never be greater than
1 for a diffraction grating.)
2. A grating with 12 660 lines/cm is illuminated by light with a wavelength
of 589.6 nm. What is the highest order number that can be observed
with this grating? (Hint: Remember that sin q can never be greater than
1 for a diffraction grating.)
3. A diffraction grating is calibrated by using the 546.1 nm line of hydro-
gen gas. The first-order maximum is found at an angle of 75.76°. Calcu-
late the number of lines per centimeter on this grating.
4. A diffraction grating is calibrated by using the 447.1 nm line of mercury
vapor. The first-order maximum is found at an angle of 40.25°. Calcu-
late the number of lines per centimeter on this grating.
5. A diffraction grating with 1950 lines/cm is used to examine the lithium
spectrum. Find the angles at which one would observe the first-order and
second-order maxima for the blue-green line of lithum (l = 497.3 nm).
Holt Physics Problem BankCh. 16-2
NAME ______________________________________ DATE _______________ CLASS ____________________
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6. A diffraction grating with 7500 lines/cm is used to examine the mercury
spectrum. Calculate the second-order angular separation of the two
closely spaced yellow lines of mercury (579.0 nm and 577.0 nm).
7. Infrared light passes through a diffraction grating of 3600 lines/cm. If the
angle of the third-order maximum is 76.54°, what is the wavelength of
the light?
8. A diffraction grating of 9550 lines/cm is used to examine the carbon
spectrum. If the angle of the second-order maximum is 54.58°, what is
the wavelength of the light?
9. A diffraction grating of 12 500 lines/cm is used to examine the iron spec-
trum. If the angle of the first-order maximum is 38.77°, what is the
wavelength of the light?
10. Ultraviolet light passes through a diffraction grating of 2400 lines/cm. If
the angle of the second-order maximum is 26.54°, what is the wave-
length of the light?
Problem 16B Ch. 16-3
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 16BDIFFRACTION GRATINGS
P R O B L E M
A grating with 13 250 lines/cm is illuminated by light with a wavelengthof 725 nm. What is the highest order number that can be observed withthis grating? (Hint: Remember that sin q can never be greater than 1 for adiffraction grating.)
S O L U T I O N
Given: 13 250 lines/cm l = 725 nm = 7.25 × 10–7 m q < 90°
Unknown: m = ?
Choose the equation(s) or situation: Use the equation for a diffraction grating.
m = 1: q = sin−1[ml /d] = sin−1[(1)(7.25 × 10−7 m) ÷ (1 325 000 lines/m)−1] = 73.9°
m = 2: q = sin−1[ml /d] = sin−1[(2)(7.25 × 10−7 m) ÷ (132 500 lines/m)−1] = ∞
Therefore, 1 is the highest-order number that can be observed.
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1. A double-slit interference experiment is performed using red light from a
helium discharge tube (l = 587.5 nm). The second-order bright fringe in
the interference pattern is 0.130° from the central maximum. How far
apart are the two slits separated?
2. A double-slit interference experiment is performed using red light from a
laser pointer pen (l = 656.3 nm). The fourth-order bright fringe in the
interference pattern is 0.626° from the central maximum. How far apart
are the two slits separated?
3. A ruby laser was developed by T. H. Maiman in 1960. Suppose a double-
slit interference experiment is performed using red light from a ruby
laser (l = 693 nm). The third-order bright fringe in the interference pat-
tern is 0.578° from the central maximum. How far apart are the two slits
separated?
4. Light falls on a double slit with slit separation of 8.04 × 10−6 m, and the
third bright fringe is seen at an angle of 13.1° relative to the central max-
imum. Find the wavelength of the light.
5. A double-slit interference experiment is performed with green light from
a mercury discharge tube. The separation between the slits is 4.43 ×10−6 m, and the third-order maximum of the interference pattern is at an
angle of 21.7° from the center of the pattern. What is the wavelength of
argon laser light?
ADDITIONAL PRACTICE
Holt Physics Problem BankCh. 16-4
NAME ______________________________________ DATE _______________ CLASS ____________________
6. Light shines on a double slit with slit separation of 3.92 × 10−6 m, and
the second bright fringe is seen at an angle of 20.4° relative to the central
maximum. What is the wavelength of the light?
7. If the distance between the two slits is 0.0220 cm, find the angle at which
a first-order bright fringe is observed for green light with a wavelength of
527 nm.
8. Blue light with a wavelength of 430.8 nm shines on two slits 0.163 mm
apart. What is the angle at which a first-order bright fringe is observed?
9. Orange light (l = 583 nm) passes through two slits 0.329 mm apart. Cal-
culate the angle at which a first-order bright fringe is observed.
10. If two slits are 0.267 mm apart, find the angle between the first-order
and second-order bright fringes for red light with a wavelength of
687 nm.
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Problem 17A Ch. 17–1
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NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 17A
P R O B L E MTwo electrostatic point charges of +20.0 µC and –30.0 µC exert attractiveforces on each other of –145 N. What is the distance between the twocharges?
S O L U T I O N
Given: q1 = 2.00 × 10−5 C q2 = −3.00 × 10−5 C
Felectric = −145 N kC = 8.99 × 109 N•m2/C2
Unknown: r = ?
Choose the equation(s) or situation:
Use Coulomb’s law, given on page 634.
Felectric = kC
r
q21q2
Rearrange the equation(s) to solve for the unknown(s): Rearrange Coulomb’s
law to solve for the distance between the two charges.
r = k
FCel
qec
1t
q
r
i
2
c =
r = 0.193 m = 19.3 cm
(8.99 × 109 N•m2/C2)(−3.0 × 10–5 C)(2.0 × 10–5 C)
−145 N
ADDITIONAL PRACTICE
1. Two electrostatic point charges of –13.0 mC and –16.0 mC exert repul-
sive forces on each other of 12.5 N. What is the distance between the
two charges?
2. Two electrostatic point charges of 99.9 mC and 33.3 mC exert repulsive
forces on each other of 87.3 N. What is the distance between the two
charges?
3. Two electrostatic point charges of –43.2 mC and 22.4 mC exert attrac-
tive forces on each other of –6.5 N. What is the distance between the
two charges?
4. A glass rod rubbed against silk gains a charge of –5.3 mC. What is the
electric force between the rod and the silk when the two are separated
by a distance of 4.2 cm? (Assume that the charges are located at a
point.)
5. A glass rod rubbed against your hair gains a charge of –14.0 nC. What
is the electric force between the balloon and your hear when the two
are separated by a distance of 7.1 cm? (Assume that the charges are lo-
cated at a point.)
Holt Physics Problem BankCh. 17–2
NAME ______________________________________ DATE _______________ CLASS ____________________
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6. A dog’s fur is combed and the comb gains a charge of 8.0 nC. What is
the electric force between the fur and the comb when the two are sepa-
rated by 2.0 cm?
7. Suppose two pions are separated by 8.3 × 10–10 m. If the magnitude of
the electric force between the charges is 3.34 × 10–10 N, what is the
value of q?
8. Suppose two muons having equal but opposite charge are separated by
6.4 × 10–8 m. If the magnitude of the electric force between the charges
is 5.62 ×10–14 N, what is the value of q?
9. Suppose two delta particles are separated by 9.3 × 10–11 m. If the mag-
nitude of the electric force between the charges is 2.66 × 10–8 N, what
is the value of q?
10. Suppose two equal charges are separated by 6.5 × 10–11 m. If the mag-
nitude of the electric force between the charges is 9.92 × 10–4 N, what
is the value of q?
Problem 17B Ch. 17–3
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 17B
P R O B L E M
Consider three point charges on the x-axis: q1 = 4.92 10–9 C is at theorigin, q2 = –6.99 10–8 C is at x = –3.60 10–1 m, and q3 = 5.65 10–9 Cis at x = 1.44 m. Find the magnitude and direction of the resultant forceon q1.
S O L U T I O N
Given: q1 = 4.92 × 10–9 C r1,2 = –3.60 × 10–1 m
q2 = –6.99 × 10–8 C r1,3 = 1.44 m
q3 = 5.65 × 10–9 C kC = 8.99 × 109 N•m2/C2
Unknown: F1,tot = ?
Calculate the magnitude of the forces with Coulomb’s law:
F1,2 = kC
r1
q
,2
12q2 = = –2.39 × 10–5 N
F1,3 = kC
r1
q
,3
12q3 = = 1.21 × 10–5 N
The forces are all along the x-axis, so add up the x-components:
F1,tot = F1,2 + F1,3 = −2.39 × 10–5 N + 1.21 × 10–5 N = −1.18 × 10–5 N
(8.99 × 109 N•m2/C2)(4.92 × 10–9 C)(5.65 × 10–9 C)
(1.44 m)2
(8.99 × 109 N•m2/C2)(4.92 × 10–9 C)(–6.99 × 10–8 C)
(–3.60 × 10–1 m)2
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ed. 1. Suppose four protons were at the corners of a square. The length of
each side of the square is 1.52 × 10–9 m. If q1 is on the upper right cor-
ner, calculate the magnitude and direction of the resultant force on q1.
2. Consider three point charges, q1 = 4.50 C, q2 = 4.50 C, and q3 = 6.30 C,
located at the corners of an isosceles triangle. The charges q1 and q2 are
5.00 m apart and form the base. The triangle is 3.50 m high, and q3 is
located at the top. Calculate the magnitude and direction of the resul-
tant force on q3.
3. Imagine three point charges on the corners of a triangle: q1 = –9.00 nC
is at the origin, q2 = –8.00 nC is at x = 2.00 m, and q3 = 7.00 nC is at y =3.00 m. Find the magnitude and direction of the resultant force on q1.
4. Suppose three point charges are on the y-axis: q1 = –2.34 × 10–8 C is at
the origin, q2 = 4.65 × 10–9 C is at y = 0.500 m, and q3 = –2.99 × 10–10 C
is at y = 1.00 m. What is the magnitude and direction of the resultant
force on q1?
5. Consider four electrons at the corners of a square. Each side of the
square is 3.02 × 10–5 m. Find the magnitude and direction of the resul-
tant force on q3 if it is at the origin.
ADDITIONAL PRACTICEADDITIONAL PRACTICE
NAME ______________________________________ DATE _______________ CLASS ____________________
6. Imagine three point charges at the corners of an isosceles triangle: q1 =2.22 × 10–10 C, q2 = 3.33 × 10–9 C, and q3 = 4.44 × 10–8 C. The charges
q1 and q2 are 1.00 m apart and form the triangle’s base. The triangle is
0.250 m tall. If q3 is at the top, what is the magnitude and direction of
the resultant force on q3?
7. Consider three 2.0 nC point charges at the following locations: at (0 m,
0 m), at (1.0 m, 2.0 m), and at (1.0 m, 0 m). Find the magnitude and
direction of the resultant force on the charge at the origin.
8. Consider three point charges on the corners of a triangle, where q1 =–4.0 mC at the origin; q2 = –8.0 mC at (2.0 m, 0 m); and q3 = 2.0 mC
at (0 m, 2.0 m). Calculate the magnitude and direction of the resultant
force on q1.
9. Suppose three point charges are on the corners of a triangle: q1 =9.00 mC is at the origin, q2 = 6.00 mC is at the point (1.00 m, 1.00 m),
and q3 = 3.00 mC is at (–1.00 m, 1.00 m). Find the magnitude and di-
rection of the resultant force on q1.
10. Consider three equal point charges of 4.00 nC on a line. All charges are
4.00 m apart. Calculate the magnitude and direction of the resultant
force on the charge in the middle.
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Holt Physics Problem BankCh. 17–4
Problem 17C Ch. 17–5
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 17C
P R O B L E MThree charges lie along the x-axis. One charge, q1 = –12 µC is at the origin. Another charge, q2 = 24 µC is at x = 1.0 m. A third charge,q3 = –36 µC, is placed so that q2 is in electrostatic equilibrium with q1 and q3. How large must the electrostatic force between q2 and q3 be to balance the force of q1 and q2?
S O L U T I O NGiven: q1 = −12 mC r1,2 = 1.0 m
q2 = 24 mC kC = 8.99 × 109 N•m2/C2
q3 = −36 mC
Unknown: r3,2 = ? F3,2 = ?
Diagram: y
q1 q2 x
x = 0 x = 1.0 m
Choose the equation(s) or situation: To solve for the electrostatic force, r3,2 is
needed. To solve for r3,2, determine where q3 must be placed in order to achieve
electrostatic equilibrium. To have electrostatic equilibrium, the direction of the
force of the charge on q2 must be opposite the direction of the force of the first
charge. The electrostatic force due to q1 points to q2. If q3 were between q1 and
q2, both forces on q2 would point in the same direction, so it would not be in
electrostatic equilibrium. So q3 can not be between the charges. Because q3 is
greater than q1, we would expect the location of q1 to be closer to q2 in order for
the forces to balance. So q3 is opposite the position of q1. The electric forces must
be equal and opposite to have electrostatic equilibrium.
F3,2 = –F2,1 = (x
k
–C
1
q
.03q
m2
)2 = –kC
x
q22q1
q2 x2 = –q1(x – 1.0 m)2 = – q1x2 + 2q1x – (1.0 m2)q1
(q2 + q1)x2 – 2q1x + (1.0 m2)q1 = 0
Use the quadratic formula to solve for x.
x =2q1 ±
√(2q1)2 – 4(q2+ q1)(1.0 m2) q1
2(q2 + q1)
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x = = 2.0 m ± 1.4 m
To achieve electrostatic equilibrium, q3 must be opposite q1, so x = 3.4 m
from the origin.
2(24 mC) ±√
4(24 mC)2 – 4(–12 mC + 24mC)(24 mC)(1.0m2)2(–12 mC + 24 mC)
NAME ______________________________________ DATE _______________ CLASS ____________________
Substitute the values into the equation(s) and solve: Calculate the electro-
static force between q2 and q3.
r3,2 = x – r1,2 = 3.4 m – 1.0 m = 2.4 m
F3,2 = kC
r3
q
,
2
22q3 = = 0.67 m
(8.99 × 109 N•m2/C2)(–1.2 × 10–5 C)(–3.6 × 10–5 C)
(2.4 m)2
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Ch. 17–6
ADDITIONAL PRACTICE
1. Three charges are on the y-axis. One positive charge, q1 = 9.0 mC, is at
y = 3.0 m. A second charge of –19 mC is located at the origin. A third
charge of 9.0 mC is placed so that q2 is in electrostatic equilibrium with
q1 and q3. How large must the electrostatic force between q2 and q3 be
to balance the force between q1 and q2?
2. Three charges lie along the x-axis. One positive charge, q1 = 25 mC is at
x = 0.25 m. Another charge, q2 = –5.0 mC, is located at the origin. A
third charge, q3 = –35 mC, is placed so that q2 is in electrostatic equilib-
rium with q1 and q3. How large must the electrostatic force between q2
and q3 be to balance the force between q1 and q2?
3. Three charges are on the y-axis. One positive charge, q1 = 6.0 mC is
at y = 5.0 cm. Another charge, q2 = –12 mC, is at the origin. A third,
6.0 mC charge is placed so that q2 is in electrostatic equilibrium with
q1 and q3. How large must the electrostatic force between q2 and q3
be to balance the force between q1 and q2?
4. A charge of 7.2 nC and a charge of 6.7 nC are separated by 32 cm. Find
the equilibrium position for a –3.0 nC charge.
5. A charge of 5.5 nC and a charge of 11 nC are separated by 88 cm. Find
the equilibrium position for a –22 nC charge.
6. A charge of –2.5 nC and a charge of –7.5 nC are separated by 20.0 cm.
Find the equilibrium position for a 5.0 nC charge.
7. Three charges are on the y-axis. A –2.3 C charge is at the origin and an-
other, unknown charge is at y = 2.0 m. A third charge of –4.6 C is
placed at y = –2.0 m so that it is in electrostatic equilibrium with the
first two charges. What is the charge on q2?
8. Three charges are on the x-axis. At the origin is an 8.0 C charge. An un-
known charge, q2, is at x = 1.0 m. A third charge, q3 = –4.0 C, is placed
at x = –1.0 m, where it is in electrostatic equilibrium with the other two
charges. What is the charge on q2?
ADDITIONAL PRACTICE
Holt Physics Problem Bank
Problem 17C Ch. 17–7
NAME ______________________________________ DATE _______________ CLASS ____________________
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9. Three charges are on the y-axis. At the origin is a charge, q1 = 49 C, an
unknown charge, q2, is at y = 7.0 m. A third charge, q3 = –7.0 C, is
placed at y = –18 m, where it is in electrostatic equilibrium with q1 and
q2. What is the charge on q2?
10. Three charges are on the y-axis. At the origin is a charge, q1 = 72 C, an
unknown charge, q2, is at y = 15 mm. A third charge, q3 = –8.0 C, is
placed at y = –9.0 mm, so that it is in electrostatic equilibrium with q1
and q2. What is the charge on q3?
NAME ______________________________________ DATE _______________ CLASS ____________________
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Holt Physics Problem BankCh. 17–8
Holt Physics
Problem 17D
P R O B L E M
A typical thundercloud has an electric field of about 3.0 105 N/C. If theelectric field is directed toward the ground. What is the electric force onan 18 nC charge in this field?
S O L U T I O N
Given: E = 3.0 × 105 N/C q = 18 nC = 1.8 × 10–8 C
Unknown: Felectric = ?
Choose the equation(s) or situation: Use the definition of electric field and
rearrange it to solve for Felectric.
E = Fele
qctric
Felectric = Eq = (3.0 × 105 N/C)(1.8 × 10–8 C)
Felectric = 5.4 × 10–3 N, directed toward the ground
ADDITIONAL PRACTICE
1. An electric field of 9.0 N/C is directed along the positive x-axis. What is
the electric force on a –6.0 C charge in this field?
2. An electric field of 1500 N/C is directed along the positive y-axis. What
is the electric force on a 5.0 nC charge in this field?
3. Millikan’s experiment measures the charge of an electron by suspend-
ing charged oil droplets in an electric field. If an oil droplet with a mass
of 3.35 × 10–15 kg has the same charge as an electron. What electric
force is required to balance the weight of the oil droplet?
4. Two equal charges of 3.00 mC lie along the x-axis: one is at the origin,
and the other is 2.0 m from the origin. Find the magnitude and direc-
tion of the electric field at a point on the y-axis 0.25 m from the origin.
5. A charge, q1 = 15.0 mC, is at the origin, and a second charge, q2 =5.00 mC, is on the y-axis 0.500 m from the origin. Find the magnitude
and direction of the electric field at a point on the y-axis 1.00 m from
the origin.
6. An electric field of 1663 N/C is directed along the positive x-axis. If the
electric force on a charge is 8.42 × 10–9 N, what is the charge?
7. An electric field of 4.0 × 103 N/C is directed downward. If the electric
force on a charge is 6.43 × 10–9 N, what is the charge?
ADDITIONAL PRACTICE
NAME ______________________________________ DATE _______________ CLASS ____________________
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Problem17D Ch. 17–9
8. One electron is at the point (2.00 × 10–10 m, 0 m) and another electron
is at the point (3.00 × 10–10 m, 0 m). If the field point is at the origin,
how far from the origin (along the x-axis) could a proton be placed so
that the strength of the resultant electric field would be zero?
9. One charge, q1 = –9.00 C, is at the point (1.50 mm, 0 mm). Another
charge, q2 = 6.00 C, is at the point (–1.50 mm, 0 mm). If the field point
is at the origin, how far from the origin (along the x-axis) could a
3.00 C charge be placed so that the strength of the resultant electric
field would be zero?
10. One charge, q1 = –55.0 nC, is at the point (–5.00 × 10–7 m, 0 m). An-
other charge, q2 = 11.0 nC, is at the point (5.00 × 10–7 m, 0 m). If the
field point is at the origin, how far from the origin (along the x-axis)
could a 5.00 nC charge be placed so that the strength of the resultant
electric field would be zero?
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Problem 18A Ch. 18-1
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 18AELECTRICAL POTENTIAL ENERGY
P R O B L E MWhat if, instead of an apple, Sir Isaac Newton noticed a charge of 6.0 nCfall on his head from a branch 2.5 m above him. What would have beenthe change in the electrical potential energy if Earth’s electric fieldstrength has a magnitude of 95 N/C and a downward direction?
S O L U T I O NGiven: q = 6.0 nC = 6.0 × 10–9 C E = 225 N/C d = 2.5 m
Unknown: PEelectric = ?
Choose the equation(s) or situation:
Use the equation for electrical potential energy in a uniform electric field, given
on page 667.
PEelectric = –qEd = – (6.0 × 10–9 C) (95 N/C) (2.5 m) = –1.4 10–6 J
1. As a child tries to attach a string to a helium-filled balloon, 14.5 nC of
charge collects on the balloon’s surface as it rubs against her wool
sweater. Suppose the child lets go of the balloon before the string is at-
tached. The yellow balloon rises 290 m. What is the change in the elec-
trical potential energy if Earth’s electric field strength has a magnitude
of 105 N/C and a downward direction?
2. After brushing your hair many times, you notice that your hair is at-
tracted to the brush. Suppose a charge of +64 nC is on the brush and a
charge of +64 nC is on your hair. What is the electrical potential energy
associated with your hair and the brush when you hold the brush
0.95 m from your hair?
3. Sparks fly when free atmospheric electrons move fast enough to knock
electrons off of an atom with which they collide, causing an electron
“avalanche.” Suppose the electric field in the air is 3.0 × 106 N/C. If the
electron moves 7.3 × 10–7 m, what is the electrical potential energy as-
sociated with the electron?
4. Sometimes wearing woolen socks while walking on carpet can cause
charge to accumulate. You might have noticed this if you felt a spark as
you tried to open a door. Suppose a charge of –42 µC exists on a door-
knob, and your hand has a charge of 63 µC. The electrical potential en-
ergy associated with the hand and the doorknob is –6.92 × 10–4 J. How
far is your hand from the knob?
ADDITIONAL PRACTICE
Holt Physics Problem BankCh. 18-2
NAME ______________________________________ DATE _______________ CLASS ____________________
5. Suppose you have a charge of 16 nC and your friend has a charge of
14 nC. As you greet each other, the electrical potential energy associ-
ated with you and your friend is 2.1 × 10–6 J. How far are you standing
from your friend?
6. A magician holds a charged wand over scattered confetti and the con-
fetti moves to the wand. Suppose the confetti has a charge of −55 nC
and the wand has a charge of 77 nC. When the electrical potential en-
ergy associated with the wand and the confetti is −1.3 × 10–2 J, how far
is the wand from the confetti?
7. The largest electric field measured in a thundercloud is 3.4 × 105 N/C.
Suppose the airplane that measured this field flew into this thunder-
cloud at an altitude of 7.3 km. If the electrical potential energy associ-
ated with the cloud and the plane is –1.39 × 1011 J, what is the charge
on the plane?
8. Suppose you would like to develop a hat that wouldn’t fall off some-
one’s head by putting enough charge on the hat so that it would stay
no matter how much the wind blew. You begin by rubbing a silk hat
against your head so that the same amount of charge collects on the
hat as the charge on your head. When you wear the hat (r = 1.25 × 10–3 m), the electrical potential energy associated with the hat and
your head is 1.25 × 10–4 J. The hat stays as you walk into the wind.
a. How much charge was on the hat?
b. How much charge was on your head?
c. Suppose your friend approaches and puts the hat on. Is the hat
likely to stay on your friend’s head by charge alone? Why or why
not?
9. When you remove clothes from the dryer, a sock clings to a towel
(r = 9.4 × 10–4 m). The electrical potential energy associated with the
sock and the towel is 8.89 × 10–10 J. Suppose the sock and the towel
have equal and opposite charge. Calculate the charges on both.
10. After rubbing a balloon against your hair, the balloon sticks to the wall
because they have equal and opposite charge. The electrical potential
energy associated with your hair and the wall is 6.3 × 10–6 J when they
are 1.25 × 10–6 m apart. What are the charges on the balloon and your
hair?
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Problem 18B Ch. 18-3
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 18BPOTENTIAL DIFFERENCE
P R O B L E MA giraffe is the tallest animal on Earth. With a height of 5.3 m, a giraffecan eat leaves high in acacia trees. Suppose an ion sits atop a giraffe’shead. If the potential difference across the giraffe is 90.0 V, what is thecharge on the ion?
S O L U T I O N
Given: ∆V = 90.0 V PEelectric = 1.18 × 10–15 J r = 5.3 m
Unknown: q = ?
Choose the equation(s) or situation: Use the equation for the potential differ-
ence near a point charge, given on page 672.
q = r∆kc
V = = 5.3 10–8 C
(5.3 m)(90.0 V)8.99 × 109 N•m2/C2
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1. The Nile is 6695 km long, making it the longest river in the world. Sup-
pose an ion placed at the Nile River’s remotest stream in Burundi. If
the potential difference across the Nile is 114.0 V, what is the charge on
the ion?
2. The Akashi Kaikyo Bridge is the world’s longest bridge. Built in 1998,
this bridge spans the Japanese islands of Honshu and Awaji—a total of
1991 m across the islands. Suppose a support at one end of the bridge
is charged. If the potential difference across the bridge is 18.6 kV, what
is the charge on the ion?
3. How far from an electron is the electric potential 1.0 V?
4. How far from a charge of 94 nC is the electric potential 9.0 V?
5. Find the potential difference between a point infinitely far away and a
point 3.95 cm from a carbon atom having a charge of 1.28 × 10−18 C.
6. Cars start by generating a 3.0 × 106 N/C electric field that causes a
spark to cross a gap in a spark plug. The gap of a standard spark plug is
6.25 × 10–4 m. What minimum potential difference must you apply to
the spark plug?
7. Gustave Alexandre Eiffel designed and built the 0.30 km-tall Eiffel
Tower in 1889 for the World’s Fair in Paris. Suppose the atmospheric
electric field of the Earth is 95 N/C directed downward. What is the
electrical potential difference between the ground and the tip of the
Eiffel Tower?
ADDITIONAL PRACTICE
8. A water molecule is V-shaped, with oxygen in the center, and hydrogen
on either side, as shown below. The oxygen nucleus is 9.58 × 10−11 m
away from each of the hydrogen nuclei. The angle between the two hy-
drogen atoms is 105°. Find the electrical potential produced by the
protons at the point P, the midpoint between the hydrogen nuclei.
[Hint: Oxygen has eight protons and hydrogen has one proton.]
Holt Physics Problem BankCh. 18-4
NAME ______________________________________ DATE _______________ CLASS ____________________
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9. Some molecules, such as KBr, occur in a straight line in nature. There-
fore, you can imagine KBr as being on an imaginary x-axis. Potassium
has a charge of 3.04 × 10–18 C and is located 1.89 × 10–10 m from the
origin. Bromine has a charge of 5.60 × 10–18 C and is located –9.30 ×10–11 m from the origin. Calculate the electrical potential at the origin.
10. Salt forms when Cl– and Na+ ions combine into a cube. Suppose you
look at one face of the cube so that you see a square with charges on
each corner, as shown below. Each side of the square is 2.82 × 10–10 m
long. Calculate the electrical potential at the center of the square.
[Hint: Treat chlorine atoms as electrons and sodium atoms as
protons.]
P
0
H
H
1e
8e
1e
9.58
× 10
–ll m9.58 × 10 –
llm
105°
Problem 18C Ch. 18-5
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 18CCAPACITANCE
P R O B L E MIce has a dielectric constant of 96.5. Suppose you were going to make a1.07 nF capacitor by placing an ice cube between two aluminum sheetswith an area of 6.25 cm2. What is the plate separation?
S O L U T I O N
Given: C = 1.07 × 10−9 F k = 96.5 A = 6.25 cm2 = 6.25 10−4 m2
Unknown: d = ?
Choose the equation(s) or situation:
Use the equation for constructive a parallel-plate capacitor in a vacuum, given on
page 677 and multiply it by the dielectric constant of ice, k.
d = κe
C0A = = 5.00 10–4 m2= 5.00 cm2(96.5)(8.85 × 10–12 C2/N•m2)(6.25 10–4 m2)
1.07 × 10–9 F
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ADDITIONAL PRACTICE
1. Suppose a high-voltage filter uses a parallel-plate capacitor where air
(κ = 1) is in the gap between the plates. The 18 m F capacitor is placed
across a potential difference of 9.9 kV. If each plate has an area of 4.8 ×10–3 m2, what is the distance between the plates? Assume that this ca-
pacitor is in air, with a dielectric constant of 1.
2. Suppose a transmitter uses a 4550 pF parallel-plate capacitor in a vac-
uum. Each plate has an area of 6.4 × 10–3 m2. The capacitor is placed
across a potential difference of 36 kV. What is the distance between the
plates?
3. The radius of the Earth is 6.4 × 106 m. What is the capacitance of the
Earth, regarded as a conducting sphere?
4. What is the capacitance of your head if it is a conducting sphere of ra-
dius 0.10 m?
5. A TV receiver contains a capacitor of 14 mF charged to a potential dif-
ference of 1.5 × 104 V. How much charge and electrical potential en-
ergy does this capacitor store?
6. Some glass capacitors can store as much as 1000 pF when placed across
a potential difference as high as 600 V. What is the maximum amount
of charge and electrical potential energy a glass capacitor can store?
7. A 0.50 pF capacitor is connected across a 1.5 V battery. How much
charge can this capacitor store?
Holt Physics Problem BankCh. 18-6
NAME ______________________________________ DATE _______________ CLASS ____________________
8. Capacitors made of ceramic are very popular because of their small
size. Some ceramic capacitors can have a 1 m F capacity, allowing them
to store as much as 3 × 10–2 C. Across what potential difference can a
ceramic capacitor be placed?
9. Teflon, the same material used to coat some “non-stick” pots and pans,
is also used to build capacitors. Teflon capacitors can have a 2.0 m F ca-
pacitance, allowing them to store as much as 4.0 × 10–4 C. Across what
potential difference can a Teflon capacitor be placed?
10. Polyester capacitors can have a capacitance of 5.0 × 10–5 F. What po-
tential difference is required to store 6.0 × 10–4 C?
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Problem 19A Ch. 19-1
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 19ACURRENT
P R O B L E MA total charge of 55 mC passes through a cross-sectional area of a copperwire in 0.5 s. What is the current in the wire?
S O L U T I O NGiven: ∆Q = 5.5 × 10−2 C ∆t = 0.5 s
Unknown: I = ?
Choose the equation(s) or situation:
Use the equation for electric current, given on page 694.
I = ∆∆Q
t =
5.5 ×0.
1
5
0
s
−2 C = 0.11 A
1. A total charge of 76 C passes through a cross-sectional area of a copper
wire in 19 s. What is the current in the wire?
2. A total charge of 114 µC passes through a cross-sectional area of an
aluminum wire in 0.36 s. What is the current in the wire?
3. A total charge of 29 mC passes through a cross-sectional area of a
nichrome wire in 11 s. What is the current in the wire?
4. If a current in a wire of a telephone is 1.4 A, how long would it take for
98 C of charge to pass a point in this wire?
5. If a current in a wire of a vacuum cleaner is 9.65 A, how long would it
take for 30.9 C of charge to pass a point in this wire?
6. If a current in a wire of a blender is 7.8 A, how long would it take for
56 C of charge to pass a point in this wire?
7. A photocopy machine draws 9.3 A when it starts up. If the start-up
time is 15 s, how much charge passes a cross-sectional area of the cir-
cuit in this time?
8. A computer draws 3.0 A when it starts up. If the start-up time is
2.0 min, how much charge passes a cross-sectional area of the circuit in
this time?
9. A printer draws 0.70 A when it starts up. If the start-up time is
2.0 s, how much charge passes a cross-sectional area of the circuit in
this time?
10. A space heater draws 5.6 A when it starts up. If the start-up time is 4.3 s,
how much charge passes a cross-sectional area of the circuit in this
time?
ADDITIONAL PRACTICE
Holt Physics Problem BankCh. 19-2
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 19BRESISTANCE
P R O B L E MA clothes dryer is equipped with an electric heater. The heater works bypassing air across an electric wire that is heated by the electricity passingthrough it. If the wire’s resistance is 10.0 and the current through thewire equals 24 A, what is the potential difference across the heater wire?
S O L U T I O NGiven: R = 10.0 Ω I = 24 A
Unknown: ∆V = ?
Choose the equation(s) or situation: Rearrange the definition for resistance on
page 700 to solve for the potential difference across the wire.
∆V = IR = (24 A) (10.0 Ω) = 240 V
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ADDITIONAL PRACTICE
1. You have probably heard that high-voltages are more dangerous than
low voltages. To understand this, assume that your body has a resis-
tance of 1.0 × 105 Ω. What potential difference would have to be across
your body to produce a current of 1.0 mA (which would cause a tin-
gling feeling) and 15 mA (a fatal amount of current)?
2. A battery-powered electric lantern is used as a light source for camp-
ing. The light bulb in the lantern has a resistance of 6.4 Ω. Assume that
the rest of the lantern’s circuit has no resistance and that the current
through the circuit is 0.75 A. Calculate the potential difference across
the lantern’s battery.
3. Some kitchen sinks are equipped with electric garbage disposals. These
are units with rapidly rotating steel blades, which are able to crush and
chop up food so that it can be washed down the drain instead of taking
up space as solid garbage. The motor of a garbage disposal has a resis-
tance of about 25.0 Ω. If the current through the motor equals 4.66 A,
what is the potential difference across the motor’s terminals?
4. A washing machine motor carries a current through a circuit with a re-
sistance of 12.2 Ω. If the washing machine is plugged into a 120 V out-
let, what is the current in the motor?
5. If you were to swim in the Atlantic Ocean off the coast of Brazil, the re-
sistance of your body could drop as low as 1.0 × 102 Ω. An electric eel
in Brazil can have a potential difference of up to 650 V across it. If you
came into contact with this eel while swimming, what current would
be delivered to your body?
Problem 19B Ch. 19-3
NAME ______________________________________ DATE _______________ CLASS ____________________
6. When traveling to another country, you should always find out the
voltage that is used in that country before you plug in an appliance. To
understand the reason for this precaution, calculate the current that a
laptop computer would draw from a 120 V outlet in the United States
if the computer has a resistance of 40.0 Ω. Then, calculate the current
that the same computer would draw if you plugged it into a 240 V out-
let in the United Kingdom.
7. A television set is plugged into a 120 V outlet. The television circuit
carries a current equal to 0.75 A. What is the overall resistance of the
television set?
8. An electric car is equipped with an electric motor that can deliver
50 horsepower. The voltage across the motor’s terminals equals 5.00 ×102 V, and the current through the motor is 0.89 A. What is the resis-
tance in the motor’s circuit?
9. A medium-sized household oscillating fan draws 545 mA of current
when the potential difference across its motor is 120 V. How large is the
fan’s resistance?
10. A refrigerator’s circuit carries a current equal to 0.65 A when the volt-
age across the circuit equals 117 V. How large is the resistance of the re-
frigerator’s circuit?
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Holt Physics Problem BankCh. 19-4
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 19CELECTRIC POWER
P R O B L E MIf an alarm clock is plugged into a 120 V outlet, the electric current in theclock’s circuit is 4.2 102 A. How much power does the alarm clock use?
S O L U T I O N
Given: ∆V = 120 V I = 4.2 × 10–2 A
Unknown: P = ?
Choose the equation(s) or situation:
Because the current and potential difference are given but the power is un-
known, use the third form of the power equation on page 709, which in-
cludes these three variables.
P = I∆V = (4.2 × 10−2 A) (120 V) = 5.0 W
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1. A generator at a central electric power plant produces electricity with a
potential difference of 2.5 × 104 V across power lines which carry a
current of 20.0 A. How much power does the generator produce?
2. An electric sports car was developed several years ago at Texas A&M
University in College Station, Texas. If the potential difference across
the car’s motor is 720 V and the resistance was 0.30 Ω, how much
power was needed for the car to run?
3. A light bulb has a filament with a resistance of 144 Ω, while a second
bulb has a filament with a resistance of 240 Ω. Both bulbs are con-
nected across a 120 V outlet. Which light bulb is brighter? [Hint: The
brightest bulb uses the most power.]
4. A microwave oven requires 1750 W of power to cook food. If the oven
is plugged into a 120 V outlet, what is the resistance in the oven’s
circuit?
5. A waffle iron requires 650 W of power to operate. If the waffle iron is
plugged into a 120 V outlet, what is the resistance in the waffle iron’s
circuit?
6. An electric kettle requires 370 W of power to boil water. If the kettle is
plugged into a 120 V outlet, what is the resistance in the kettle’s circuit?
7. A blender requires 350 W to power the rotating blades that chop food.
If the blender has a resistance of 75 Ω, how much current passes
through the blender’s circuit?
ADDITIONAL PRACTICE
Problem 19C Ch. 19-5
NAME ______________________________________ DATE _______________ CLASS ____________________
8. A computer with a power input of 230.0 W has a resistance of
91.0 Ω. Find the current in the computer.
9. A laser was developed in 1995 at the University of Rochester, New York,
that produced a beam of light that lasted for about a billionth of a sec-
ond. The power output of this beam was 6.0 × 1013 W. If all of the elec-
trical power was converted into light and 8.0 × 106 A of current was
needed to produce this beam, what was the potential difference across
the circuit of the laser?
10. Fuel cells are chemical cells that combine hydrogen and oxygen gas to
produce electrical energy. In recent years, a fuel cell has been developed
that can generate 1.06 × 104 W of power. If this fuel cell has a current
of 16.3 A, what is the potential difference across the fuel cell?
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Holt Physics Problem BankCh. 19-6
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 19DCOST OF ELECTRICAL ENERGY
P R O B L E MSuppose you woke up one morning and realized that you had forgotten toturn off your front porch light the night before. If you had used .540 kW•hof energy over a period of 12 h, what is power emitted by the light bulb?
S O L U T I O NGiven: Energy = 0.540 kW•h ∆t = 12 h
Unknown: P = ?
Choose the equation(s) or situation:
Use the equation relating energy and power on page 712.
Energy = P∆t
Rearrange the equation(s) to isolate the unknown(s):
Rearrange the equation to solve for the total power.
P = En
∆er
t
gy =
0.54
1
0
2
k
h
W•h = 0.045 kW = 45 W
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ADDITIONAL PRACTICE
1. Suppose that you’ve just returned to work from your lunch hour.
When you reach your desk, you realize that you had forgotten to turn
off your computer. Fortunately, your computer was in its energy-
conserving “sleep” mode. What is the power of a computer which
consumed 2.7 × 108 J of energy?
2. Suppose you’ve just returned to the parking lot from a 3.0 h shopping
spree at the mall and realized that you had forgotten to turn off the
headlights to your car. If 4.86 × 108 J of energy was spent, what is the
power of the headlights?
3. A bread machine requires 1200 W to bake bread. How much time is re-
quired for it to use 1.512 × 1010 J of energy?
4. Calculate how much time is required for a 600 W air conditioner to use
8.64 × 109 J of energy?
5. As an incentive to conserve electricity, some electric companies charge
lower prices for electricity up to a certain number of kilowatt-hours,
and then raise the cost of electricity for each kilowatt-hour you use
over that number. Suppose your local electric company charges
$0.0650/kW•h for the first 200.0 kW•h, and then drastically raises the
price for every kilowatt-hour more. What is the maximum your family
would pay if they wanted to pay only the cheaper rate?
Problem 19D Ch. 19-7
NAME ______________________________________ DATE _______________ CLASS ____________________
6. Suppose you receive an electric bill which states the following:
Final Meter Reading 24422 kW•h
Previous Meter Reading 24204 kW•h
Cost of Electricity Used for 20 Days $0.078/kW•h
How much must you pay the electric company?
7. How much energy does a 125 W computer use in an 8.0 h workday?
8. Calculate how many joules of energy a 750 W refrigerator uses in 24 h.
9. Calculate how many joules of energy a 550.0 W hair dryer uses in
10.0 min.
10. Calculate how many joules of energy an 850 W toaster uses in 3.0 min.
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Problem 20A Ch. 20–1
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NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 20ARESISTORS IN SERIES
P R O B L E MA 18.0-Ω resistor is connected in series with another resistor across a 1.55-V battery. The current in the circuit is 25 mA. Calculate the value ofthe unknown resistance.
S O L U T I O NGiven: ∆V = 1.55 V R1 = 18 Ω I = 0.025 A
Unknown: R2 = ?
Choose the equation(s) or situation:
Use the equation relating the potential difference across the load for resistors in
series, given on page 737.
∆V = IR1 + IR2
Rearrange the equation(s) to isolate the unknown(s): Rearrange to solve for R2.
R2 = ∆
I
V − R1 =
0
1
.
.
0
5
2
5
5
V
A – 18 Ω = 62 Ω – 18 Ω = 44 Ω
1. A 16- Ω resistor is connected in series with another resistor across a
12-V battery. The current in the circuit is 0.42 A. Calculate the value of
the unknown resistance.
2. A 24-Ω resistor is connected in series with another resistor across a 3.0-V
battery. The current in the circuit is 62 mA. Calculate the value of the
unknown resistance.
3. A 9-Ω resistor is connected in series with another resistor across a 9.0-V
battery. The current in the circuit is 0.33 A. Calculate the value of the
unknown resistance.
4. A string of holiday lights has 73 light bulbs in series. Each light bulb
has a resistance of 3.0 Ω. Calculate the equivalent resistance.
5. A movie theater has 25 surround-sound speakers wired in series. Each
speaker has a resistance of 12.0 Ω. What is the equivalent resistance?
6. In case of an emergency, a corridor on an airplane has 57 lights wired
in series. Each light bulb has a resistance of 2.0 Ω. Find the equivalent
resistance.
7. A quadraphonic car stereo operates on electricity provided by the car’s
12-V battery and is connected in series. Each channel of the stereo,
which feeds the electric signal to one of the stereo’s four speakers, has a
resistance of 4.1 Ω. How much current is in the circuit of each stereo
channel?
ADDITIONAL PRACTICE
Holt Physics Problem BankCh. 20–2
NAME ______________________________________ DATE _______________ CLASS ____________________
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8. A chandelier has 10 sockets wired in series, each of which holds a 10 Ωlight bulb. If the voltage across the chandelier’s circuit is 100 V, what is
the current drawn by the chandelier?
9. A portable lamp has three bulbs wired in series: one bulb has a resistance
of 96 Ω, one bulb has a resistance of 48 Ω, and one bulb has a resistance
of 29 Ω. If the voltage across the lamp is 115 V, what is the current
through the lamp’s circuit?
10. Three bulbs are wired in series: one bulb has a resistance of 56 Ω, one
bulb has a resistance of 82 Ω, and one bulb has a resistance of 24 Ω. If
the voltage across the circuit is 9.0 V, what is the current through the
circuit?
Problem 20B Ch. 20–3
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 20BRESISTORS IN PARALLEL
P R O B L E M
A 42.0 resistor is connected in parallel with another resistor across a9.0 V battery. The current in the circuit is 0.41 A. Calculate the value ofthe unknown resistance.
S O L U T I O NGiven: ∆V = 9.0 V R1 = 42.0 Ω I = 0.41 A
Unknown: R2 = ?
Choose an equation(s) or situation:
Use the equation relating potential difference and equivalent resistance for resis-
tors in parallel, given on page 742.
∆V = IReq
I = R
∆
e
V
q =
∆R
V
1 +
∆R
V
2
Rearrange the equation(s) to isolate the unknown(s): Rearrange to solve for R2.
∆R
V
2 = I –
∆R
V
1
R2 = = = = 45 Ω9.0 V
[0.41 A − 0.21 A]
9.0 V
0.41 A − 9
4
.
2
0
ΩV
∆V
I – ∆R
V
1
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1. A 3.3 Ω resistor is connected in parallel with another resistor across a
3.0 V battery. The current in the circuit is 1.41 A. Calculate the value of
the unknown resistance.
2. A 56 Ω resistor is connected in parallel with another resistor across a
12 V battery. The current in the circuit is 3.21 A. Calculate the value of
the unknown resistance.
3. An 18 Ω resistor is connected in parallel with another resistor across a
1.5 V battery. The current in the circuit is 103 mA. Calculate the value
of the unknown resistance.
4. A 39 Ω resistor, an 82 Ω resistor, a 12 Ω resistor and a 22 Ω resistor are
connected in parallel across a potential difference of 3.0 V. Calculate
the equivalent resistance.
5. A 10 Ω resistor, a 12 Ω resistor, a 15 Ω resistor and an 18 Ω resistor are
connected in parallel across a potential difference of 12 V. What is the
equivalent resistance?
Holt Physics Problem BankCh. 20–4
NAME ______________________________________ DATE _______________ CLASS ____________________
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6. A 33 Ω resistor, a 39 Ω resistor, a 47 Ω resistor and a 68 Ω resistor are
connected in parallel across a potential difference of 1.5 V. Find the
equivalent resistance.
7. A refrigerator and an oven are wired in parallel across a potential differ-
ence of 120 V. The refrigerator has a resistance of 75 Ω and the oven has
a resistance of 91 Ω. How much current is in the circuit of each
appliance?
8. A computer and a printer are wired in parallel across a potential differ-
ence of 120 V. The computer has a resistance of 82 Ω and the printer
has a resistance of 24 Ω. How much current is in the circuit of each
machine?
9. A lamp and a stereo are wired in parallel across a potential difference of
120 V. The lamp has a resistance of 11 Ω and the stereo has a resistance
of 36 Ω. How much current is in the circuit of each load?
10. Two bulbs are wired in parallel: one bulb has a resistance of 3.3 Ω, and
the other bulb has a resistance of 4.3 Ω. If the voltage across the circuit
is 1.5 V, what is the current through each bulb?
Problem 20C Ch. 20–5
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 20CANGULAR ACCELERATION
P R O B L E MDetermine the unknown resistance in thecomplex circuit shown at right. The currentin the circuit is 0.36 A.
S O L U T I O N1. Redraw the circuit as a group of resistors along one side of the circuit.
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12.0 V
8.0 ΩR = ?
9.0 Ω
5.0 Ω 3.0 Ω2.0 Ω
3.0 Ω5.0 Ω
5.0 Ω 3.0 Ω2.0 Ω
3.0 Ω5.0 Ω
9.0 Ω
(a)
(c)
(b)
R = ? 8.0 Ω
2. Identify components in series, and calculate their equivalent resistance.
Resistors in groups (a), (b), and (c) are in series.
For group (a): Req = 9.0 Ω + R
For group (b): Req = 2.0 Ω + 5.0 Ω + 3.0 Ω = 10.0 Ω = R1 below
For group (c): Req = 5.0 Ω + 3.0 Ω = 8.0 Ω = R2 below
3. Identify components in parallel, and calculate their equivalent resistance.
Groups (b) and (c) combine to become group (d).
For group (d):
R
1
eq =
R
1
1 +
R
1
2 =
10.
1
0 Ω +
8.0
1
Ω =
0
1
.1
Ω0
+ 0
1
.1
Ω3
= 0
1
.2
Ω3
R = 4.4 Ω
4. Repeat steps 2 and 3 until the resistors in the circuit are reduced to a single
equivalent resistance. All resistors combine to become group (e). The remainder of
the resistors in group (e) are in series.
For group (e): Req = 9.0 Ω + R + 4.4 Ω + 8.0 Ω = R + 21.4 Ω
5. Choose the equation(s) or situation:
Use the equation relating equivalent resistance to potential difference and current.
Req = R − 21.4 Ω = ∆
I
V − 21.4 Ω =
1
0
2
.3
.0
6
V
A – 21.4 Ω = 33 Ω – 21.4 Ω = 12 Ω
NAME ______________________________________ DATE _______________ CLASS ____________________
2. Determine the un-
known resistance
in the complex cir-
cuit shown at right.
The current in the
circuit is 375 mA.
3. Determine the un-
known resistance in
the complex circuit
shown at right. The
current in the circuit
is 185 mA.
4. Determine the equivalent re-
sistance of the complex circuit
shown at right.
5. Determine the equivalent re-
sistance of the complex circuit
shown at right.
6. Determine the equiv-
alent resistance of the
complex circuit
shown at right.
7. What will be the net current
for the circuit shown at right? Cop
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Holt Physics Problem BankCh. 20–6
18.0 Ω8.00 Ω
3.0 Ω 4.0 Ω
20.0 Ω
R = ?
22.0 Ω
6.0 Ω9.00 V
2.0 Ω 3.0 Ω8.00 Ω 8.00 Ω
3.0 Ω 4.0 Ω
2.0 Ω
R = ?8.0 Ω 8.0 Ω
3.0 Ω
12.0 V
3.0 Ω 4.0 Ω2.0 Ω
5.0 Ω 6.0 Ω 3.0 Ω 4.0 Ω4.0 Ω
10.0 Ω
10.0 Ω
10.0 Ω
10.0 Ω
10.0 Ω
10.0 Ω
10.0 Ω
10.0 Ω
10.0 Ω10.0 Ω
10.0 Ω 10.0 Ω
10.0 Ω
10.0 Ω
10.0 Ω
10.0 Ω
2.0 Ω 2.0 Ω
2.0 Ω
2.0 Ω
2.0 Ω
2.0 Ω
2.0 Ω
2.0 Ω
2.0 Ω
2.0 Ω
2.0 Ω 2.0 Ω
10.0 Ω 10.0 Ω
10.0 Ω
10.0 Ω10.0 Ω 10.0 Ω10.0 Ω
10.0 Ω
10.0 Ω 10.0 Ω 10.0 Ω
10.0 Ω
10.0 Ω
10.0 Ω 10.0 Ω 10.0 Ω
10.0 Ω
10.0 Ω
10.0 Ω
10.0 Ω 10.0 Ω
5.0 Ω
2.0 Ω
5.0 Ω
6.0 Ω4.0 Ω
1.5 Ω3.0 Ω
12.0 V
ADDITIONAL PRACTICE
1. Determine the unknown resistance
in the complex circuit shown at
right. The current in the circuit is
680 mA.
15.0 V
11 ΩR = ?12.0 Ω
15 Ω
7.0 Ω6.0 Ω
NAME ______________________________________ DATE _______________ CLASS ____________________
8. What will be the net current for
the circuit shown at right?
9. What will be the net current for
the circuit shown at right?
10. What will be the net current for
the circuit shown at right?
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Problem 20C Ch. 20–7
5.0 Ω
5.0 Ω
5.0 Ω
3.0 Ω
5.0 Ω
5.0 Ω3.0 Ω
5.0 Ω
15.0 V
4.0 Ω 3.0 Ω 1.0 Ω
4.0 Ω 2.0 Ω 2.0 Ω
24.0 V
4.0 Ω 2.0 Ω2.0 Ω
4.0 Ω
4.0 Ω
8.0 Ω3.0 Ω
4.0 Ω
2.0 Ω
24.0 V
3.0 Ω
8.0 Ω
4.0 Ω5.0 Ω 2.0 Ω
Holt Physics Problem BankCh. 20–8
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 20DCURRENT IN AND POTENTIAL DIFFERENCE ACROSS A RESISTOR
P R O B L E M
Determine the current in and the potential difference across the 5.0 re-sistor in the circuit diagram at right.
S O L U T I O N1. Determine the equivalent resistance in the circuit.
For group (a): Req,a = 5.0 Ω + 2.0 Ω = 7.0 Ω = R2
For group (b): Re
1
q,b =
R
1
1 +
R
1
2 =
4.0
1
Ω +
7.0
1
Ω
Re
1
q,b =
0
1
.2
Ω5
+ 0
1
.1
Ω4
= 0
1
.3
Ω9
Req,b = 2.6 Ω
For group (c): Req,c = 3.0 Ω + 2.6 Ω + 6.0 Ω = 11.6 Ω
2. Calculate the total current in the circuit, which is the current in group (c).
I = ∆R
V
e
t
q
ot = 1
1
1
2
.
.
6
0
ΩV
= 1.0 A
3. Determine a path from the equivalent resistance found in step 1 to the
5.0 resistor. Review the path taken to find the equivalent resistance and
work backward through this path.
4. Follow the path determined in step 3, and calculate the current in and the
potential difference across each equivalent resistance. Repeat this process
until the desired values are found.
Regroup, evaluate, and calculate. The circuit’s equivalent resistance is that of
group (c), as found in step 1 above. The resistors in group (c) are in series; there-
fore, the current in each resistor is the same as the current in the equivalent re-
sistance, which equals 1.0 A. The potential difference across group (b), which is
represented by the 2.6 Ω resistor in group (c), can be replaced with ∆V = IR.
Given: I = 1.0 A R = 2.6 Ω
Unknown: ∆V = ?
∆V = IR = (1.0 A)(2.6 Ω) = 2.6 V
Regroup, evaluate, and calculate. Replace the center resistor with group (b).
The resistors in group (b) are in parallel; therefore, the potential difference
across each resistor is the same as the potential difference across the 2.6 Ωequivalent resistance, which equals 2.6 V. The current in the 7.0 Ω resistor in
group (b) can be calculated using I = ∆V/R.
Given: ∆V = 2.6 V R = 7.0 Ω
Unknown: I = ?
I = ∆R
V =
7
2
.
.
0
6
ΩV
= 0.37 A
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4.0 Ω
2.0 Ω6.0 Ω
5.0 Ω3.0 Ω
(c)
(a)
(b)
12 V
NAME ______________________________________ DATE _______________ CLASS ____________________
Regroup, evaluate, and calculate. Replace the 7.0 Ω resistor with group (a).
The resistors in group (a) are in series; therefore, the current in each resistor
is the same as the current in the 7.0 Ω equivalent resistance, which equals
0.37 A.
The potential difference across the 5.0 Ω resistor can be calculated using
∆V = IR.
Given: I = 0.37 A R = 5.0 Ω
Unknown: ∆V = ?
∆V = IR = (0.37 A)(5.0 Ω) = 1.85 V
∆V = 1.85 V
I = 0.37 A
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Problem 20D Ch. 20–9
ADDITIONAL PRACTICE
1. Determine the current in and the potential
difference across the 4.0 Ω resistor in the
circuit diagram at right.
2. Determine the current in and the potential
difference across the 9.0 Ω resistor in the
circuit diagram at right.
3. Determine the current in and the
potential difference across the 6.0 Ωresistor in the circuit diagram at right.
4. Determine the current in and the potential
difference across the 10.0 Ω resistor in the
circuit diagram at right.
5. Determine the current in and the potential
difference across the 4.0 Ω resistor in the
circuit diagram at right.
3.0 Ω 12.0 V
10.0 Ω
10.0 Ω
10.0 Ω
4.0 Ω
1.5 V
2.0 Ω
9.0 Ω
8.0 Ω6.0 Ω 4.0 Ω
3.0 Ω7.0 Ω 6.0 Ω
4.0 Ω 5.0 Ω
2.0 Ω9.0 V
7.0 Ω 9.0 V
4.0 Ω 3.0 Ω5.0 Ω
10.0 Ω 6.0 Ω2.0 Ω
3.0 V
4.0 Ω 6.0 Ω2.0 Ω
5.0 Ω
3.0 Ω1.0 Ω
NAME ______________________________________ DATE _______________ CLASS ____________________
6. Determine the current in and the
potential difference across the
3.0 Ω resistor in the circuit dia-
gram at right.
7. Determine the current in and the
potential difference across the 2.0 Ωresistor in the circuit diagram at
right.
8. Determine the current in and the
potential difference across the 7.0 Ωresistor in the circuit diagram at right.
9. Determine the cur-
rent in and the
potential difference
across the 12.0 Ωresistor in the circuit
diagram at right.
10. Determine the current in and the
potential difference across the 15.0 Ωresistor in the circuit diagram at
right.
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Holt Physics Problem BankCh. 20–10
2.0 Ω 3.0 Ω
2.0 Ω 4.0 Ω
3.0 V
2.0 Ω
2.0 Ω
5.0 Ω5.0 Ω
12.0 V
2.0 Ω
5.0 Ω
5.0 Ω
5.0 Ω5.0 Ω
3.0 Ω
2.0 Ω
7.0 Ω
3.0 Ω
12.0 V
4.0 Ω
5.0 Ω
3.0 Ω 3.0 Ω3.0 Ω5.0 Ω
1.5 V 4.0 Ω
4.0 Ω6.0 Ω
12.0 Ω 6.0 Ω6.0 Ω
15.0 Ω
6.0 Ω
30.0 Ω
5.0 Ω 12.0 V
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Problem 21A Ch. 21–1
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 21APARTICLE IN A MAGNETIC FIELD
P R O B L E MA bubble chamber is a device used to make the paths of charged particlesvisible in collision experiments. Suppose a proton travels in one of thesebubble chambers at a speed of 1.9 × 107 m/s. A magnetic force of 4.3 ×10–12 N acts on the proton. What is the magnitude of the magnetic field atright angles to the proton’s path?
S O L U T I O N
Given: q = 1.60 × 10–19 C v = 1.9 × 107 m/s
Fmagnetic = 4.3 × 10–12 N
Unknown: B = ?
Use the equation for the magnetic field from page 773.
B = Fma
qg
vnetic = = 1.4 Τ
4.3 × 10–12 N(1.60 × 10–19 C)( 1.9 × 107 m/s)
ADDITIONAL PRACTICE
1. An electron in a cathode ray tube (found in many televisions) moving at a speed of
1.2 × 106 m/s has a magnetic force of 1.2 × 10–17 N acting on it. What is the magnitude
of the magnetic field perpendicular to the electron’s path?
2. Humans emit a magnetic field because ions move through the blood. Suppose an elec-
tron whizzes horizontally by a standing person at a speed of 3.9 × 106 m/s. If the elec-
tron is affected by a magnetic force of 1.9 × 10–22 N, how much of a magnetic field does
this person emit?
3. An electron moves perpendicular to a sunspot at a speed of 7.8 × 106 m/s. A magnetic
force of 3.7 × 10–13 N is exerted on the electron. What is the magnitude of the magnetic
field emitted by the sunspot?
4. In the Fermilab accelerator, a proton maintains circular motion about a radius of
1.0 km. The proton travels at right angles to a uniform magnetic field of 3.3 T. What is
the speed of the proton? (Hint: The magnetic force exerted on the proton is the force
that maintains circular motion.)
5. Aurora Borealis, the “northern lights” found at high latitudes, happens when charged
particles fall into Earth’s atmosphere near the poles. The magnetic field near the poles is
5.0 × 10–5 T. Suppose the particle has a charge of 1.60 × 10–19 C. If the magnetic force
exerted on the particle is 6.1 × 10–17 N, at what speed is the particle moving? (Assume
that the particle moves perpendicular to Earth’s magnetic field.)
6. The magnetic field in the Crab Nebula is about 1 × 10–8 T. If an electron moving per-
pendicular to this field is affected by a magnetic force of 3.2 × 10–22 N, what is the elec-
tron’s speed?
Holt Physics Problem BankCh. 21–2
NAME ______________________________________ DATE _______________ CLASS ____________________
7. Suppose an electron moves to the right at a velocity of 6 × 106 m/s at an
angle of 45° to a uniform magnetic field. If the magnetic field is 3 × 10–4 T
upward, what is the magnitude and direction of the magnetic force ex-
erted on the electron? (Hint: Only the component perpendicular to the
magnetic field contributes to the magnetic force.)
8. A proton moves at right angles to a uniform magnetic field of 0.8 T. If
the speed of the proton is 3.0 × 107 m/s, what is the magnetic force ex-
erted on the proton?
9. Suppose an electron moves at a speed of 2.2 × 106 m/s on the surface of
the Sun. Suppose the Sun has a magnetic field of 1.1 × 10–2 T perpendic-
ular to the electron’s path. What is the magnetic force exerted on the
electron by the Sun?
10. Suppose an electron moves at a speed of 9.3 × 105 m/s. If it moves at
right angles to a uniform magnetic field of 4.1 × 10–10 T in interstellar
galactic space, what is the magnetic force exerted on the electron?
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Problem 21B Ch. 21–3
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 21BFORCE ON A CURRENT CARRYING CONDUCTOR
P R O B L E MSuppose a wire carries a current of 7.32 A. If a magnetic field of 0.0038 Tproduces a force that has a magnitude of 4.2 × 10–3 N, how long is thewire?
S O L U T I O NGiven: I = 7.32 A B = 0.0038 T
Fmagnetic = 4.2 × 10–3 N
Unknown: l = ?
Use the equation for the force on a current-carrying conductor perpendicular to
a magnetic field, shown on page 776.
l = Fma
Bg
Inetic =
(0.0
4
0
.
3
2
8
×T
1
)
0
(
–
7
3
.3
N
2 A) = 0.15 m
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ADDITIONAL PRACTICE
1. Suppose a wire is placed perpendicular to a uniform magnetic field of 4.6 × 10–4 T. A
magnetic force of 2.9 × 10–3 N is exerted on the wire. If the current in the wire is 10.0 A,
how long is the wire?
2. An electric eel in Brazil can produce 1 A of current and up to 650 V. If the eel swims
perpendicular to the Earth’s magnetic field of 2.8 × 10–5 T and is affected by a magnetic
force of 5.6 × 10–5 N, how long is the eel?
3. A 12 m wire carrying a current of 12 A is placed at right angles to a uniform magnetic
field. A magnetic force of 7.3 × 10–2 N acts on this wire. What is the magnitude of the
magnetic field?
4. Suppose a lightning bolt strikes the ground in a straight path and a magnetic force of
7.8 × 105 N acts on the bolt. The bolt carries 1.8 × 104 A of current. The cloud from
which the lightning bolt came is 12 km above the ground. What is the magnitude of the
magnetic field perpendicular to this lightning bolt?
5. Suppose a 14.32 A current moves through a 15.0 cm wire. If a magnetic force of 6.62 ×10–4 N acts on the wire, what would be the magnitude of the magnetic field at right an-
gles to the wire?
6. A magician wants to build a device that will allow him to levitate for his magic act. His
device has a wire 10 m long bent into a loop. If the magician’s mass is 75 kg, and Earth’s
magnetic field (perpendicular to the wire) is 4.8 × 10–4 T, how much current must
move through the wire for the magician to levitate?
Holt Physics Problem BankCh. 21–4
NAME ______________________________________ DATE _______________ CLASS ____________________
7. A printer is connected to a 1.0 m cable. If the magnetic force is 9.1 × 10–5 N,
and the magnetic field is 1.3 × 10–4 T, how much current passes through
the wire?
8. Suppose a high-voltage cable carries 1.5 × 103 A of current 15 km along a
straight city road. The Earth’s magnetic field at this latitude is at a 45°
angle to the power line and has a magnitude of 2.3 × 10–5 T. What is the
magnetic force exerted on the cable?
9. A fax machine requires 1.4 A of current to pass through a 2 m cable in
order to work. If the magnetic field perpendicular to the cable is 3.6 ×10–4 T, what is the magnetic force exerted on the cable?
10. Light bulb filaments can weaken partially because of the magnetic forces
acting on them. If a filament weakens enough, it breaks, and the light
bulb “burns out.” Suppose a current of 0.5 A moves through a 5 cm-long
wire filament and that a magnetic field of 1.3 × 10–4 T acts on the wire.
What is the magnetic force acting on the filament? (Assume that the fila-
ment is a straight wire.)
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Problem 22A Ch. 22–1
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 22AINDUCED EMF AND CURRENT
P R O B L E MA coil with 245 turns and an area of 0.51 m2 is placed 45° to a uniform ex-ternal magnetic field. The coil is 90.0° to the magnetic field 0.63 s later. Ifthe average induced emf is 45 V, what is the strength of the magnetic field?
S O L U T I O N
Given: N = 245 turns A = 0.51 m2 qi = 45°
qf = 90.0° ∆t = 0.63 s emf = 45 V
Unknown: B = ?
Choose the equation(s) or situation: Use Faraday’s law of magnetic induction
to find the induced emf in the coil.
emf = –N ∆[A
∆B
t
(cos q)]
Rearrange the equation(s) to isolate the unknown(s): In this example, only the
angle between the magnetic field and the coil changes with time. The other com-
ponents (the coil area and the magnetic field strength) remain constant.
B = –N
e
A
m
∆f
[c
∆o
t
s q] =
B =
B = = 0.32 T(45 V)(0.63 s)
–(245)(0.51 m2)(– 0.71)
(45 V)(0.63 s)– (245)(0.51 m2)[cos(90.0°) – cos(45°)]
emf ∆t–NA [cos qf – cos qi]
ADDITIONAL PRACTICE
1. A coil with 540 turns and an area of 0.016 m2 is placed parallel to a uniform external
magnetic field. The coil is perpendicular to the magnetic field 0.05 s later. If the average
induced emf is 3.0 V, what is the strength of the magnetic field?
2. A 320-turn coil with an area of 0.068 m2 is placed parallel to a uniform external mag-
netic field. Exactly 0.25 s later, the coil is at right angles to the magnetic field. This in-
duces an average emf of 4.0 V. How strong is the magnetic field?
3. A coil with an area of 0.93 m2 and 628 turns begins parallel to a uniform external mag-
netic field. It takes 0.30 s for the coil to rotate so that it makes an angle of 30.0° with the
magnetic field. If the average induced emf is 62 V, what is the strength of the magnetic
field?
Holt Physics Problem BankCh. 22–2
NAME ______________________________________ DATE _______________ CLASS ____________________
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4. Credit cards have a magnetic tape along one edge of the card. The mag-
netic fields on the tape correspond with the unique number of the card.
When a credit card slides through a credit-card reader, the changes in the
magnetic fields on the tape induce an emf in a coil that are converted to
an electrical signal. Suppose a 550-turn coil with an area of 5.0 × 10–5 m2
is positioned parallel to a magnetic field that changes by 2.5 × 10–4 T in
2.1 × 10–5 s. What is the induced emf in the coil?
5. A VCR plays movies by placing a video head over a magnetic tape. The
changing magnetic fields in the tape induce a current in the coil, which is
converted into an electrical signal, which goes to a television set. Suppose
a VCR has a video head made of a 220-turn coil wrapped around an iron
ring with an area of 6.0 × 10–6 m2. The head is parallel to a magnetic
field that changes by 9.7 × 10–4 T in 1.7 × 10–6 s. What is the induced
emf in the coil?
6. Computers store memory by spinning a stack of magnetic-coated plates
beneath an electromagnetic head. Suppose a head consisting of a 148-
turn coil wrapped around an iron ring with an area of 1.25 × 10–8 m2.
The head is parallel to a magnetic field that changes by 5.2 × 10–4 T in
8.5 × 10–9 s. What is the induced emf in the coil?
7. Suppose an emf of 220 V is induced in a coil that has a resistance of
120 Ω. What is the induced current in the coil?
8. At your job, you use a security card that unlocks certain doors. A mag-
netic tape along the edge of the card induces a current in a coil when you
insert and remove the card from a slot. Suppose the coil designed to read
your card has 180 turns and an area of 5.0 × 10–5 m2. The coil is parallel
to the magnetic field that changes by 5.2 × 10–4 T in 1.9 × 10–5 s. What is
the induced current if the resistance in the coil is 1.0 × 102 Ω?
9. A 246-turn coil with an area of 0.40 m2 is parallel to a magnetic field. Ini-
tially, the magnetic field is 0.237 T, but it increases to 0.320 T in 0.9 s. If
this induces an emf of 9.1 V, what is the time interval necessary to induce
this emf?
10. To measure earthquakes, seismologists suspend a bar magnet (using very
sensitive springs) within a coiled wire housing that is parallel to the mag-
netic field. Suppose a coil has 785 turns of wire and an area of 7.3 ×10–2m2. Suppose an earthquake occurs, causing the magnetic field to
change by 6.9 × 10–3 T. Each time the magnet oscillates (moves up and
down), it induces an emf of 2.8 V. What is the time interval for each os-
cillation? If there are a total of 120 oscillations during this earthquake,
how long did the earthquake last?
Problem 22B Ch. 22–3
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 22BINDUCTION IN GENERATORS
P R O B L E M
A generator is made of a 200-turn coil with an area of 0.040 m2 that ro-tates in a magnetic field of 6.9 × 10–3 T. What is the angular speed neededto produce a maximum emf of 25 V?
S O L U T I O NGiven: maximum emf = 25 V N = 200 turns
A = 0.040 m2 B = 6.9 × 10−3 T
Unknown: w = ?
Choose the equation(s) or situation: Use the maximum emf for a generator.
maximum emf = NABw
Rearrange the equation(s) to isolate the unknown(s): Rearrange the equation
for maximum emf of a generator to solve for the angular frequency.
w = maxim
NA
um
B
emf
Substitute the values into the equation(s) and solve:
w = maxim
NA
um
B
emf = = 4.5 × 102 rad/s
25 V(200)(0.040 m2)((0.0069 T)
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ADDITIONAL PRACTICE
1. A generator consisting of a coil with 220 turns and an area of 0.080 m2
rotates in a magnetic field of 4.8 × 10–3 T. What is the angular speed
needed to produce a maximum emf of 12 V?
2. A generator is made of a 140-turn coil with a diameter of 0.33 m that ro-
tates in a magnetic field of 9.3 × 10–2 T. What is the angular speed needed
to produce a maximum emf of 150 V?
3. A generator is made of a 195-turn coil with an area of 0.052 m2 that ro-
tates in a magnetic field of 3.2 × 10–3 T. What is the angular speed needed
to produce a maximum current of 1.2 A on a 16 Ω load?
4. A generator is made of a 385-turn coil with an area of 0.38 m2 that ro-
tates in a magnetic field of 9.4 × 10–3 T. If the coil rotates at 45 Hz, what
is the maximum emf produced by the generator?
5. A generator produces a maximum current of 14 A on a 5 Ω load. What is
the maximum emf produced?
Holt Physics Problem BankCh. 22–4
NAME ______________________________________ DATE _______________ CLASS ____________________
6. A generator is made of a 119-turn coil with an area of 4.9 × 10–2 m2 that
rotates in a magnetic field of 9.4 × 10–3 T. If the coil rotates at 345 rad/s,
what is the maximum emf produced by the generator?
7. A generator produces a maximum emf of 40 V on an 8 Ω load. What is
the maximum current produced by the generator?
8. A generator is made of a 425-turn coil with an area of 2.16 × 10–2 m2
that rotates in a magnetic field of 3.9 × 10–2 T. If the coil rotates at 33 Hz,
what is the maximum current produced by the generator if the load is
25 Ω?
9. A generator is made of a coil with an area of 1.20 × 10–2 m2 that rotates
in a magnetic field of 6.0 × 10–2 T. The coil rotates at 393 rad/s and in-
duces a maximum emf of 213 V. How many turns of wire are there in the
coil?
10. A generator is made of a coil with an area of 0.60 m2 that rotates in a
magnetic field of 0.012 T. The coil rotates at 44 Hz and induces a maxi-
mum emf of 320 V. How many turns of wire are there in the coil?
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Problem 22C Ch. 22–5
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 22CRMS CURRENTS POTENTIAL DIFFERENCES
P R O B L E MA generator with a rms potential difference of 124 V is connected to a 85 resistor. Calculate the maximum output emf. Find the rms currentand the maximum ac current in the circuit.
S O L U T I O NGiven: ∆Vrms = 124 V R = 85 Ω
Unknown: ∆Vmax = ? Irms = ? Imax = ?
Choose the equation(s) or situation: Use the equation for the rms potential dif-
ference to find the rms potential difference.
∆Vrms = 0.707 ∆Vmax
Use the equation for rms current to find Imax.
Irms = 0.707 Imax
Rearrange the equation(s) to isolate the unknown(s): Rearrange the definition
for resistance to calculate the rms current.
Irms = ∆V
Rrms
Rearrange the equation relating rms current to maximum current so that maxi-
mum current is calculated.
Imax = 0
I
.
r
7
m
0
s
7
Rearrange the equation relating rms potential difference to maximum potential
difference so that maximum potential difference is calculated.
∆Vmax = ∆0
V
.7r
0m
7s
Substitute the values into the equation(s) and solve:
∆Vmax = ∆0
V
.7r
0m
7s =
1
0
2
.7
4
0
V
7 =
Irms = ∆V
Rrms =
1
8
2
5
4
ΩV
=
Imax = 0
I
.r
7m
0s
7 =
0
2
.
.
7
1
0
A
7 =
Evaluate: The maximum values for potential difference and current are a little
less than three-halves (1.5 times) the rms values, as expected.
3.0 A
2.1 A
175 V
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Holt Physics Problem BankCh. 22–6
NAME ______________________________________ DATE _______________ CLASS ____________________
1. Suppose you build a hydroelectric turbine in your back yard along a
river. The turbine is connected to a generator which can provide a rms
potential difference of 320 V onto a 100 Ω load. Calculate the maximum
output emf. Find the rms current and the maximum ac current in the
circuit.
2. Some wind turbines in the United States can provide an rms current of
1.3 A. Calculate the maximum ac current.
3. In Iceland, people harness geothermal energy by placing pipes deep into
the Earth in order to tap hot-water aquifers. Steam from these aquifers
spins a turbine that is connected to a generator. Suppose such a generator
can provide a rms current of 2.5 A and an rms potential difference of
22 kV. Calculate the maximum ac current. Find the resistance of the
load.
4. In Denmark, farmers and people living in rural areas harness wind en-
ergy to spin a turbine that is coupled to a generator. The generator can
provide an rms current of 1.7 A and an rms potential difference of 220 V.
Calculate the maximum ac current and the maximum output emf.
5. A generator can provide a maximum ac current of 1.2 A and a maximum
output emf of 211 V. Calculate the rms current, the rms potential differ-
ence, and the resistance in the circuit.
6. A generator can provide a maximum output emf of 170 V. Calculate the
rms potential difference.
7. Suppose you measure a circuit with a voltmeter. You measure a rms po-
tential difference of 115 V and a resistance of 50.0 Ω. Find the maximum
ac current and the rms current in the circuit.
8. Off the western coast of Scotland, turbines are placed in deep ocean
water so that ocean waves can spin them. The turbines are connected to a
generator, which produces electricity for the locals. These turbines can
supply an rms current of 2.1 A on a 16 kΩ load. Calculate the maximum
ac current. How much power is supplied?
9. A wind turbine is built to service farmers in England. The turbine is con-
nected to a generator, which can supply a rms current of 1.3 A on a
12 kΩ load. If the rms potential difference is 15.6 kV, find the maximum
output emf provided by the wind turbine. Calculate the maximum ac
current. How much power is supplied?
10. The total world potential for hydroelectric generation is 1.2 million MW,
however, not all of it has been harnessed. Suppose the world’s rms cur-
rent is 2.2 × 1010 A and it supplies a load of 6.1 × 10–10 Ω. Calculate the
total maximum ac current that is harnessed. How much total power can
be supplied with this current?
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ADDITIONAL PRACTICE
Problem 22D Ch. 22–7
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 22DTRANSFORMERS
P R O B L E MAn ac generator at a central power station uses a step-up transformer toprovide a potential difference of 68 kV across the secondary coil. If theprimary coil has 125 turns and the secondary coil has 625 turns, what isthe potential difference across the primary coil?
S O L U T I O N
Given: ∆V2 = 68 kV = 6.8 × 104 V N1 = 125 turns
N2 = 625 turns
Unknown: ∆V1 = ?
Choose the equation(s) or situation: Use the transformer equation.
∆∆
V
V1
2 =
N
N
2
1
Rearrange the equation(s) to isolate the unknown(s): Rearrange the equation
for a transformer to solve for the potential difference in the primary coil.
∆V1 = ∆V
N2
2
N1
Substitute the values into the equation(s) and solve:
∆V1 = ∆V
N2
2
N1 = (6.8 × 1
(
0
6
4
25
V
)
)(125) =
Evaluate: The potential difference across the primary should be 14 kV. The step-
up factor for the transformer is 1:5.
1.4 × 104 V = 14 kV
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ADDITIONAL PRACTICE
1. A transmission line to a city has a potential difference of 6.9 kV across
the secondary coil. If the primary coil has 1400 turns and the secondary
coil has 140 turns, what is the potential difference across the primary
coil?
2. A power line has a potential difference of 3.4 kV across the secondary
coil. If the primary coil has 9.0 × 101 turns and the secondary coil has
2250 turns, what is the potential difference across the primary coil?
3. A transmission line to a city has a potential difference of 46 kV across the
primary coil. If the primary coil has 1250 turns and the secondary coil
has 250 turns, what is the potential difference across the secondary coil?
4. A high-voltage cable has a potential difference of 5.6 kV across the pri-
mary coil. If the primary coil has 140 turns and the secondary coil has
840 turns, what is the potential difference across the secondary coil?
Holt Physics Problem BankCh. 22–8
NAME ______________________________________ DATE _______________ CLASS ____________________
5. A high-voltage cable has a potential difference of 9.2 kV across the pri-
mary coil. If the primary coil has 120 turns and the secondary coil has
1200 turns, what is the potential difference across the secondary coil?
6. A ac power station has a potential difference of 36 kV across the primary
coil and a potential difference of 7.2 kV across the secondary coil. If the
primary coil has 55 turns, how many turns does the secondary coil have?
7. The plug to your zip drive has a step-down transformer in it. The poten-
tial difference across the primary coil is 240 V and a potential difference
of 5.0 V across the secondary. What is the step-down ratio? (Hint: the
step-down ratio is the ratio of N1:N2.)
8. A transmission line has a potential difference of 3.6 kV across the pri-
mary coil and a potential difference of 1.8 kV across the secondary coil. If
the primary coil has 58 turns, how many turns does the secondary coil
have?
9. A central ac power station has a potential difference of 49 kV across the
primary coil and a potential difference of 4.9 kV across the secondary
coil. If the secondary coil has 480 turns, how many turns does the pri-
mary coil have?
10. A ac generator central power station can produce 1380 kW of power. The
secondary coil has a potential difference of 3.4 kV. If the primary coil has
340 turns, and the secondary coil has 17 turns, what is the potential dif-
ference across the primary coil? What is the current in the primary coil?
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Problem 23A Ch. 23–1
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 23AQUANTUM ENERGY
P R O B L E MFind the energy of a photon of red light, having a wavelength of 690 nm.
S O L U T I O N
Given: l = 6.90 × 10–7 m
Unknown: f = ? E = ?
Choose the equation(s) or situation: Use the equation for the energy of a light
quantum, given on page 832.
E = hf
Use the equation relating frequency to wavelength, given on page 522.
c = lf
Rearrange the equation(s) to isolate the unknown(s): Rearrange the equation
to solve for the frequency.
f = lc
= 6
3
.
.
9
0
0
××1
1
0
0
8
–m7 m
/s = 4.3 × 1014 Hz
E = hf = = 1.8 MeV(6.63 × 10–34 J • s)(4.3 × 1014 Hz)
1.60 × 10–19 J/eV
ADDITIONAL PRACTICE
1. Determine the energy of a photon of green light, having a wavelength of 527 nm.
2. What is the energy of a photon of blue light, having a wavelength of 430.8 nm?
3. Calculate the frequency of ultraviolet (UV) light, having the energy of 20.7 eV.
4. Microwave ovens work by vibrating the water molecules in food, causing the food to
warm up. Microwaves can reach energies as high as 1.24 × 10–3 eV. At what frequency is
this?
5. Calculate the frequency of infrared (IR) light, having the energy of 1.78 eV.
6. An X ray can have an energy of 12.4 MeV. To what wavelength does this correspond?
7. A neutron has 939.57 MeV of energy. If a photon had the same energy as a neutron,
what would be the photon’s wavelength? The visible part of the spectrum ranges from
700 nm – 400 nm. Would this wavelength lie within the visible spectrum?
8. Calculate the wavelength of a radio wave that has an energy of 3.1 × 10–6 eV.
Holt Physics Problem BankCh. 23–2
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 23BPHOTOELECTRIC EFFECT
P R O B L E M
Light of wavelength 3.5 × 10–7 m shines on a cesium surface. Cesium has awork function of 2.14 eV. What is the maximum kinetic energy of thephotoelectrons?
S O L U T I O N
Given: l = 3.5 × 10–7 m hft = 2.14 eV
Unknown: KEmax = ?
Choose the equation(s) or situation: Use the equation for the maximum kinetic
energy of a photoelectron, given on page 835.
KEmax = hf – hft = h
lc – hft
KEmax = – 2.14 eV
KEmax = 1.41 eV
(6.63 × 10–34 J • s)(3.0 × 108 m/s)(1.60 × 10–19 J/eV)(3.5 × 10–7 m)
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1. Light of wavelength 240 nm shines on a potassium surface. Potassium
has a work function of 2.3 eV. What is the maximum kinetic energy of
the photoelectrons?
2. Light of wavelength 519 nm shines on a rubidium surface. Rubidium has
a work function of 2.16 eV. What is the maximum kinetic energy of the
photoelectrons?
3. Light of frequency 6.5 × 1014 Hz illuminates a lithium surface. The
ejected photoelectrons are found to have a maximum kinetic energy of
0.20 eV. Find the threshold frequency of this metal.
4. Light of frequency 9.89 × 1014 Hz illuminates a calcium surface. The
ejected photoelectrons are found to have a maximum kinetic energy of
0.90 eV. Find the threshold frequency of this metal.
5. The threshold frequency of platinum is 1.36 × 1015 Hz. What is the work
function of platinum?
6. The threshold frequency of copper is 1.1 × 1015 Hz. What is the work
function of copper?
7. Manganese has a work function of 4.1 eV. What is the wavelength of the
photon that will just have the threshold energy for manganese?
ADDITIONAL PRACTICE
Problem 23B Ch. 23–3
NAME ______________________________________ DATE _______________ CLASS ____________________
8. Cobalt has a work function of 5.0 eV. What is the wavelength of the pho-
ton that will just have the threshold energy for cobalt?
9. Light shines on a photoelectric metal and the maximum kinetic energy is
measured to be 0.6 eV. What is the speed of the photoelectrons?
10. Light shines on a photoelectric metal and the maximum kinetic energy is
measured to be 1.2 eV. What is the speed of the photoelectrons?
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Holt Physics Problem BankCh. 23–4
NAME ______________________________________ DATE _______________ CLASS ____________________
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Holt Physics
Problem 23CDE BROGLIE WAVES
P R O B L E MA grain of sand blows along a seashore at a velocity of 5.2 m/s. If it has a deBroglie wavelength of 5.8 × 10–29 m, what is the mass of the sand grain?
S O L U T I O N
Given: v = 5.2 m/s l = 5.8 × 10–32 m
Unknown: m = ?
Choose the equation(s) or situation: Use the equation for the de Broglie wave-
length, given on page 849.
l = m
h
v
Rearrange the equation(s) to isolate the unknown(s): Rearrange the equation
relating wavelength, mass, and velocity to solve for mass.
m = lh
v = = 2.2 × 10–6 kg
6.63 × 10–34 J • s(5.2 m/s)(5.8 × 10–29 m)
ADDITIONAL PRACTICE
1. A cheetah can run as fast as 28 m/s. If the cheetah has a de Broglie wavelength of
8.97 × 10–37 m, what is the cheetah’s mass?
2. A Boeing 747 jet airliner has a maximum airspeed of 7.1 × 102 m/s. If the airliner has a
de Broglie wavelength of 5.8 × 10–42 m, what is the mass of the jet?
3. The smallest known virus is a potato spindle. Suppose a potato spindle moves across a
Petri dish at 5.6 × 10–6 m/s and has a de Broglie wavelength of 2.96 × 10–8 m. What is
the mass of a potato spindle?
4. Suppose a raindrop falls from the sky at a velocity of 12 m/s and has a de Broglie wave-
length of 2.6 × 10–29 m. What is the mass of the raindrop?
5. Calculate the de Broglie wavelength of an electron orbiting the hydrogen atom at a
velocity of 2.19 × 106 m/s.
6. The ship Queen Elizabeth has a mass of 7.6 × 107 kg. Calculate the de Broglie wave-
length if this ship sails at 35 m/s.
7. Earth has a mass of 5.94 × 1024 kg and orbits the sun at a velocity of 3.0 × 104 m/s.
Calculate Earth’s de Broglie wavelength.
8. Our solar system is within the Milky Way galaxy. Astronomers estimate that our galaxy
has a mass of 4.0 × 1041 kg. Calculate the de Broglie wavelength of our galaxy if it were
to move at a velocity of 1.7 × 104 m/s.
9. What is the speed of an electron with a de Broglie wavelength of 9.87 × 10–14 m?
10.What is the speed of a neutron with a de Broglie wavelength of 5.6 × 10–14 m?
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Problem 25A Ch. 25-1
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 25ABINDING ENERGY
P R O B L E MGiven that the atomic mass of iron –55 is 54.938 297 u, calculate the bind-ing energy of 55
26Fe.
S O L U T I O NGiven: Z = 26 atomic mass of iron −55 = 54.938 297 u
N = 29 atomic mass of H = 1.007 825 u
mn = 1.008 665 u
Unknown: Ebind = ?
Choose the equation(s) or situation:
First find the mass defect with the following relationship:
∆m = Z(atomic mass of H) + Nmn − atomic mass
Then find the binding energy by converting the mass defect to rest energy.
Substitute the values into the equation(s) and solve:
∆m = 26(1.007 825 u) + 29(1.008 665 u) − 54.938 297 u
∆m = 0.51644 u
Ebind = (0.51644 u)(931.50 MeV/u) = 481.06 MeV
ADDITIONAL PRACTICE
1. Calculate the binding energy of 3919K.
2. Compute the binding energy of 12050Sn.
3. Determine the difference in the binding energy of 10747Ag and 63
29Cu.
4. Calculate the difference in the binding energy of 126C and 16
8O.
5. What is the binding energy of 3517Cl?
6. Calculate the binding energy of deuterium, 21H.
7. Find the mass defect of 5828Ni.
8. What is the mass defect of 6430Zn?
9. Calculate the binding energy of 9040Zr.
10. Find the mass defect of 3216S.
Holt Physics
Problem 25BNUCLEAR DECAY
P R O B L E MBromine-80 decays by emitting a positron and a neutino. Write the com-plete decay formula for this process.
S O L U T I O NGiven: The decay can be written symbolically as follows:
8035Br → X + 00
1e + v
Unknown: the daughter element (X)
The mass numbers and atomic numbers on the two sides of the expression must
be the same so that both charge and nucleon number are conserved during the
course of a particular decay.
Mass number of X = 80 − 0 = 80
Atomic number of X = 35 − (1) = 348035Br → 80
34X + 01e + v
The periodic table (Appendix F) shows that the nucleus with an atomic number
of 34 is selenium, Se. Thus, the process is as follows:
8035Br → 80
34Se + 01e + v
Holt Physics Problem BankCh. 25-2
NAME ______________________________________ DATE _______________ CLASS ____________________
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ADDITIONAL PRACTICE
1. Complete this radioactive-decay formula: 21084Po → ? + 42He
2. Complete this radioactive-decay formula: 167N → ? + –1
0e + v
3. Complete this radioactive-decay formula: 14762Sm → 143
60Nd + ?
4. Complete this radioactive-decay formula: 1910Ne → ? + 01e + v
5. Complete this radioactive-decay formula: ? → 13154Xe + –1
0e + v
6. Complete this radioactive-decay formula: ? → 9039Y + –1
0e + v
7. Complete this radioactive-decay formula: 16074W → 156
72Hf + ?
8. Complete this radioactive-decay formula: ? → 10752Te + 42He
9. Complete this radioactive-decay formula: 15772Hf → 153
70Yb + ?
10. Complete this radioactive-decay formula: 14158Ce → ? + –1
0e + v
Problem 25C Ch. 25-3
NAME ______________________________________ DATE _______________ CLASS ____________________
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Holt Physics
Problem 25CMEASURING NUCLEAR DECAY
P R O B L E MSuppose you start with 9.36 × 10–3 g of a pure radioactive substance anddetermine 4 h later that only 2.34 × 10–3 g of the substance is left unde-cayed. What is the half-life of this substance?
S O L U T I O N
Given: mi = 9.36 × 10–3 g mf = 2.34 × 10–3 g ∆t = 4 h
Unknown: T1⁄2 = ?
Choose the equation(s) or situation: Take the ratio of the final mass to the ini-
tial mass to find what fraction of the sample remains.
fraction of remaining sample = m
m
i
f = 2
9
.
.
3
3
4
6
××
1
1
0
0
–
–
3
3g
g =
1
4
If one-quarter of the sample remains after 4 h, then half of the sample must have
remained after 2.00 h, or T1⁄2 = .2.00 h
ADDITIONAL PRACTICE
1. Suppose you start with 5.25 × 10–3 g of a pure radioactive substance and
determine 12 h later that only 3.28 × 10–4 g of the substance is left unde-
cayed. What is the half-life of this substance?
2. Suppose you start with 3.29 × 10–3 g of a pure radioactive substance and
determine 30.0 s later that only 8.22 × 10–4 g of the substance is left un-
decayed. What is the half-life of this substance?
3. Suppose you start with 4.14 × 10–4 g of a pure radioactive substance and
determine 1.25 days later that only 2.07 × 10–4 g of the substance is left
undecayed. What is the half-life of this substance?
4. How long will it take a sample of lead-212 with a half-life of 10.64 h to
decay to one-eighth its original strength?
5. How long will it take a sample of cadmium-109 with a half-life of 462
days to decay to one-fourth its original strength?
6. The half-life of 5526Fe is 2.7 years. An iron-55 sample contains 3.2 × 109
nuclei. What is the decay constant for the decay?
7. A sample of a radioactive isotope is measured to have an activity of
765.3 mCi. If the sample has a half-life of 22 h, how many nuclei of the
isotope are there at this time?
Holt Physics Problem BankCh. 25-4
NAME ______________________________________ DATE _______________ CLASS ____________________
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8. The half-life of 4824Cr is 21.6 h. A chromium-48 sample contains 6.5 × 106 nuclei. Calcu-
late the decay constant and the activity of the sample in mCi.
9. The half-life of 31H is 12.33 years. A tritium sample contains 4.8 × 108 nuclei. What is
the decay constant for the decay?
10. A sample of a radioactive isotope is measured to have an activity of 360.0 mCi. If the
sample has a half-life of 17.2 s, how many nuclei of the isotope are there at this time?
1ChapterThe Science of Physics
1. mass = 6.0 × 103 kga. mass = 6.0 × 103 kg ×
1
1
0
k
3
g
g ×
1
1
0
m−3
g
g =
b. mass = 6.0 × 103 kg × 1
1
0
k
3
g
g ×
1
10
M6 g
g = 6.0 Mg
6.0 × 109 mg
Additional Practice 1A
Givens Solutions
2. Volume = 6.4 × 104 cm3
a. volume = 6.4 × 104 cm3 × 10
12m
cm
3= 6.4 × 104 cm3 ×
10
16m
cm
3
3 =
b. volume = 6.4 × 104 cm3 × 11
0
c
m
m
m
3= 6.4 × 104 cm3 ×
10
1
3
c
m
m
m3
3
= 6.4 = 107 mm3
6.4 × 10−2 m3
3. energy = 4.2 × 109 J a. energy = 4.2 × 109 J × 1
10
M6 J
J =
b. energy = 4.2 × 109 J × 1
1
0
G9
J
J = 4.2 GJ
4.2 × 103 MJ
4. distance = 1 parsec = 3.086 × 1016 m a. distance = 1 parsec = 3.086 × 1016 m ×
1
1
0
k3m
m =
b. distance = 1 parsec = 3.086 × 1016 m × 1
1
01E8m
m = 3.086 × 10–2 Em
3.086 × 1013 km
5. area = 1 acre = 4.0469 × 103 m2 a. area = 1 acre = 4.0469 × 103 m2 × 1
1
0
k3m
m2
area = 4.0469 × 103 m2 × 1
1
0
k6m
m
2
2 =
b. area = 1 acre = 4.0469 × 103 m2 × 10
1
2
m
cm
2
area = 4.0469 × 103 m2 × 10
1
4
m
cm2
2
= 4.0469 × 107 cm2
4.0469 × 10−3 km2
6. electric charge = 15 Ca. electric charge = 15 C ×
10
1
3
C
mC =
b. electric charge = 15 C × 1
1
0
k3C
C = 1.5 × 10–2 kC
1.5 × 104 mC
Section Two—Problem Workbook Solutions V Ch. 1–1
V
Holt Physics Solution ManualV Ch. 1–2
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7. depth = 1.168 × 103 cma. depth = 1.168 × 103 cm ×
10
12m
cm =
b. depth = 1.168 × 103 cm × 10
12m
cm ×
1
1
0−m6
m
m = 1.168 × 107 mm
1.168 × 101 m = 11.68 m
Givens Solutions
8. area = 0.344 279 km2
a. area = 0.344 279 km2 × 110
k
3
m
m
2= 0.344 279 × 106 m2 =
b. area = 0.344 279 km2 × 110
k
3
m
m
2× 10
1
3
m
mm
2= 0.344 279 × 1012 mm2
area = 3.442 79 × 1011 mm2
3.442 79 × 105 m3
9. time = 4.50 × 109 years ×
365
1
.2
y
5
ea
d
r
ays ×
1
24
da
h
y ×
36
1
0
h
0 s
= 1.42 × 1017 s
a. time = 1.42 × 1017 s × 1
1
0
G9
s
s =
b. time = 1.42 × 1017 s × 1
1
01P5s
s = 1.42 × 102 Ps = 142 Ps
1.42 × 108 Gs
10. time = 6.7 × 10−17 s a. time = 6.7 × 10−17 s × 10
1
6
s
ms =
b. time = 6.7 × 10−17 s × 10
1
18
s
as = 6.7 = 101 as = 67 as
6.7 × 10−11 ms
Section Five—Problem Bank V Ch. 2–1
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2ChapterMotion in One Dimension
1. ∆x = 3.33 km forward∆t = 30.0 s vavg =
∆∆
x
t =
3.33
3
×0.
1
0
0
s
3 m =
vavg = (111 m/s)(3600 s/h)(10−3 km/m) = 4.00 × 102 km/h forward
111 m/s forward
Additional Practice 2A
Givens Solutions
2. ∆x = 15.0 km west∆t = 15.3 s
vavg = ∆∆
x
t = =
4.2
1
5
5
×.0
1
k
0
m−3 h
= 3.53 × 103 km/h west15.0 km
(15.3 s)36
1
0
h
0 s
3. ∆x = 4.0 m∆t = 5.0 min vavg =
∆∆
x
t = = 48 m/h
4.0 m
(5.0 min)60
1
m
h
in
4. ∆x = 3.20 × 104 km south∆t = 122 days vavg =
∆∆
x
t =
3.2
1
0
2
×2
1
d
0
a
4
ys
km = 262 km/day south
5. ∆x1 = 1.70 × 104 km south= +1.70 × 104 km
∆x2 = 6.0 × 102 km north= −6.0 × 102 km
∆x3 = 1.44 × 104 km south= +1.44 × 104 km
∆t = 122 days
d = total distance traveled = magnitude ∆x1 + magnitude ∆x2 + magnitude ∆x3
d = 1.70 × 104 km + 6.0 × 102 km + 1.44 × 104 km
d = (1.70 + 0.060 + 1.44) × 104 km
d = 3.20 × 104 km
average speed = ∆d
t =
3.2
1
0
2
×2
1
d
0
a
4
ys
km =
vavg = ∆∆xt
tot =
∆x1 + ∆∆x
t2 + ∆x3
vavg =
vavg =
vavg = 3.0
1
8
2
×2
1
d
0
a
4
ys
km = +252 km/day = 252 km/day south
(1.70 − 0.060 + 1.44) × 104 km
122 days
(1.70 × 104 km) + (−6.0 × 102 km) + (1.44 × 104 km)
122 days
262 km/day
6. ∆x1 = 20.0 km east = + 20.0 km
∆x2 = 20.0 km west = − 20.0 km
∆x3 = 0 km
∆x4 = 40.0 km east = +40.0 km
∆t = 60.0 min
a. vavg = =
vavg =
vavg = = + 40.0 km/h = 40.0 km/h east40.0 km
(60.0 min) 60
1
m
h
in
(20.0 km) + (−20.0 km) + (0 km) + (40.0 km)
60.0 min
∆x1 + ∆x2 + ∆x3 + ∆x4∆t
∆xtot∆t
Holt Physics Solution ManualV Ch. 2–2
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b. d = total distance traveled
d = magnitude ∆x1 + magnitude ∆x2 + magnitude ∆x3 + magnitude ∆x4
d = 20.0 km + 20.0 km + 0 km + 40.0 km = 80.0 km
average speed = = = 80.0 km/h80.0 km
(60.0 min)60
1
m
h
in
d∆t
Givens Solutions
7. v = 89.5 km/h north
vavg = 77.8 km/h north
∆trest = 22.0 min
∆x = vavg ∆t = v(∆t − ∆trest)
∆t(vavg − v) = −v∆trest
= =
∆t = 2.80 h = 2 h, 48 min
(89.5 km/h)(22.0 min)60
1
m
h
in
11.7 km/h
(89.5 km/h)(22.0 min)60
1
m
h
in
89.5 km/h − 77.8 km/h
∆t = v∆restv − vavg
8. v = 6.50 m/s downward = −6.50 m/s
∆t = 34.0 s
∆x = v∆t = (−6.50 m/s)(34.0 s) = −221 m = 221 m downward
9. vt = 10.0 cm/s
vh = 20 vt = 2.00 × 102 cm/s
∆trace = ∆tt
∆th = ∆tt − 2.00 min
∆xt = ∆xh + 20.0 cm = ∆xrace
∆xt = vt∆tt
∆xh = vh∆th = vh (∆tt − 2.00 min)
∆xt = ∆xrace = ∆xh + 20.0 cm
vt ∆tt = vh (∆tt − 2.00 min) + 20.0 cm
∆tt (vt − vh) = −vh (2.00 min) + 20.0 cm
∆tt =
∆trace = ∆tt =
∆trace = =
∆trace = 126 s
−2.40 × 104 cm−1.90 × 102 cm/s
20.0 cm − 2.40 × 104 cm
−1.90 × 102 cm/s
20.0 cm − (2.00 × 102 cm/s)(2.00 min)(60 s/min)
10.0 cm/s − 2.00 × 102 cm/s
20.0 cm − vh (2.00 min)
vt − vh
10. ∆xrace = ∆xt
vt = 10.0 cm/s
∆tt = 126 s
∆xrace = ∆xt = vt∆tt = (10.0 cm/s)(126 s) = 1.26 × 103 cm = 12.6 m
Additional Practice 2B
1. ∆t = 6.92 s
vf = 17.34 m/s
vi = 0 m/s
aavg = ∆∆
v
t =
vf
∆−t
vi = = 2.51 m/s217.34 m/s − 0 m/s
6.92 s
2. vi = 0 m/s
vf = 7.50 × 102 m/s
∆t = 2.00 min
aavg = = = = 6.25 m/s27.50 × 102 m/s − 0 m/s
(2.00 min)16
m
0
i
s
n
vf − vi∆t
∆v∆t
Section Five—Problem Bank V Ch. 2–3
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Givens Solutions
3. vi = 0 m/s
vf = 0.85 m/s forward
∆t = 3.7 s
aavg = = = = 0.23 m/s2 forward0.85 m/s − 0 m/s
3.7 s
vf − vi∆t
∆v∆t
4. vi = 13.7 m/s forward = +13.7 m/s
vf = 11.5 m/s backward = −11.5 m/s
∆t = 0.021 s
aavg = = = =
aavg = −1200 m/s2, or 1200 m/s2 backward
−25.2 m/s
0.021 s
(− 11.5 m/s) − (13.7 m/s)
0.021 s
vf − vi∆t
∆v∆t
5. vi = +320 km/h
vf = 0 km/h
∆t = 0.18 saavg = = =
aavg = = −490 m/s2−89 m/s
0.18 s
(0 km/h − 320 km/h)36
1
0
h
0 s11
0
k
3
m
m
0.18 s
vf − vi∆t
∆v∆t
6. aavg = 16.5 m/s2
vi = 0 km/h
vf = 386.0 km/h∆t = = =
∆t = = 6.50 s107.2 m/s16.5 m/s2
(386.0 km/h − 0 km/h)36
1
0
h
0 s11
0
k
3
m
m
16.5 m/s2
vf − viaavg
∆vaavg
7. vi = −4.0 m/s
aavg = −0.27 m/s2
∆t = 17 s
vf = aavg ∆t + vi
vf = (−0.27 m/s2)(17 s) + (−4.0 m/s) = −4.6 m/s − 4.0 m/s = −8.6 m/s
8. vi = 4.5 m/s
vf = 10.8 m/s
aavg = 0.85 m/s2
∆t = a
∆
a
v
vg = =
10.8
0
m
.8
/
5
s −m
4
/s
.52
m/s =
0
6
.8
.3
5
m
m
/
/
s
s2 = 7.4 svf − viaavg
9. vf = 296 km/h
vi = 0 km/h
aavg = 1.60 m/s2∆t =
a
∆
a
v
vg = = =
1
8
.
2
6
.
0
2
m
m
/
/
s
s2 = 51.4 s
(296 km/h − 0 km/h)36
1
0
h
0 s11
0
k
3
m
m
1.60 m/s2
vf − viaavg
10. aavg = − 0.87 m/s2
∆t = 3.85
∆v = aavg ∆t = (−0.87 m/s2)(3.85 s) = –3.35 m/s
1. ∆t = 0.910 s
∆x = 7.19 km
vi = 0 km/s
vf = 2
∆∆t
x − vi =
(2)
0
(7
.9
.1
1
9
0
k
s
m) − 0 km/s = 15.8 km/s
Additional Practice 2C
Givens Solutions
Holt Physics Solution ManualV Ch. 2–4
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2. vi = 4.0 m/s
∆t = 18 s
∆x = 135 m
vf = 2
∆∆t
x − vi =
(2)(
1
1
8
35
s
m) − 4.0 m/s = 15 m/s − 4.0 m/s =
vf = (11 m/s)36
1
0
h
0 s
1
1
0
k3m
m
vf = 4.0 × 101 km/h
11 m/s
3. ∆x = 55.0 m
∆t = 1.25 s
vf = 43.2 m/s
vi = 2
∆∆t
x − vf =
(2)
1
(5
.2
5
5
.0
s
m) − 43.2 m/s = 88.0 m/s − 43.2 m/s = 44.8 m/s
4. ∆x = 38.5 m
∆t = 5.5 s
vi = 0 m/s
vf = 2
∆∆t
x − vi =
(2)(
5
3
.
8
5
.5
s
m) − 0 m/s = 14 m/s
5. ∆x = 478 km
∆vi = 72 km/h
∆t = 5 h, 39 min
vf = 2
∆∆t
x − vi = − 72 km/h =
(
5
2
h
)(4
+7
0
8
.6
k
5
m
h
) − 72 km/h
vf = 9
5
5
.
6
65
km
h − 72 km/h = 169 km/h − 72 km/h = 97 km/h
(2)(478 km)
5 h + 39 min60
1
m
h
in
6. ∆x = 4.2 m
∆t = 3.0 s
vf = 1.3 m/s
vi = 2
∆∆t
x − vf =
(2)
3
(4
.0
.2
s
m) − 1.3 m/s = 2.8 m/s − 1.3 m/s = 1.5 m/s
7. vi = 25 m/s west
vf = 35 m/s west
∆x = 250 m west
∆t = v
2
i
∆+
x
vf =
25
(
m
2)
/
(
s
2
+50
35
m
m
)
/s =
6
5
.0
.0
××1
1
0
01
2
m
m
/s = 8.3 s
8. vi = 755.0 km/h
vf = 777.0 km/h
∆t = 63.21 s
∆x = 1
2 (vi + vf)∆t =
1
2 (755.0 km/h + 777.0 km/h)(63.21 s)36
1
0
h
0 s
∆x = 1
2 (1532.0 km/h)(1.756 × 10−2 h) = 13.45 km
9. vi = 0 m/s
vf = 30.8 m/s
∆x = 493 m
∆t = v
2
i
∆+
x
vf =
0 m
(2
/s
)(
+49
3
3
0.
m
8 m
)
/s =
3
9
0
8
.8
6
m
m
/s = 32.0 s
10. ∆x = 1220 km
vi = 11.1 km/s
vf = 11.7 km/s
∆t = v
2
i
∆+
x
vf = =
2
2
2
4
.
4
8
0
k
k
m
m
/s = 107 s
(2)(1220 km)11.1 km/s + 11.7 km/s
Section Five—Problem Bank V Ch. 2–5
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.Givens Solutions
2. ∆t = 1.5 s
vi = 2.8 km/h
vf = 32.0 km/ha = =
a = = 5.4 m/s2
(29.2 km/h)36
1
0
h
0 s11
0
k
3
m
m
1.5 s
(32.0 km/h − 2.8 km/h)36
1
0
h
0 s10
k
3
m
m
1.5 s
vf − vi∆t
3. ∆x = 18.3 m
∆t = 2.74 s
vi = 0 m/s
Because vi = 0 m/s, a = 2
∆∆t
x2 =
(2
(
)
2
(
.
1
7
8
4
.3
s)
m2
) = 4.88 m/s2
4. vi = 2.3 m/s
vf = 46.7 m/s
∆t = 7.0 s
a = = 46.7 m/
7
s
.0
−s
2.3 m/s =
44
7
.4
.0
m
s
/s = 6.3 m/s2vf − vi
∆t
5. vi = 6.23 m/s
∆x = 255 m
∆t = 82 s
a = 2(∆x
∆−t2vi ∆t) =
a = = (
6
2
.
)
7
(−×2
1
5
0
53m
s2)
= −7.6 × 10−2 m/s2(2)(255 m − 510 m)
6.7 × 103 s2
(2)[255 m − (6.23 m/s)(82 s)]
(82 s)2
6. vi = 11 km/h
vf = 55 km/h
∆ = 4.1 sa = =
a = = 3.0 m/s2
(44 km/h)36
1
0
h
0 s11
0
k
3
m
m
4.1 s
(55 km/h − 11 km/h)36
1
0
h
0 s11
0
k
3
m
m
4.1 s
vf − vi∆t
7. vi = 42.0 m/s southeast
∆t = 0.0090 s
∆x = 0.020 m/s southeast
a = 2(∆x
∆−t2vi ∆t) =
a = =
a = −8.9 × 103 m/s2, or 8.9 × 103 m/s2 northwest
(2)(−0.36 m)8.1 × 10−5 s2
(2)(0.020 m/s − 0.38 m)
8.1 × 10−5 s2
(2)[0.020 m − (42.0 m/s)(0.0090 s)]
(0.0090 s)2
8. ∆t = 28 s
a = 0.035 m/s2
vi = 0.76 m/s
vf = a∆t + vi = (0.035 m/s2)(28.0 s) + 0.76 m/s = 0.98 m/s + 0.76 m/s = 1.74 m/s
Additional Practice 2D
1. ∆x = 12.4 m upward
∆t = 2.0 s
vi = 0 m/s
Because vi = 0 m/s, a = 2
∆∆t2x
= (2)
(
(
2
1
.0
2.
s
4
)2m)
= 6.2 m/s2 upward
Givens Solutions
Holt Physics Solution ManualV Ch. 2–6
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4. ∆x = 2.00 × 102 m
vi = 9.78 m/s
vf = 10.22 m/s
a = vf
2
2∆−
x
vi2
= =
a = 4.
8
0
.
0
8
×m
1
2
0
/2s2
m = 2.2 × 10−2 m/s2
104.4 m2/s2 − 95.6 m2/s2
4.00 × 102 m
(10.22 m/s)2 − (9.78 m/s)2
(2)(2.00 × 102 m)
10. vi = +4.42 m/s
vf = 0 m/s
a = −0.75 m/s2
∆t = 5.9 s
a. ∆t = vf −
a
vi = 0 m
−/
0
s
.7
−5
4
m
.42
/s2m/s
= −−0
4
.
.
7
4
5
2
m
m
/
/
s
s2 =
b. ∆x = vi∆t + 1
2a∆t2 = (4.42 m/s)(5.9 s) +
1
2(−0.75 m/s2)(5.9 s)2
∆x = 26 m − 13 m = 13 m
5.9 s
1. vi = 1.8 km/h
vf = 24.0 km/h
∆x = 4.0 × 102 ma =
vf2
2∆−
x
vi2
=
a =
a = = 5.5 × 10−2 m/s2(573 km2/h2)36
1
0
h
0 s
2
1
1
0
k
3
m
m
2
8.0 × 102 m
(576 km2/h2 − 3.2 km2/h2)36
1
0
h
0 s
2
1
1
0
k
3
m
m
2
8.0 × 102 m
[(24.0 km/h)2 − (1.8 km/h)2]36
1
0
h
0 s
2
1
1
0
k
3
m
m
2
(2)(4.0 × 102 m)
2. vf = 0 m/s
vf = 8.57 m/s
∆x = 19.53 m
a = vf
2
2
−∆x
vi2
= = 7
3
3
9
.4
.0
m
6
2
m
/s2
= 1.88 m/s2(8.57 m/s)2 − (0 m/s)2
(2)(19.53 m)
9. vi = 0 m/s
vf = 72.0 m/s north
a = 1.60 m/s2 north
∆t = 45.0 s
a. ∆t = vf −
a
vi = 72.0
1.
m
60
/s
m
−/
0
s2m/s
=
b. ∆x = vi∆t + 1
2a∆t2 = (0 m/s)(45.0 s) +
1
2 (1.60 m/s2)(45.0 s)2 = 0 m + 1620 m
∆x = 1.62 km
45.0 s
Additional Practice 2E
3. vi = 7.0 km/h
vf = 34.5 km/h
∆x = 95 ma =
vf2
2
−∆x
vi2
=
a =
a = = 0.46 m/s2
(1140 km2/h2)36
1
0
h
0 s
2
1
1
0
k
3
m
m
2
190 m
(1190 km2/h2 − 49 km2/h2)36
1
0
h
0 s
2
1
1
0
k
3
m
m
2
190 m
[(34.5 km/h)2 − (7.0 km/h)2]36
1
0
h
0 s
2
1
1
0
k
3
m
m
2
(2)(95 m)
Section Five—Problem Bank V Ch. 2–7
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6. vi = 50.0 km/h forward= +50.0 km/h
vf = 0 km/h
a = 9.20 m/s2 backward= −9.20 m/s2
Givens Solutions
∆x = vf
2
2
−a
vi2
=
∆x =
∆x = 10.5 m = 10.5 m forward
−(2.50 × 103 km2/h2)36
1
0
h
0 s
2
110
k
3
m
m
2
−18.4 m/s2
[(0 km/h)2 − (50.0 km/h)2]36
1
0
h
0 s
2
110
k
3
m
m
2
(2)(−9.20 m/s2)
7. a = 7.56 m/s2
∆x = 19.0 m
vi = 0 m/s
vf =√
vi2+ 2a∆x =√
(0 m/s)2 + (2)(7.56m/s2)(19.0m)
vf =√
287m2/s2 = ±16.9 m/s = 16.9 m/s
8. vi = 1.8 m/s
vf = 9.4 m/s
a = 6.1 m/s2
∆x = vf
2
2
−a
vi2
= =
∆x = (2)
8
(
5
6.
m
1
2
m
/s
/
2
s2) = 7.0 m
88 m2/s2 − 3.2 m2/s2
(2)(6.1 m/s2)
(9.4 m/s)2 − (1.8 m/s)2
(2)(6.1 m/s2)
9. vi = 1.50 m/s to the right = +1.50 m/s
vf = 0.30 m/s to the right = +0.30 m/s
a = 0.35 m/s2 to the left = −0.35 m/s2
∆x = vf
2
2
−a
vi2
=
∆x =
∆x = −−2
0
.
.
1
7
6
0
m
m
2
/
/
s
s2
2
= +3.1 m = 3.1 m to the right
9.0 × 10−2 m2/s2 − 2.25 m2/s2
−0.70 m/s2
(0.30 m/s)2 − (1.50 m/s)2
(2)(−0.35 m/s2)
10. a = 0.678 m/s2
vf = 8.33 m/s
∆x = 46.3 m
vi =√
vf2− 2a∆x =√
(8.33m/s)2 − (2)(0.678 m/s2)(46.3m)
vi =√
69.4 m2/s2 − 62.8m2/s2 =√
6.6m2/s2 = ±2.6 m/s = 2.6 m/s
1. vi = 0 m/s
vf = 49.5 m/s downward= 49.5 m/s
a = −9.81 m/s2
∆tot = −448 m
∆xi = vf
2
2
−a
vi2
= = (2)
2
(
4
−5
9
0
.8
m
1
2
m
/s
/
2
s2) = −125 m
∆x2 = ∆xtot − ∆x1 = (−448 m) − (−125 m) = −323 m
distance from net to ground = magnitude ∆x2 = 323 m
(−49.5 m/s)2 − (0 m/s)2
(2)(−9.81 m/s2)
Additional Practice 2F
5. ∆x = +42.0 m
vi = +153.0 km/h
vf = 0 km/ha =
vf2
2∆−
x
vi2
=
a = = −21.5 m/s2
−(2.34 × 104 km2/h2)36
1
0
h
0 s
2
1
1
0
k
3
m
m
2
(84.0 m)
[(0 km/h)2 − (153.0 km/h)2]36
1
0
h
0 s
2
1
1
0
k
3
m
m
2
(2) (42.0 m)
Givens Solutions
Holt Physics Solution ManualV Ch. 2–8
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.
2. vi = 0 m/s
a = −9.81 m/s2
∆t1 = 1.00 s
∆t2 = 2.00 s
∆t3 = 3.00 s
Because vi = 0 m/s, ∆x1 = 1
2a∆t1
2 = 1
2(−9.81 m/s2)(1.00 s)2 =
∆x2 = 1
2a∆t2
2 = 1
2(−9.81 m/s2)(2.00 s)2 =
∆x3 = 1
2a∆t3
2 = 1
2(−9.81 m/s2)(3.00 s)2 = −44.1 m
−19.6 m
−4.90 m
3. vi = 0 m/s
∆t = 2.0 s
a = −9.81 m/s2
Because vi = 0 m/s, ∆x = 1
2a∆t2 =
1
2(−9.81 m/s2)(2.0 s)2 = −2.0 × 101 m
distance of bag below balloon = 2.0 × 101 m
4. vi = +17.5 m/s
a = −9.81 m/s2
∆ttot = 3.60 s
∆xtot = 0 m
To find the ball’s maximum height, calculate ∆xup. The time for this distance to betraveled is ∆tup = 1
2 ∆xtot
∆xup = vi ∆tup + 1
2a∆tup
2 = vi 1
2∆ttot +
1
2a
1
2∆ttot
2
∆xup = (17.5 m/s)3.6
2
0 s +
1
2(−9.81 m/s2)
3.6
2
0 s
2= (31.5 m) + (−15.9 m) = +15.6 m
softball’s maximum height = 15.6 m
5. vi = 0 m/s
vf = 11.4 m/s downward= −11.4 m/s
a = 3.70 m/s2 downward= −3.70 m/s2
∆x = vf
2
2
−a
vi2
= = = −17.6 m
∆x = 17.6 m downward
1.30 × 102 m2/s2
−7.40 m/s2
(−11.4 m/s)2 − (0 m/s)2
(2)(−3.70 m/s2)
6. ∆ttot = 5.10 s
∆tdown = 1
2 ∆ttop
vi = 0 m/s
a = −9.81 m/s2
To find the ball’s maximum height, calculate the displacement from that height to itsoriginal position. The time interval for this free-fall is 1
2 ∆ttot, and vi = 0 m/s.
∆xdown = 1
2 a∆tdown
2 = 1
2a
1
2∆ttot
2=
1
2(−9.81 m/s2)
5.1
2
0 s
2= −31.9 m
ball’s maximum height = 31.9 m
7. ∆ttot = 5.10 s
a = −9.81 m/s2
∆xtot = 0 m
∆ttot = vi ∆ ttot + 1
2a∆ttot
2
Because ∆xtot = 0,
vi = − 1
2a∆ttot = −
1
2(−9.81 m/s2)(5.10 s) = +25.0 m/s = 25.0 m/s upward
8. vi = 85.1 m/s upward= + 85.1 m/s
a = −9.81 m/s2
∆x = 0 m
Because ∆x = 0 m, vi ∆t + 1
2a∆t 2 = 0
∆t = − −2
a
vi = − −(
(
2
−)
9
(
.
8
8
5
1
.1
m
m
/s
/2s
)
) = 17.3 s
Section Five—Problem Bank V Ch. 2–9
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.
Givens Solutions
10. aup = 3.125 m/s2, upward= +3.125 m/s2
∆tup = 4.00 s
vi = 0 m/s
vf = aup ∆tup + vi = (3.125 m/s2) (4.00 s) + 0 m/s = 2.5 m/s
When the cable brakes, the upward-moving elevator undergoes free-fall acceleration.
vi = +12.5 m/s
a = −9.81 m/s2
∆t1 = 0.00 s
∆t2 = 1.00 s
∆t3 = 2.00 s
∆t4 = 3.00 s
vf, 1 = a∆t1 + vi = (−9.81 m/s2)(0.00 s) + 12.5 m/s =
vf, 2 = a∆t2 + vi = (−9.81 m/s2)(1.00 s) + 12.5 m/s = −9.81 m/s + 12.5 m/s =
vf, 3 = a∆t3 + vi = (−9.81 m/s2)(2.00 s) + 12.5 m/s = −19.6 m/s + 12.5 m/s =
vf, 4 = a∆t4 + vi = (−9.81 m/s2)(3.00 s) + 12.5 m/s = −29.4 m/s + 12.5 m/s = − 16.9 m/s
−7.1 m/s
+2.7 m/s
+12.5 m/s
9. ∆td = 4.00 s
∆tp = ∆td + 3.00 s= 7.00 s
a = −9.81 m/s2
∆xd = ∆xp
vi, p = 0 m/s
∆xd = vi, d ∆td + 1
2a∆td
2
∆xp = vi, p ∆tp + 1
2a∆tp
2
vi, d ∆td + 1
2a∆td
2 = vi, p∆tp + 1
2a∆tp
2
vi, d = + vi,
∆p
t
∆
d
tp
Because vi, p = 0 m/s,
vi, d = =
vi, d = − = − =
vf, d = a ∆td + vi, d = (−9.81 m/s2)(4.00 s) + (−40.5 m/s) = (−39.2 m/s) + (−40.5 m/s)
vf, d = −79.7 m/s
−40.5 m/s(9.81 m/s2)(33.0 s2)
8.00 s
(9.81 m/s2)(49.0 s2 − 16.0 s2)
8.00 s
1
2(−9.81 m/s2)[(7.00 s)2 − (4.00 s)2]
4.00 s
1
2a(∆tp
2 − ∆td2)
∆td
1
2a(∆tp
2 − ∆td2)
∆td
Section Five—Problem Bank V Ch. 3–1
Chapter 3Two -Dimensional Motion and Vectors
V
1. ∆x1 = 8 m to the left = +8 m
∆x2 = 8 m to the right = −8 m
∆x3 = 8 m to the left = +8 m
a. distance traveled = 8 m + 8 m + 8 m =
b. d = ∆x1 + ∆x2 + ∆x3 = 8 m + (−8 m) + 8 m = 8 m
24 m
Additional Practice 3A
Givens Solutions
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2. hi = 0.91 m
hf = (0.90)hi
∆x = 0.11 m
∆y = hf − hi = (0.90 − 1.00)hi
∆y = (−0.10)(0.91 m) = −9.1 × 10−2 m
d =√
∆x2+ ∆y2 =√
(0.11m)2 + (−9.1× 10−2m)2 =√
1.2× 10−2m2+ 8.3 × 10−3m2
d = 2.0 × 10−2 m2 =
q = tan−1∆∆
x
y = tan−1 −9.1
0
×.11
10
m
−2 m
q = −4.0 × 101° = 4.0 × 101° below the horizontal
0.14 m
3. ∆x = 165 m
∆y = −45 m
d =√
∆x2+ ∆y2 =√
(165 m)2 + (−45 m)2√
2.72 × 104 m2+ 2.0 × 103 m2 =√
2.92 × 104 m2
d =
q tan−1 ∆∆
x
y = tan−1−16
4
5
5
m
m
q = −15° = 15° below the horizontal
171 m
4. ∆y = −13.0 m
∆x = 9.0 m
d =√
∆x2+ ∆y2 =√
(9.0 m)2 + (−13.0 m)2 =√
81 m2+ 169 m2 =√
2.50 × 102 m2
d =
q = tan−1 ∆∆
x
y tan−1−9
1
.
3
0
.0
m
m
q = −55° = 55° below the horizontal
15.8 m
5. ∆x = 36.0 m, east
∆y = 42.0 m, north
∆z = 17.0 m, up
d =√
∆x2+ ∆y2+ ∆z2 =√
(36.0m)2 + (42.0 m)2 + (17.0 m)2
d =√
1.30 × 103 m2+ 1.76× 103 m2+ 289 m2 =√
3.35 × 103 m2
d =
horizontal direction = qh = tan−1 ∆∆
x
y = tan−1 432
6
.
.
0
0
m
m
qh =
vertical direction = qv = tan−1 ∆x 2∆+z
∆y249.4° north of east
57.9 m
9. d = 3.88 km
∆x = 3.45 km
h1 = 0.8 km
d2 = ∆x2 + ∆y2
∆y =√
d2− ∆x2 =√
(3.88km)2 − (3.45 km)2 =√
15.1 km2− 11.9km2 =√
3.2km2
∆y = 1.8 km
height of mountain = h = ∆y + h1 = 1.8 km + 0.8 km
h = 2.6 km
Holt Physics Solution ManualV Ch. 3–2
V
qv = tan−1 = tan−1 = tan−1 √3.06
17
×.0
10
m3 m2
qv = 17.1° above the horizontal
17.0 m√
1.30 × 103 m2+ 1.76× 103 m217.0 m
√(36.0m)2 + (42.0 m)2
Givens Solutions
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6. d = 599 m
∆y = 89 m north
d2 = ∆x2 + ∆y2
∆x =√
d2− ∆y2 =√
(599 m)2 − (89 m)2 =√
3.59 × 105 m2− 7.9 × 103 m2
∆x =√
3.51 × 105 m2
∆x =
q = sin−1 ∆d
y = sin−1 58999m
m
q = 8.5 ° north of east
592 m, east
7. d = 478 km
∆y = 42 km, south = −42 km
d2 = ∆x2 + ∆y2
∆x =√
d2− ∆y2 =√
(478 km)2 − (−42 km)2√
2.28 × 105 km2− 1.8 × 103 km2
∆x =√
2.26 × 105 km2 = −475 km
∆x =
q = sin−1 ∆d
y = sin−1 −47
4
8
2
k
k
m
m
q = 5.0° south of west
475 km, west
8. d = 7400 km
q = 26° south of west
∆y = 3200 km, south =−3200 km
d2 = ∆x2 = ∆y2
∆x =√
d2− ∆y2 = √
(7400km)2 − (−3200 km)2 =√
5.5× 107 km2− 1.0 × 107 km2
∆x =√
4.5× 107 km2 = −6700 km
∆x = 6700 km, west
Section Five—Problem Bank V Ch. 3–3
V
Givens Solutions
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10. d = 2.9 × 103 km
∆x = 2.8 × 103 km, west = −2.8 × 103 km
d2 = ∆x2 + ∆y2
∆y =√
d2− ∆x2 =√
(2.9 × 103 km)2 − (−2.8× 103)2
∆y =√
8.4× 106 km2− 7.8 × 106 km2 =√
0.6× 106 km2 = −800 km
∆y =
q = cos−1 ∆d
x = cos−1 −22.9.8×
×1
1
0
03
3
k
k
m
m
q = 15° south of west
800 km, south
1. d = 5.3 km
q = 8.4° above horizontal
∆y = d(sin q) = (5.3 km)(sin 8.4°)
∆y = 0.77 km = 770 m
the mountain’s height = 770 m
2. d = 19.1 m
q = 3.0° to the left
∆y = d(sin q) = (19.1 m)(sin 3.0°)
∆y = 1.0 m to the left
the lane’s width = 1.0 m
Additional Practice 3B
3. d = 113 m
q = 82.4° above the horizontal south
∆x = d(cos q) = (113 m)(cos 82.4°)
∆x = 14.9 m, south
4. v = 55 km/h
q = 37° below the horizontal= −37°
vy = v(sin q) = (55 km/h)[sin(−37°)]
vy = −33 km/h = 33 km/h, downward
5. d = 2.7 m
q = 13° from the table’slength
∆x = d(cos q) = (2.7 m)(cos 13°)
∆x =
∆y = d(sin q) = (2.7 m)(sin 13°)
∆y = 0.61 m along the table’s width
2.6 m along the table’s length
6. v = 1.20 m/s
q = 14.0° east of north
vx = v(sin q) = (1.20 m/s)(sin 14.0°)
vx =
vy = v(cos q) = (1.20 m/s)(cos 14.0°)
vy = 1.16 m/s, north
0.290 m/s, east
7. d = 31.2 km
q = 30.0° west of south
∆x = d (sin q) = (31.2 km)(sin 30.0°)
∆x =
∆y = d (cos q) = (31.2 km)(cos 30.0°)
∆y = 27.0 km, south
15.6 km, west
Holt Physics Solution ManualV Ch. 3–4
V
8. v = 165.2 km/s
q = 32.7°
vforward = v(cos q) = (165.2 km/s)(cos 32.7°)
vforward =
vside = v(sin q) = (165.2 km/s)(sin 32.7°)
vside = 89.2 km/s to the side
139 km/s, forward
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9. v = 55.0 km/h
q = 13.0° above horizontal
vy = v(sin q) = (55.0 km/h)(sin 13.0°)
vy =
vx = v(cos q) = (55.0 km/h)(cos 13.0°)
vx = 53.6 km/h, forward
12.4 km/h, upward
10. v = 13.9 m/s
qh = 24.0° east of north
qv = 26.0° above the horizontal
vz = v(sin qv) = (13.9 m/s)(sin 26.0°)
vz =
horizontal velocity = vh = v(cos qv)
vy = vh(cos qh) = v(cos qv)(cos qh) = (13.9 m/s)(cos 26.0°)(cos 24.0°)
vy =
vx = vh(sin qh) = v(cos qv)(sin qh) = (13.9 m/s)(cos 26.0°)(sin 24.0°)
vx = 5.08 m/s, east
11.4 m/s, north
6.09 m/s, upward
1. d1 = 55 km
q1 = 37 north of east
d2 = 66 km
q2 = 0.0° (due east)
∆x1 = d1(cos q1) = (55 km)(cos 37°) = 44 km
∆y1 = d1(sin q1) = (55 km)(sin 37°) = 33 km
∆x2 = d2(cos q2) = (66 km)(cos 0.0°) = 66 km
∆y2 = d2(sin q2) = (66 km)(sin 0.0°) = 0 km
∆xtot = ∆x1 + ∆x2 = 44 km + 66 km = 110 km
∆ytot = ∆y1 + ∆y2 = 33 km + 0 km = 33 km
d =√
(∆xtot)2 + (∆ytot)2 =√
(110 km)2 + (33 km)2
=√
1.21 × 104 km2+ 1.1 × 103 km2=√
1.32 × 104 km2
d =
q = tan−1 ∆∆x
yt
t
o
o
t
t = tan−1 13130k
k
m
m
q = 17° north of east
115 km
Additional Practice 3C
Section Five—Problem Bank V Ch. 3–5
V
2. d1 = 4.1 km
q1 = 180° (due west)
d2 = 17.3 km
q2 = 90.0° (due north)
d3 = 1.2 km
q3 = 24.6° west of north = 90.0° + 24.6° = 114.6°
∆x1 = d1(cos ∆1) = (4.1 km)(cos 180°) = −4.1 km
∆y1 = d1(sin q1) = (4.1 km)(sin 180°) = 0 km
∆x2 = d2(cos q2) = (17.3 km)(cos 90.0°) = 0 km
∆y2 = d2(sin q2) = (17.3 km)(sin 90.0°) = 17.3 km
∆x3 = d3(cos q3) = (1.2 km)(cos 114.6°) = −0.42 km
Dy3 = d3(sin q3) = (1.2 km)(sin 114.6°) = 1.1 km
∆xtot = ∆x1 + ∆x2 + ∆x3 = −4.1 km + 0 km + (−0.42 km) = −4.5 km
∆ytot = ∆y1 + ∆y2 + ∆y3 = 0 km + 17.3 km + 1.1 km = 18.4 km
d =√
(∆xtot)2 + (∆ytot)2 =√
(−4.5km)2 + (18.4 km)2
=√
2.0× 101 km2+ 339 km2 =√
359km2
d =
q = tan−1∆∆x
yt
t
o
o
t
t = tan−1 −18
4
.
.
4
5
k
k
m
m = −76° = 76° north of west
18.9 km
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3. d1 = 850 m
q1 = 0.0°
d2 = 640 m
q2 = 36°
∆x1 = d1(cos q1) = (850 m)(cos 0.0°) = 850 m
∆y1 = d1(sin q1) = (850 m)(sin 0.0°) = 0 m
∆x2 = d2(cos q2) = (640 m)(cos 36°) = 520 m
∆y2 = d2(sin q2) = (640 m)(sin 36°) = 380 m
∆xtot = ∆x1 + ∆x2 = 850 m + 520 m = 1370 m
∆ytot = ∆y1 + ∆y2 = 0 m + 380 m = 380 m
d = (∆xtot)2 + (∆ytot)2 = (1370m)2 + (380m)2 = 1.9× 106 m2+ 1.4 × 105 m2
= 2.0× 106 m2
d =
q = tan−1∆∆x
yt
t
o
o
t
t = tan−11338700m
m
q = 16° to the side of the initial displacement
1400 m
Holt Physics Solution ManualV Ch. 3–6
V
4. d1 = 2.00 × 102 m
q1 = 0.0°
d2 = 3.00 × 102 m
q2 = 3.0°
d3 = 2.00 × 102 m
q3 = 8.8°
∆x1 = d1(cos q1) = (2.00 × 102 m)(cos 0.02) = 2.0 × 102 m
∆y1 = d1(sin q1) = (2.00 × 102 m)(sin 0.0°) = 0 m
∆x2 = d2(cos q2) = (3.00 × 102 m)(cos 3.0°) = 3.0 × 102 m
∆y2 = d2(sin q2) = (3.00 × 102 m)(sin 3.0°) = 16 m
∆x3 = d3(cos q3) = (2.00 × 102 m)(cos 8.8°) = 2.0 × 102 m
∆y3 = d3(sin q3) = (2.00 × 102 m)(sin 8.8°) = 31 m
∆xtot = ∆x1 + ∆x2 + ∆x3 = 2.0 × 102 m + 3.0 × 102 m + 2.0 × 102 m = 7.0 × 102 m
∆ytot = ∆y1 + ∆y2 + ∆y3 = 0 m + 16 m + 31 m = 47 m
d =√
(∆xtot)2 + (∆ytot)2 =√
(7.0 × 102 m)2 + (47 m)2 =√
4.9× 105 m2+ 2.2 × 103m2
=√
4.9× 105 m2
d =
q = tan−1 ∆∆x
yt
t
o
o
t
t = tan−1 7.0
4
×7
1
m
02 m
q = 3.8° above the horizontal
7.0 × 102 m
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5. d1 = 46 km
q1 = 15° south of east = −15°
d2 = 22 km
q2 = 13° east of south = −77°
d3 = 14 km
q3 = 14° west of south = −90.0° − 14° = −104°
∆x1 = d1(cos q1) = (46 km)[cos(−15°)] = 44 km
∆y1 = d1(sin q1) = (46 km)[sin(−15°)] = −12 km
∆x2 = d2(cos q2) = (22 km)[cos(−77°)] = 4.9 km
∆y2 = d2(sin q2) = (22 km)[sin(−77°)] = −21 km
∆x3 = d3(cos q3) = (14 km)[cos(−104°)] = −3.4 km
∆y3 = d3(sin q3) = (14 km)[sin(−104°)] = −14 km
∆xtot = ∆x1 + ∆x2 + ∆x3 = 44 km + 4.9 km + (−3.4 km) = 46 km
∆ytot = ∆y1 + ∆y2 + ∆y3 = −12 km + (−21 km) + (−14 km) = −47 km
d =√
(∆xtot)2 + (∆ytot)2 =√
(46km)2 + (−47 km)2 =√
2.1× 103 km2+ 2.2 × 103 km2
=√
4.3× 103 km2
d =
q = tan−1 ∆∆x
yt
t
o
o
t
t = tan−1 −4467k
k
m
m = −46°
q = 46° south of east
66 km
Section Five—Problem Bank V Ch. 3–7
V
6. d1 = 6.3 × 108 km
q1 = 0.0°
d2 = 9.4 × 108 km
q2 = 68°
d3 = 3.4 × 109 km
q3 = 94° + 68° = 162°
∆x1 = d1(cos q1) = (6.3 × 108 km)(cos 0.0°) = 6.3 × 108 km
∆y1 = d1(sin q1) = (6.3 × 108 km)(sin 0.0°) = 0 km
∆x2 = d2(cos q2) = (9.4 × 108 km)(cos 68°) = 3.5 × 108 km
∆y2 = d2(sin q2) = (9.4 × 108 km)(sin 68°) = 8.7 × 108 km
∆x3 = d3(cos q3) = (3.4 × 109 km)(cos 162°) = −3.2 × 109 km
∆y3 = d3(sin q3) = (3.4 × 109 km)(sin 162°) = 1.1 × 109 km
∆xtot = ∆x1 + ∆x2 + ∆x3 = 6.3 × 108 km + 3.5 × 108 km + (−3.2 × 109 km) = −2.2 × 109 km
∆ytot = ∆y1 + ∆y2 + ∆y3 = 0 km + 8.7 × 103 km + 1.1 × 109 km = 2.0 × 109 km
d =√
(xtot)2+ (ytot)2 =√
(−2.2× 109 km)2 + (2.0× 109 km)2
=√
4.8× 1018km2+ 4.0 × 1018km2 =√
8.8× 1018km2
d =
q = tan−1 ∆∆x
yt
t
o
o
t
t = tan−1 −22.0.2×
×1
1
0
0
9
9k
k
m
m = − 42° 180.0° − 42° = 138°
q = 138° from the probe’s initial direction
3.0 × 109 km
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7. di = 2.50 × 103 m
q1 = 58.5° north of east
d2 = 375 m
q2 = 21.8° north of east
d3 = 875 m
q3 = 21.5° east of north
∆x1 = d1(cos q1) = (2.50 × 103 m)(cos 58.5°) = 1310 m
∆y1 = d1(sin q1) = (2.50 × 103 m)(sin 58.5°) = 2130 m
∆x2 = d2(cos q2) = (375 m)(cos 21.8°) = 348 m
∆y2 = d2(sin q2) = (375 m)(sin 21.8°) = 139 m
∆x3 = d3(sin q3) = (875 m)(sin 21.5°) = 321 m
∆y3 = d3(cos q3) = (875 m)(cos 21.5°) = 814 m
∆xtot = ∆x1 + ∆x2 + ∆x3 = 1310 m + 348 m + 321 m = 1.98 × 103 m
∆ytot = ∆y1 + ∆y2 + ∆y3 = 2130 m + 139 m + 814 m = 3.08 × 103 m
d =√
(∆xtot)2 + (∆ytot)2 =√
(1.98× 103 m)2 + (3.08 × 103 m)2
=√
3.92 × 106 m2+ 9.49× 106 m2 =√
13.41× 106 m2
d =
q = tan−1 ∆∆x
yt
t
o
o
t
t = tan−1 31.
.
0
9
8
8
××
1
1
0
0
3
3m
m
q = 57.3° north of east
3.66 × 103 m
Holt Physics Solution ManualV Ch. 3–8
V
8. d1 = 5.0 km
q1 = 36.9° south of east = −36.9°
d2 = 1.5 km
q2 = 90.0° due south = −90.0°
d3 = 8.5 km
q3 = 42.2° south of east
= −42.2°
d4 = 0.8 km
q4 = 0° (due east)
∆x1 = d1(cos q1) = (5.0 km)[cos(−36.9°)] = 4.0 km
∆y1 = d1(sin q1) = (5.0 km)[sin(−36.9°)] = −3.0 km
∆x2 = d2(cos q2) = (1.5 km)[cos(−90.0°)] = 0 km
∆y2 = d2(sin q2) = (1.5 km)[sin(−90.0°)] = −1.5 km
∆x3 = d3(cos q3) = (8.5 km)[cos(−42.2°)] = 6.3 km
∆y3 = d3(sin q3) = (8.5 km)[sin(−42.2°)] = −5.7 km
∆x4 = d4(cos q4) = (0.8 km)(cos 0°) = 0.8 km
∆y4 = d4(sin q4) = (0.8 km)(sin 0°) = 0 km
∆xtot = ∆x1 + ∆x2 + ∆x3 + ∆x4 = 4.0 km + 0 km + 6.3 km + 0.8 km = 11.1 km
∆ytot = ∆y1 + ∆y2 + ∆y3 + ∆y4 = (− 3.0 km) + (−1.5 km) + (−5.7 km) + 0 km = −10.2 km
d =√
(∆xtot)2 + (∆ytot)2 =√
(11.1km)2 + (−10.2 km)2 =√
123km2+ 104 km2
=√
227km2
d =
q = tan−1 ∆∆x
yt
t
o
o
t
t = tan−1 −1110.1.2k
k
m
m = −42.6°
q = 42.6° south of east
15.1 km
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9. d1 = 1.41 m
q1 = 45.0° west of north = 90.0° + 45.0° = 135.0°
d2 = 1.98 m
q2 = 45.0° east of north = 45.0°
d3 = 0.42 m
q3 = 45.0° west of north = 135.0°
d4 = 1.56 m
q4 = 45.0° south of west = 180.0° + 45.0° = 225.0°
∆x1 = d1(cos q1) = (1.41 m)(cos 135.0°) = −0.997 m
∆y1 = d1(sin q1) = (1.41 m)(sin 135.0°) = 0.997 m
∆x2 = d2(cos q2) = (1.98 m)(cos 45.0°) = 1.40 m
∆y2 = d2(sin q2) = (1.98 m)(sin 45.0°) = 1.40 m
∆x3 = d3(cos q3) = (0.42 m)(cos 135.0°) = −0.30 m
∆y3 = d3(sin q3) = (0.42 m)(sin 135.0°) = 0.30 m
∆x4 = d4(cos q4) = (1.56 m)(cos 225.0°) = −1.10 m
∆y4 = d4(sin q4) = (1.56 m)(sin 225.0°) = −1.10 m
∆xtot = ∆x1 + ∆x2 + ∆x3 + ∆x4 = (−0.997 m) + 1.40 m + (−0.30 m) + (−1.10 m)
= −0.997 m
∆ytot = ∆y1 + ∆y2 + ∆y3 + ∆y4 = 0.997 m + 1.40 m + 0.30 m + (−1.10 m) = 1.60 m
d =√
(∆xtot)2 + (∆ytot2 =√
(−0.997m)2 + (1.60 m)2 =√
0.994m2+ 2.56m2
=√
3.55 m2
d =
q = tan−1 ∆∆x
yt
t
o
o
t
t = tan−1 −1
0
.
.
6
9
0
97
m
m = −58.1°
q = 58.1° north of west
1.88 m
Section Five—Problem Bank V Ch. 3–9
V
10. d1 = 79 km
q1 = 18° north of west 180.0° − 18° = 162°
d2 = 150 km
q2 = 180.0° due west
d3 = 470 km
q3 = 90.0° due north
d4 = 240 km
q4 = 15° east of north 90.0° − 15° = 75°
∆x1 = d1(cos q1) = (790 km)(cos 162°) = −750 km
∆y1 = d1(sin q1) = (790 km)(sin 162°) = 24 km
∆x2 = d2(cos q2) = (150 km)(cos 180.0°) = −150 km
∆y2 = d2(sin q2) = (150 km)(sin 180.0°) = 0 km
∆x3 = d3(cos q3) = (470 km)(cos 90.0°) = 0 km
∆y3 = d3(sin q3) = (470 km)(sin 90.0°) = 470 km
∆x4 = d4(cos q4) = (240 km)(cos 75°) = 62 km
∆y4 = d4(sin q4) = (240 km)(sin 75°) = 230 km
∆xtot = ∆x1 + ∆x2 + ∆x3 + ∆x4 = (−750 km) + (−150 km) + 0 km + 62 km = −840 km
∆ytot = ∆y1 + ∆y2 + ∆y3 + ∆y4 = 240 km + 0 km + 470 km + 230 km = 940 km
d =√
(∆xtot)2 + (∆ytot)2 =√
(−840km)2 + (940km)2 =√
7.1× 105 km2+ 8.8 × 105 km2
=√
15.9 × 105 km2
d =
q = tan−1 ∆∆x
yt
t
o
o
t
t = tan−1 −984400k
k
m
m = −48°
q = 48° north of west
1260 km
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Additional Practice 3D
1. vx = 430 m/s ∆t = ∆vx
x
∆x = 4020 m∆y = − 1
2g∆t2
g = 9.81 m/s2
∆y = − 12
g∆vx
x
2
= − 12
(9.81 m/s2)443
0
0
20
m
m
/s
2
= −430 m
2. ∆x = 101 m ∆t = ∆vx
x
vx = 14.25 m/s∆y = − 1
2g∆t2
g = 9.81 m/s2
y = − 12
g∆vx
x
2
= − 12
(9.81 m/s2)14
1
.
0
2
1
5
m
m/s
2
= 246 m
3. vx = 1.30 × 102 km/h ∆t = ∆vx
x
∆x = 135 m∆y = − 1
2g∆t2
g = 9.81 m/s2
∆y = − 12
g∆vx
x
2
= − 12
(9.81 m/s2)1.30 ×13
1
5
0
m2 km/h
2
130630
m
0
/
s
k
/h
m
2
= −68.6 m
airship’s altitude = 68.6 m
height of building = 246 m
height of ridge = 430 m
Holt Physics Solution ManualV Ch. 3–10
V
4. vx = 9.37 m/s ∆t = ∆vx
x
∆x = 85.0 m∆y = − 1
2g∆t2
g = 9.81 m/s2
∆y = − 12
g∆vx
x
2
= − 12
(9.81 m/s2)98.357
.0
m
m
/s
2
= − 404 m
5. vx = 6.32 cm/s ∆t = ∆vx
x
∆x = 1.00 m∆y = − 1
2g∆t2
g = 9.81 m/s2
∆y = − 12
g∆vx
x
2
= − 12
(9.81 m/s2)6.32
1
×.0
1
0
0
m−2 m/s
2
= 1230 m
6. vx = 10.0 cm/s ∆t = ∆vx
x
∆x = 18.6 cm∆y = − 1
2g∆t2
g = 9.81 m/s2
∆y = − 12
g∆vx
x
2
= − 12
(9.81 m/s2)1108.0.6c
c
m
m
/s
2
= −17.0 m
7. vx = 1.50 m/s ∆t = ∆vx
x
∆x = 3.50 m∆y = − 1
2g∆t2
g = 9.81 m/s2
∆y = − 12
g∆vx
x
2
= − 12
(9.81 m/s2)13.5.500m
m
/s
2
= −26.7 m
8. ∆y = −2.50 × 102 m ∆x = vx∆t
vx = 1.50 m/s ∆y = − 12
g∆t2
g = 9.81 m/s2
∆t = 2
−∆g
y
∆x = vx2
−∆g
y = (1.50 m/s) (2)(−
−2
9
..
5
80
1 ×m 1
/0
s22
m)
∆x = 10.7 m
the lunch pail falls 26.7 m
squirrel’s height = 17.0 m
building’s height = 1230 m
mountain’s height = 404 m
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Section Five—Problem Bank V Ch. 3–11
V
9. vx = 1.50 m/s vy,f2 = −2g∆y + vy,i
2
∆y = −2.50 × 102 m vy,i = 0 m/s, so
g = 9.81 m/s2 vy,f = vy =√
−2g∆y =√
−(2)(9.81 m/s2)(−2.50 × 102 m)vy = 70.0 m/s
v =√
vx2+ vy2 =
√(1.50m/s)2 + (70.0 m/s)2 =
√2.25 m2/s2 + 4.90× 103 m2/s2
=√
4.90 × 103 m2/s2
v =
q = tan−1 v
vx
y = tan−117.
0
5
.
0
0
m
m
/
/
s
s
q =
10. vx = 85.3 m/s ∆t = 2
−∆g
y =
∆vx
x
∆y = −1.50 m
∆x = vx 2
−∆g
y = (85.3 m/s) (2
−)
9
(.
−81
1
. 5
m0
/m
s2) = 47.2 m
g = 9.81 m/s2
range of arrow = 47.2 m
1.23° from the vertical
70.0 m/s
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1. vi = 15.0 m/s ∆t = vi(c
∆o
x
s q)
∆x = 17.6 m ∆y = vi(sin q)∆t − 12
g∆t2 = 0
vi(sin q) = 12
g∆t = 12
gvi(c
∆o
x
s q)
2(sin q)(cos q) = g
v
∆
i2x
Using the identity 2 (sin q)(cos q) = sin (2 q),
sin (2 q) = g
v
∆
i2x
q = =
q =
2. vi = 23.1 m/s vy,f2 − vy,i
2 = −2g∆y
∆ymax = 16.9 m At maximum height, vy,f
g = 9.81 m/s2 vy,i = vi(sin q) =√
2g∆ymax
q = sin−1 = sin−1 q = 52.0°
√(2)(9.81 m/s2)(16.9m)
23.1 m/s
√2g∆ymax
vi
25.1°
sin − (9.81
(1
m
5.
/
0
s2
m
)(
/
1
s
7
)
.26 m)
2
sin−1g
v
∆
i2x
2
Additional Practice 3E
Holt Physics Solution ManualV Ch. 3–12
V
3. ∆x = 7.49 m Using the form of the equation derived in problem 1,
vi = 9.50 m/s 2(sin q)(cos q) = sin(2 q) = g
v
∆
i2x
g = 9.81 m/s2
q = sin−1 = sin−1
q =
4. vi = 141 cm/s Using the form of the equation derived in problem 1,
∆x = 18.5 cm 2(sin q)(cos q) = sin(2 q) = g
v
∆
i2x
g = 9.81 m/s2
q = sin−1
sin−1 2
q =
5. vi = 6.03 m/s vy,f2 − vy,i
2 = −2g∆y
hi = 10.0 m At maximum height, vy,f = 0.
hmax = hf = 11.7 m vy,i = vi(sin q) =√
2g∆ymax∆x = 3.62 m For the diver, hf is the maximum height above the diving board.
g = 9.81 m/s2 ∆y = hf − hi
q = sin−1 = sin−1 q =
6. vi = 10.0 m/s ∆x = vi(cos q)∆t = (10.0 m/s)(cos 37.0°)(2.5 s)
q = 37.0° ∆x =
∆t = 2.5 s ∆y = vi(sin q)∆t − 12
g∆t2 = (10.0 m/s)(sin 37.0°)(2.5 s) − 12
(9.81 m/s2)(2.5 s)2
g = 9.81 m/s2= 15 m − 31 m
∆y =
7. vi = 250 m/s At the maximum height
q = 35° vy,f = vy,i − g∆t = 0
g = 9.81 m/s2 vy,i = vi(sin q) = g∆t
∆t = vi(si
g
n q) =
(250
9
m
.8
/
1
s)
m
(s
/
i
s
n235°)
∆t = 15 s
−16 m
2.0 101 m
73.3°
√(2)(9.81)m/s2)(11.7m − 10.0m)
6.03 m/s
√2g(hf − hi)
vi
33.0°
(9.81 m/s2)(18.5 × 10−2 m)
(141 × 10−2 m/s)2
g
v
∆
i2x
2
27.3°
(9.81
(
m
9.5
/s
0
2)
m
(7
/s
.429 m)
2
g
v
∆
i2x
2
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Section Five—Problem Bank V Ch. 3–13
V
8. ∆x = 73.0 m ∆t = vi(c
∆o
x
s q)
∆y = −52.8 m
∆y = vi(sin q)∆t − 12
g∆t 2 = vi(sin q)vi(c
∆o
x
s q) − 1
2gvi(c
∆o
x
s q)
2
q = −8.00°
∆y = ∆x(tanq) − 2 vi
2g
(
∆co
x
s
2
q)2g = 9.81 m/s2
2 vi
2g
(
∆co
x2
s q)2 = ∆x(tan q) − ∆y
2vi2(cos q)2 =
[∆x(ta
g
n
∆qx
)
2
− ∆y]
vi = vi = vi = vi = vi =
9. q = −30.0° ∆y = vi(sin q)∆t − 12
g∆t2
vi = 2.0 m/s 2
g∆t2 − [vi(sin q)]∆t + ∆y = 0
∆y = −45 mSolving for ∆t using the quadratic equation,
g = 9.81 m/s2
∆t =
∆t =
∆t = =
∆t =
∆t must be positive, so the positive root must be chosen.
∆t = 9.
2
8
9
1
m
m
/
/
s
s2 = 3.0 s
−1.0 m/s ± 3.0 × 101 m/s
9.81 m/s2
−1.0 m/s ±√
8.8× 102 m2/s29.81 m/s2
−1.0 m/s ±√
1.0m2/s2 + 8.8 × 102 m2/s29.81 m/s2
(2.0 m/s)[sin(−30.0°)] ±√
[(−2.0m/s)[sin(−30.0°)]2 − (2)(9.81m/s2)(−45 m)9.81 m/s2
vi(sin q) ± [−vi(sin q)]2− 42
g(∆y)
22
g
25.0 m/s
(9.81 m/s2)(73.0 m)2
(2)[−cos(–8.00°)]2 (42.5 m)
(9.81 m/s2)(73.0 m)2
(2)[cos(−8.00°)]2 (−10.3 m + 52.8 m)
(9.81 m/s2)(73.0 m)2
(2)[cos(−8.00°)]2 [(73.0 m)(tan[−8.00°]) − (−52.8 m)]
g∆x2
2(cos q)2 [∆x(tan q) − ∆y]
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Holt Physics Solution ManualV Ch. 3–14
V
10. ∆x1 = 0.46 m ∆xtot = ∆x1 + ∆x2
∆x2 = 4.00 m ∆t = vi(c
∆o
x
s q)
q = 41.0° ∆y = vi(sin q)∆t − 12
g∆t2 = vi(sin q)∆vi
x
(i
c
+os
∆qx
)2 − 1
2g∆vx
i(1
co
+s
∆qx
)2
2
∆y = − 0.35 m ∆y = (∆x1 + ∆x2)(tan q) − g
2
(∆v
x
i21
(c
+o
∆s
x
q2
)2)2
g = 9.81 m/s2 vi = vi = vi = vi = vi = 6.36 m/s
(9.81 m/s2)(4.46 m)2
(2)(cos 41.0°)2 (4.23 m)
(9.81 m/s2)(4.46 m)2
(2)(cos 41.0°)2(3.88 m + 0.35 m)
(9.81 m/s2)(0.46 m + 4.00 m)2
(2)(cos 41.0°)2 [(0.46 m + 4.00 m)(tan 41.0°) − (−0.35 m)]
g(∆x1 + ∆x2)2
2(cos q)2[(∆x1 + ∆x2)(tan q) − ∆y]
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1. vbw = 58.0 km/h, forward vbe = vbw + vwe = + 58.0 km/h + (−55.0 km/h) = +3.0 km/h= +58.0 km/h
∆t = v
∆
b
x
e =
3
1
.0
.4
k
k
m
m
/h
vwe = 55.0 km/h, backward
∆t == −55.0 km/h
∆x = 1.4 km
0.47 h = 28 min
Additional Practice 3F
2. vwe = 1.50 m/s, west vme = vmw + vwe = 4.20 m/s + 1.50 m/s = 5.70 m/s, west
vmw = 4.20 m/s, west time of travel with walkway:
∆x = 8.50 × 102m∆t1 =
v
∆
m
x
e =
8.5
5
0
.7
×0
1
m
0
/
2
s
m = 149 s
time of travel without walkway:
∆t2 = v
∆
m
x
w =
8.5
4
0
.2
×0
1
m
0
/
2
s
m = 202 s
time saved = ∆t2 − ∆t1 = 202 s − 149 s =
3. v1e = 286 km/h, forward v12 + v2e = v1e
v2e = 252 km/h, forward v12 = v1e − v2e
∆x = 0.750 km v12 = v1e − v2e = 286 km/h − 252 km/h = 34 km/h
∆t = v
∆
1
x
2 −
0
3
.
4
75
k
0
m
k
/
m
h = 2.2 × 10−2 h
∆t = (2.2 × 10−2 h)36
1
0
h
0 s = 79 s
53 s
Section Five—Problem Bank V Ch. 3–15
V
4. vma = 1.10 m/s, east vme = vma + vae
vae = 5.0 km/h at 35° Find the mosquito’s speed with respect to Earth in the x direction.west of south
vx, me = vx, ma + vx, ae = vma + vae(cos qae)∆x = 540 m
qae = –90.0° − 35° = −125°
vx, me = 1.10 m/s + (5.0 km/h)(103 m/km)36
1
0
h
0 s[cos(−125°)] = 1.10 m/s
+ (−0.80 m/s) = 0.30 m/s
∆t = vx
∆
,
x
me =
0
5
.3
4
0
0
m
m
/s
∆t =
5. vga = 150 km/h at 7.0° vge = vga + vaebelow horizontal
Find the glider’s speed with respect to Earth in the y (vertical) direction.vae = 15 km/h upward
vy, ge = vga + vy, ae = vga(sin qga) + vae∆y = −165 m
qga = −7.0°
vy, ge = (150 km/h)[sin(−7.0°)] + 15 km/h = −18 km/h + 15 km/h = −3 km/h
Time of descent with updraft:
∆t = v
∆
y,
y
ge =
∆t =
Time of descent without updraft:
∆t′ = v
∆
y,
y
ga =
∆t′ =
6. vfc = 87 km/h, west vfe = vfc + vce
vce = 145 km/h, north vx, fe = vx, fc + vx, ce = vfc = −87 km/h
∆t = 0.45 s vy, fe = vy, fc + vy, ce = vce = +145 km/h
∆x = vx, fe ∆t = (−87 km/h)(103 m/km)(1 h/3600 s)(0.45 s) = −11 m
∆x =
∆y = vy, fe ∆t = (145 km/h)(103 m/km)(1 h/3600 s)(0.45 s)
∆y = 18 m, north
11m, west
33 s
−166 m(−18 km/h)(103 m/km)(1h/3600 s)
200 s
−166 m(−3 km/h)(103 m/km)(1 h/3600 s)
1800 s = 3.0 × 101 min
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Holt Physics Solution ManualV Ch. 3–16
V
7. vaw = 55.0 km/h, north vae = vaw + vwe
vwe = 40.0 km/h at 17.0° vx, ae = vx, aw + vx, we = vwe (cos qwe)
north of west vy, ae = vy, aw + vy, we = vaw + vwe(sin qwe)
qwe = 180.0° − 17.0° = 163.0°
vx, ae = (40.0 km/h)(cos 163.0°) = −38.3 km/h
vy, ae = 55.0 km/h + (40.0 km/h)(sin 163.0°) = 55.0 km/h + 11.7 km/h = 66.7 km/h
vae =√
vx, ae)2+(vy, ae02 =√
(−38.3 km/h)2 + (66.7 km/h)2
vae =√
1.47 × 103 km2/h2+ 4.45× 103 km2/h2 =√
5.92 × 103 km2/h2
vae =
q = tan−1 v
v
x
y,
,
a
a
e
e = tan−1 = −60.1°
q =
8. vae = 76.9 km/h at 29.9° ∆x = vae(cos qae)∆twest of north ∆y = vae(sin qae)∆t
∆t = 15.0 min qae = 90.0° + 29.9° = 119.9°
∆x = (76.9 km/h)(cos 119.9°)(15.0 min)(1 h/60 min) = −9.58 km
∆y = (76.9 km/h)(sin 119.9°)(15.0 min)(1 h/60 min) = 16.7 km
∆x =
∆y =
9. vtc = 51 km/h, east vte = vtc + vce
vce = 4.0 km/h, south vx, te = vx, tc + vx, ce = vtc = 51 km/h
∆t = 14 s vy, te = vy, tc + vy, ce = vce = −4.0 km/h
∆x = vx, tc∆t = (51 km/h)(103 m/km)(1 h/3600 s)(14 s) = 2.0 × 102 m
∆y = vy, te∆t = (4.0 km/h)(103 m/km)(1 h/3600 s)(14 s) = 16 m
the torpedo must be fired 16 m north of the target
the target is 2.0 × 102 m away
16.7 km, north
9.58 km, west
60.1° west of north
66.7 km/h−38.3 km/h
76.9 km/h
Givens Solutions
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Section Five—Problem Bank V Ch. 3–17
V
10. vbw = 12.0 km/h, south vbe = vbw + vwe
vwe = 4.0 km/h at 15.0° vx, be = vx, bw + vx, we = vwe(cos qwe)south of east vy, be = vy, bw + vy, we = vbw + vwe(sin qwe)
qwe = −15.0°
vx, be = (4.0 km/h)[cos(−15.0°)] = 3.9 km/h
vy, be = (−12.0 km/h) + (4.0 km/h)[sin(−15.0°)] = (−12.0 km/h) + (−1.0 km/h) = −13.0 km/h
vbe =√
(vx, be)2+ vy,be)2 =√
(3.9 km/h)2 + (−13.0 km/h)2
vbe =√
15 km2/h2+ 169 km2/h2 =√
184km2/h2
vbe =
q = tan−1vvx
y,
,
b
b
e
e = tan−1−3
1
.
3
9
.0
km
km
/h
/h = −73°
q = 73° south of east
13.6 km/h
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Section Five—Solution Manual V Ch. 4–1
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4ChapterForces and the Laws of Motion
1. F1 = 7.5 × 104 N north
F2 = 9.5 × 104 N at 15.0°north of west
q1 = 90.0°
q2 = 180.0° − 15.0° = 165.0°
Fx,net = Fx = F1(cos q1) + F2(cos q2)
Fx,net = (7.5 × 104 N)(cos 90.0°) + (9.5 × 104 N)(cos 165.0°)
Fx,net = −9.2 × 104 N
Fy,net = Fy = F1(sin q1) + F2(sin q2)
Fy,net = (7.5 × 104 N)(sin 90.0°) + (9.5 × 104 N)(sin 165.0°)
Fy,net = 7.5 × 104 N + 2.5 × 104 N = 10.0 × 104 N
q = tan−1 = tan−1 = −47°
q = 47° north of west
10.0 × 104 N−9.2 × 104
Fy,netFx,net
Additional Practice 4A
Givens Solutions
2. F1 = 6.00 × 102 N north
F2 = 7.50 × 102 N east
F3 = 6.75 × 102 N at 30.0°south of east
q1 = 90.0°
q2 = 0.00°
q3 = −30.0°
Fx,net = Fx = F1(cos q1) + F2(cos q2) + F3(cos q3) = (6.00 × 102 N)(cos 90.0°)
+ (7.50 × 102 N)(cos 0.00°) + (6.75 × 102 N)[cos(−30.0°)]
Fx,net = 7.50 × 102 N + 5.85 × 102 N = 13.35 × 102 N
Fy,net = Fy = F1(sin q1) + F2(sin q2) + F3(sin q3) = (6.00 × 102 N)(sin 90.0°)
+ (7.50 × 102 N)(sin 0.00°) + (6.75 × 102 N)[sin (−30.0°)]
Fy,net = 6.00 × 102 N + (−3.38 × 102 N) = 2.62 × 102 N
q = tan−1 = tan−1 q = 11.1° north of east
2.62 × 102 N13.35 × 102 N
Fy,netFx,net
3. F1 = 2280.0 N upward
F2 = 2250.0 N downward
F3 = 85.0 N west
F4 = 12.0 N east
Fy,net = Fy = F1 + F2 = 2280.0 N + (−2250.0 N) = 30.0 N
Fx,net = Fx = F3 + F4 = −85.0 N + 12.0 N = −73.0 N
q = tan−1 = tan−1 = −22.3°
q = 22.3° up from west
30.0 N−73.0 N
Fy,netFx,net
4. F1 = 6.0 N
F2 = 8.0 N
Fmax = F1 + F2 = 6.0 N + 8.0 N
Fmax =Fmin = F2 − F1 = 8.0 N − 6.0 N
Fmin = 2.0 N
14.0 N
Holt Physics Solution ManualV Ch. 4–2
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Fx,net = F1 + F3(cos q) = 0F3(cos q) = −F1 = −3.0 N
Fy,net = F2 + F3(sin q) = 0F3(sin q) = −F2 = −(−4.0 N) = 4.0 N
F3 = √
[F3(cosq)]2+ [F3(sin q)]2 = √
(−3.0 N)2 + (4.0 N)2 = √
9.0N2+ 16N2 = √
25 N2
F3 =
q = tan−1 = tan−1 = −53°
q = 53° north of west
4.0 N−3.0 N
F3(sin q)F3(cos q)
5.0 N
Givens Solutions
5. F1 = 3.0 N east
F2 = 4.0 N south
6. F1 = 4.00 × 103 N east
F2 = 5.00 × 103 N north
F3 = 7.00 × 103 N west
F4 = 9.00 × 103 N south
Fx,net = F1 + F3 = 4.00 × 103 N + (−7.00 × 103 N) = −3.00 × 103 N
Fy,net = F2 + F4 = 5.00 × 103 N + (−9.00 × 103 N) = −4.00 × 103 N
Fnet = √
(Fx,net)2+ (Fy,net)2 = √
(−3.00 × 103 N)2 + (−4.00 × 103 N)2
Fnet = √
9.00 × 106 N2+ 16.0× 106 N2 = √
25.0 × 106 N2
Fnet =
q = tan−1 = tan−1 q = 53.1° south of west
−4.00 × 103 N−3.00 × 103 N
Fy,netFx,net
5.00 × 103 N
7. F1 = 15.0 N
q = 55.0°
Fy = F(sin q) = (15.0 N)(sin 55.0°)
Fy =Fx = F(cos q) = (15.0 N)(cos 55.0°)
Fx = 8.60 N
12.3 N
8. F = 76 N
q = 40.0°
Fx = F(cos q) = (76 N)(cos 40.0°)
Fx =Fy = F(sin q) = (76 N)(sin 40.0°)
Fy = 49 N
58 N
9. F1 = 350 N
q1 = 58.0°
F2 = 410 N
q2 = 43°
Fy,net = F1(sin q1) + F2(sin q2) = (350 N)(sin 58°) + (410 N)(sin 43)
Fy,net = 3.0 × 102 N + 2.8 × 102 N
Fy,net = 580 N
10. F1 = 7.50 × 102 N
q1 = 40.0°
F2 = 7.50 × 102 N
q2 = −40.0°
Fy,net = Fg = F1(cos q1) + F2(cos q2)
Fy,net = (7.50 × 102 N)(cos 40.0°) + (7.50 × 102 N) [cos(−40.0°)]
Fg = 575 N + 575 N = 1.150 × 103 N
Section Five—Solution Manual V Ch. 4–3
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1. Fnet = 2850 N
vf = 15 cm/s
vi = 0 cm/s
∆t = 5.0 s
anet = = = 3.0 cm/s2 = 3.0 × 10−2 m/s2
Fnet = m anet
m = =
m = 9.5 × 104 kg
2850 N3.0 × 10−2 m/s2
Fnetanet
15 cm/s − 0 cm/s
5.0 s
vf − vi∆t
2. ∆t = 1.0 m/s
∆t = 5.0 s
Fdownhill = 18.0 N
Fuphill = 15.0 N
anet = = = 0.20 m/s2
Fnet = m anet = Fdownhill − Fuphill = 18.0 N − 15.0 N = 3.0 N
m = = = 15 kg3.0 N0.20 m/s2
Fnetanet
1.0 m/s
5.0 s
∆v∆t
3. Fmax = 4.5 × 104 N
anet = 3.5 m/s2
g = 9.81 m/s2
Fnet = m anet = Fmax − mg
m(anet + g) = Fmax
m = = = = 3.4 × 103 kg4.5 × 104 N13.3 m/s2
4.5 × 104 N3.5 m/s2 + 9.81 m/s2
Fmaxanet + g
4. m = 2.0 kg
∆y = 1.9 m
∆t = 2.4 s
vi = 0 m/s
∆y = vi ∆t + anet ∆t2
Because vi = 0 m/s, anet = = = 0.66 m/s2
Fnet = m anet = (2.0 kg)(0.66 m/s) = 1.32 N
Fnet = 1.32 N upward
(2)(1.9 m)
(2.4 s)22∆y∆t2
12
5. m = 8.0 kg
∆y = 20.0 cm
∆t = 0.50 s
vi = 0 m/s
g = 9.81 m/s2
∆y = vi ∆t + anet ∆t2
Because vi = 0 m/s, anet = = = 1.6 m/s2
Fnet = m anet = (8.0 kg)(1.6 m/s2) = 13 N
Fnet =
Fnet = Fupward − mg
Fupward = Fnet + mg = 13 N + (8.0 kg)(9.8 m/s2) = 13 N + 78 N = 91 N
Fupward = 91 N upward
13 N upward
(2)(20.0 × 10−2 m)
(0.50 s)22∆y∆t2
12
6. m = 75 kg
aforward = 0.15 m/s2 west
abackward = 2 × 10−2 east
anet = aforward − abackward = 0.15 m/s2 − 2 × 10−2 m/s2
anet = 0.13 m/s2 west
Fnet = m anet = (75 kg)(0.13 m/s) = 9.8 N
Fnet = 9.8 N west
Additional Practice 4B
Givens Solutions
Givens Solutions
Holt Physics Solution ManualV Ch. 4–4
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7. Fnet = −65.0 N
m = 0.145 kganet = = = −448 m/s2−65.0 N
0.145 kg
Fnetm
8. m = 214 kg
Fbuoyant = 790 N
g = 9.81 m/s2
Fnet = Fbuoyant − mg = 790 N − (214 kg)(9.81 m/s2)
Fnet = 790 N − 2.10 × 103 N = −1310 N
anet = = = −6.12 m/s2−1310 N214 kg
Fnetm
9. m = 0.080 kg
q = 37.0°
g = 9.81 m/s2
Fnet = m anet = m g(sin q)
anet = g(sin q) = (9.81 m/s2)(sin 37.0°) = 5.90 m/s2
anet = 5.90 m/s2 down the incline (37.0° below horizontal)
10. m = 0.080 kg
Fupward = 1.40 N
adownward = 5.90 m/s2
Fnet = Fupward − m adownward = 1.40 N − (0.080 kg)(5.90 m/s2) = 1.40 N − 0.47 N = 0.93 N
Fnet = 0.93 N up the incline (37.0° above the horizontal)
anet = F
mnet =
0
0
.0
.9
8
3
0
N
kg = 12 m/s2
1. Fdownward = 4.26 × 107 N
mk = 0.25
Fnet = Fdownward − Fk = 0
Fk = mk Fn = Fdownward
Fn = = = 1.7 × 108 N4.26 × 107 N
0.25
Fdownwardmk
2. Fn = 1.7 × 108 N
q = 10.0°
g = 9.81 m/s2
Fn = mg (cos q)
m = = = 1.8 × 107 kg1.7 × 108 N
(9.81 m/s2)(cos 10.0°)
Fng (cos q)
3. Fs,max = 2400 N
ms = 0.20
q = 30.0°
g = 9.81 m/s2
Fs,max = ms Fn
Fn = = = 1.2 × 104 N
Fn =
Fn = mg(cos q)
m = = = 1400 kg1.2 × 104 N
(9.81 m/s2)(cos 30.0°)
Fng (cos q)
1.2 × 104 N perpendicular to and away from the incline
2400 N
0.20
Fs,maxms
Additional Practice 4C
4. m = 60.0 kg
a = 3.70 m/s2
ms = 0.455
g = 9.81 m/s2
For the passenger to remain standing without sliding,
Fs,max ≥ F = ma
Fs,max = ms Fn = ms mg
ms mg ≥ ma
ms g ≥ a
(0.455)(9.81 m/s2) = 4.46 m/s2 > 3.70 m/s2
The passenger will be able to stand without sliding.
Section Five—Solution Manual V Ch. 4–5
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Givens Solutions
5. m = 90.0 kg
q = 17.0°
g = 9.81 m/s2
Fnet = mg(sin q) − Fk = 0
Fk = mg(sin q) = (90.0 kg)(9.81 m/s2)(sin 17.0°) = 258 N
Fk = 258 N up the slope
6. msled = 47 kg
msupplies = 33 kg
mk = 0.075
q = 15°
Fk = mk Fn = mk(msled + msupplies)g = (0.075)(47 kg + 33kg)(9.81 m/s2)
Fk = (0.075)(8.0 × 101 kg)(9.81 m/s2)
Fk =
Fk = mk Fn = mk(msled + msupplies)g(cos q) = (0.075)(47 kg + 33 kg)(9.81 m/s2)(cos 15°)
Fk = (0.075)(8.0 × 101 kg)(9.81 m/s2)(cos 15°) = 57 N
59 N
7. m = 1.8 × 103 kg
q = 15.0°
Fs,max = 1.25 × 104 N
g = 9.81 m/s2
Fs,max = ms Fn = ms mg(cos q)
ms = mg
F
(s
c,m
oa
sx
q) =
ms = 0.73
1.25 × 104 N(1.8 × 103 kg)(9.81 m/s2)(cos 15.0°)
8. m = 15.0 g
q = 2.3°
g = 9.81 m/s2
Fnet = mg(sin q) − Fk = 0
Fk = mk Fn = mk mg(cos q)
mk = m
m
g
g
(
(
c
si
o
n
s
)
) = tan q = tan 2.3°
mk = 0.040
9. vf = 88.0 km/h
vi = 0 km/h
∆t = 3.07 s
g = 9.81 m/s2
Fnet = Fapplied − Fs,max = 0
Fs,max = ms Fn = ms mg
Fapplied = m a
m a = ms mg
ms = a
g =
vf
g∆−
t
vi =
ms = 0.812
(88.0 km/h − 0 km/h)(103 m/km)(1 h/3600 s)
(9.81 m/s2)(3.07 s)
10. q = 5.0° Fnet = mg(sin q) − Fk = 0
Fk = mk Fn = mk mg(cos q)
mk = m
m
g
g
(
(
c
si
o
n
s
)
) = tan q = tan 5.0°
mk = 0.087
Holt Physics Solution ManualV Ch. 4–6
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3. m = 5.1 × 102 kg
q = 14°
Fapplied = 4.1 × 103 N
g = 9.81 m/s2
ms = =
ms = 0.40
1.00 × 103 N(266 kg)(9.81 m/s2)(cos 17°)
1760 − 760 N(266 kg)(9.81 m/s2)(cos 17°)
1. anet = 1.22 m/s2
q = 12.0°
g = 9.81 m/s2
Fnet = m anet = mg(sin q) − Fk
Fk = mk Fn = mk mg(cos q)
m anet + mk mg(cos q) = mg(sin q)
mk = g(s
g
in
(c
qo
)
s
−q
a
)net = =
mk = = 0.0850.82 m/s2
(9.81 m/s2)(cos 12.0°)
2.04 m/s2 − 1.22 m/s2
(9.81 m/s2)(cos 12.0°)
(9.81 m/s2)(sin 12.0°) − 1.22 m/s2
(9.81 m/s2)(cos 12.0°)
2. Fapplied = 1760 N
q = 17.0°
m = 266 kg
g = 9.81 m/s2
Fnet = Fapplied − mg(sin q) − Fs,max = 0
Fs,max = ms Fn = ms mg(cos q)
ms mg(cos q) = Fapplied − mg(sin q)
ms = Fappl
m
ied
g
−(c
m
os
g
q(s
)
in q) =
1760 − (266 kg)(9.81 m/s2)(sin 17°)
(266 kg)(9.81 m/s2)(cos 17°)
Fnet = Fapplied − mg(sin q) − Fs,max = 0
Fs,max = ms Fn = ms mg(cos q)
ms mg(cos q) = Fapplied − mg(sin q)
ms = Fappl
m
ied
g
−(c
m
os
g
q(s
)
in q) =
ms = =
ms = 0.60
2.9 × 103 N(5.1 × 102 kg)(9.81 m/s2)(cos 14°)
4.1 × 103 N − 1.2 × 103 N(5.1 × 102 kg)(9.81 m/s2)(cos 14°)
4.1 × 103 N − (5.1 × 102 kg)(9.81 m/s2)(sin 14°)
(5.1 × 102 kg)(9.81 m/s2)(cos 14°)
Additional Practice 4D
Givens Solutions
4. Fapplied = 5.0 N to the left
m = 1.35 kg
anet = 0.76 m/s2 to the left
Fnet = m anet = Fapplied − Fk
Fk = Fapplied − m anet
Fk = 5.0 N − (1.35 kg)(0.76 m/s2) = 5.0 N − 1.0 N = 4.0 N
Fk = 4.0 N to the right
5. mk = 0.20
g = 9.81 m/s2
Fnet = m anet = Fk
Fk = mkFh = mkmg
anet = mk
m
mg = mkg = (0.20)(9.81 m/s2)
anet = 2.0 m/s2
Section Five—Solution Manual V Ch. 4–7
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6. Fapplied = 2.50 × 102 N
m = 65.0 kg
q = 18.0°
anet = 0.44 m/s2
Givens Solutions
Fnet = m anet = Fapplied − mg(sin q) − Fk
Fk = Fapplied − mg(sin q) − manet
Fk = 2.50 × 102 N − (65.0 kg)(9.81 m/s2)(sin 18.0°) − (65.0 kg)(0.44 m/s2)
Fk = 2.50 × 102 N − 197 N − 29 N = 24 N = 24 N downhill
7. m = 65.0 kg
g = 9.81 m/s2
Fk = 24 N
q = 18.0°
Fnet = m anet = mg(sin q) − Fk
anet = g(sin q) − F
mk = (9.81 m/s2)(sin 18.0°) −
6
2
5
4
.0
N
kg = 3.03 m/s2 − 0.37 m/s2 = 2.66 m/s2
anet = 2.66 m/s2 downhill
8. Fapplied = 3.00 × 102 N
q = 20.0°
mk = 0.250
Fx,net = Fapplied(cos q) − Fk = 0Fy,net = Fn − mg + Fapplied(sin q) = 0
Fk = mkFn
Fn = Fapplie
md(
k
cos q) = =
m = Fn + Fapp
glied(sin q) =
m = 113
9
0
.8
N
1
−m
1
/s
023 N
= 9.
1
8
0
1
3
m
0 N
/s2 = 105 kg
1130 N + (3.00 × 102 N)[sin(−20.0°)]
9.81 m/s2
1130 N(3.00 × 102 N)[cos(−20.0°)]
0.25°
9. Fapplied = 590 N
Fdownhill = 950 N
ms = 0.095
q = 14.0°
Fnet = Fapplied + Fs,max − Fdownhill = 0Fs,max = ms Fn = ms mg(cos q)
ms Fn = Fdownhill − Fapplied
Fn = Fdownhill
m−
s
Fapplied = 950 N
0.0
−9
5
5
90 N =
3
0
6
.0
0
9
N
5 = 3800 N
Fn =
m = g(c
F
on
s q) = = 4.0 × 102 kg
3800 N(9.81 m/s2)(cos 14.0°)
3800 N perpendicular to and up from the ground
10. anet = 1.20 m/s2
Fapplied = 1.50 × 103 N
q = −10.0°
ms = 0.650
g = 9.81 m/s2
Fx,net = Fapplied(cos q) − Fs,max = 0Fs,max = msFn
Fn = Fapplie
md(
s
cos q) = = 2.27 × 103 N
Fn =
Fy,net = m anet = Fn − mg + Fapplied(sin q)
m(anet + g) = Fn + Fapplied(sin q)
m = Fn + F
aa
n
p
e
p
t
li
+ed(
g
sin q) =
m = = 2
1
.0
1
1
.0
×1
1
m
0
/
3
s
N2 = 183 kg
2.27 × 103 N − 2.60 × 102 N
11.01 m/s2
2.27 × 103 N + (1.50 × 103 N)[sin(−10.0°)]
1.20 m/s2 + 9.81 m/s2
2.27 × 103 N, upward
(1.50 × 103 N)[cos(−10.0°)]
0.650
Section Five—Problem Bank V Ch. 5–1
Chapter 5Work and Energy
V
1. d = 3.00 × 102 m
W = 2.13 × 106 J
q = 0°
F = d(c
W
os q) = = 7.10 × 3 N
2.13 × 106 J(3.00 × 102 m)(cos 0°)
Additional Practice 5A
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2. d = 76.2 m
Wnet = 1.31 × 103 J
q = 0°
Fnet = d(
W
con
set
q ) =
(76
1
.
.
2
31
m
×)(
1
c
0
o
3
s
J
0°) = 17.2 N
3. W = 1800 J
d1 = 1.5 m
d2 = 5.0 m
q = 0°
W = F1 d1 = (cos q) = F2 d2 (cos q)
F1 = d1 (c
W
os q) =
(1.5
1
m
8
)
0
(
0
co
J
s 0°) =
F2 = d2 (c
W
os q) =
(5.0
1
m
8
)
0
(
0
co
J
s 0°) = 3.6 × 102 N
1.2 × 103 N
4. Wnet = 4.27 × 103 J
d = 17 m
q = 0°
Fnet = d(
W
con
set
q) =
(1
4
7
.2
m
7
)
×(c
1
o
0
s
3
0
J
°) = 2.5 × 102 N
5. F = 1.6 N
d = 1.2 m
q = 180°
W = Fd(cos q) = (1.6 N) (1.2 m) (cos 180°) = −1.9 J
6. d = 15.0 m
Fapplied = 35.0 N
q1 = 20.0°
Fk = 24.0 N
q2 = 180°
Wnet = Fapplied d (cos q1) + Fk d (cos q2)
Wnet = (35.0 N) (15.0 m) (cos 20.0°) + (24.0 N) (15.0 m) (cos 180°)
Wnet = 493 J + (−3.60 × 102 J)
Wnet =
Alternatively,
Wnet = Fnet d (cos q ′)
q ′ = 0°
Fnet = Fapplied (cos q1) − Fk
Wnet = [Fapplied (cos q1) − Fk] d (cos q ′)
Wnet = [(35.0 N) (cos 20.0°) − 24.0 N] (15.0 m) (cos 0°)
Wnet = (32.9 N − 24.0 N) (15.0 m) = (8.9 N)(15.0 m) = 130 J
133J
1. m = 7.5 × 107 kg
v = 57 km/h
KE = 12
mv2 = 12
(7.5 × 107 kg) [(57 km/h)(103 m/km)(1h/3600 s)]2
KE = 9.4 × 109 J
Holt Physics Solution ManualV Ch. 5–2
V
7. v1 = 88.9 m/s
vf = 0 m/s
∆t = 0.181 s
d = 8.05 m
m = 70.0 kg
q = 180°
W = Fd(cos q)
F = ma
a = ∆∆
v
t =
vf
∆−t
vi
W = m (v
∆f
t
− vi) d (cos q) = (8.05 m)(cos 180°)
W =
W = 2.77 × 105 J
(70.0 kg)(88.9 m/s)(8.05 m)
(0.181 s)
(70.0 kg)(0 m/s − 88.9 m/s)
(0.181 s)
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8. F = 715 N
W = 2.72 × 104 J
q = 0°
d = F (c
W
os q) =
(7
2
1
.
5
72
N
×)(
1
co
0
s
4
0
J
°) = 38.0 m
9. Fnet = 7.25 × 10−2 N
Wnet = 4.35 × 10−2 J
q = 0°
d = Fnet
W
(cn
oet
s q) = = 0.600 m
4.35 × 10−2 J(7.25 × 10−2 N)(cos 0°)
10. W = 6210 J
F = 2590 N
q = 0°
d = F (c
W
os q) =
(2590
6
N
21
)
0
(c
J
os 0°) = 2.398 m
Additional Practice 5B
2. v = 15.8 km/s KE = 12
mv2 = 12
(0.20 × 10−3 kg)(15.8 × 103 m/s)2
m = 0.20 g KE =
3. v = 35.0 km/h KE = 12
mv2 = 12
(9.00 × 102 kg) [(35.0 km/h)(103 m/km)(1 h/3600 s)]2
m = 9.00 × 102 kg KE =
4. v1 = 220.0 km/h KE1 = 12
m1 v12 = 1
2 (8.84 × 105 kg)[(220.0 km/h)(103 m/km)(1 h/3600 s)]2
m1 = 8.84 × 105 kg KE1 =
v2 = 320.0 km/h KE2 = 12
m2 v22 = 1
2 (4.80 × 105 kg)[(320.0 km/h)(103 m/km)(1h/3600 s)]2
m2 = 4.80 × 105 kg KE2 =
5. KE = 2.78 × 109 J
v = 275 km/h
1.90 × 109 J
1.65 × 109 J
4.25 × 104 J
2.5 × 104 J
m = 2
v
K2E
=
m = 9.53 × 105 kg
(2)(2.78 × 109 J)[(275 km/h)(103 m/km)(1h/3600 s)]2
Section Five—Problem Bank V Ch. 5–3
V
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6. v = 850 km/h
KE = 9.76 × 109 Jm =
2
v
K2E
=
m = 3.50 × 105 kg
(2)(9.76 × 109 J)[(850 km/h)(103 m/km)(1 h/3600 s)]2
7. v = 9.78 m/s
KE = 6.08 × 104 Jm =
2
v
K2E
= (2)
(
(
9
6
.
.
7
0
8
8
m
×/
1
s
0
)2
4 J) = 1.27 × 103 kg
8. KE = 7.81 × 104 J
m = 55.0 kgv =
2
m
K
E
= (2)(7
5
.8
5
1
.0
×kg104
J) = 53.3 m/s
10. KEA,i = 12
KEB
vA,f = vA,i + 1.3 m/s
KEA,f = KEB
mA = 2.0 mB
1. vi = 8.0 m/s
vf = 0 m/s
d = 45 m
Fk = 0.12 N
q = 180°
Wnet = ∆KE = KEf − KEi = 12
mvf2 − 1
2 mvi
2
Wnet = Fnetd(cos q) = Fkd(cos q)
12
m(vf2 − vi
2) = Fkd(cos q)
m = 2 F
vk
f2d(
−co
vi
s2q)
= = −(2)(
−0
6
.1
4
2
m
N2)
/
(
s
425 m)
m = 0.17 kg
(2)(0.12 N)(45 m)(cos 180°)
(0 m/s)2 − (8.0 m/s)2
9. KE = 1433 J
m = 47.0 gv =
2
m
K
E
= 47(
.
20
)(
×1 4
13
0
3−3
J)
kg = 247 m/s
KEA,i = 12
KEB
12
mAvA,i2 = 1
21
2 mBvB
212
(2mB)vA,i2 = 1
4 mBvB
2
vA,i2 = 1
4 vB
2, or vB
2 = 4vA,i2
vA,i = 12
vB
KEA,f = KEB
12
mAvA,f2 = 1
2 mBvB
2
12
(2mB)(vA,i + 1.3 m/s)2 = 12
mBvB2
(vA,i + 1.3 m/s)2 = 12
vB2 = 1
2 (4vA,i
2) = 2vA,i2
vA,i2 + (2.6 m/s) vA,i + 1.7 m2/s2 = 2vA,i
2
vA,i2 − (2.6 m/s) vA,i − 1.7 m2/s2 = 0
Using the quadratic equation,
vA,i = =
vA,i = 2.6 m/s ±√
2
13.6 m2/s2 = 2.6 m/s ±
2
3.69 m/s =
6.3
2
m/s =
vA,i = 12
vB
vB = vA,i = (2)(3.2 m/s) = 6.4 m/s
3.2 m/s
2.6 m/s ±√
6.8m2/s2 + 6.8 m2/s22
2.6 m/s ±√
(−2.6m/s)2 − 4(−1.7m2/s2)2
Additional Practice 5C
Holt Physics Solution ManualV Ch. 5–4
V
2. vi = 15.00 km/s
vf = 14.97 km/s
Fr = 9.00 × 10–2 N
d = 500.0 km
q = 180°
Wnet = ∆KE =KEf − KEi = 12
mvf2 − 1
2 mvi
2
Wnet = Fnetd(cos q) = Fr d (cos q)
12
m(vf2 − vi
2) = Fr d (cos q)
m = 2 F
vr
f
d2 −
(c
v
o
i
s2
q) =
m = = −−9
9
×.0
1
0
0
×8
1
m
02
4
/s
J2
m = 1.00 × 10−4 kg
−(2)(9.00 × 10−2 N)(500.0 × 103 m)2.241 × 108 m2/s2 − 2.250 × 108 m2/s2
(2)(9.00 × 10−2 N)(500.0 × 103 m)(cos 180°)
(14.97 × 103 m/s)2 − (15.00 × 103 m/s)2
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3. vi = 48.0 km/h
vf = 59.0 km/h
d = 100.0 m
m = 1100 kg
q = 0°
Wnet = ∆KE = KEf − KEi = 12
mvf2 − 1
2 mvi
2
Wnet = Fnet d (cos q)
Fnet d (cos q) = 12
m(vf2 − vi
2)
Fnet = m
2
(
d
vf
(
2
co
−s
v
qi2
)
) =
Fnet =
Fnet =
Fnet = 5.01 × 102 N
(1100 kg)(1.18 × 109 m2/h2)(1h/3600 s)2
(2)(100.0 m)
(1100 kg)(3.48 × 109 m2/h2 − 2.30 × 109 m2/h2)(1h/3600 s)2
(2)(100.0 m)
(1100 kg)[(59.0 km/h)2 − (48.0 km/h)2](103 m/km)2 (1h/3600 s)2
(2)(100.0 m)(cos 0°)
4. m = 450 kg
d = 7.0 m
vf = 1.1 m/s
vi = 0 m/s
q = 0°
Wnet = ∆KE = KEf − KEi = 12
mvf2 − vi
2
Wnet = Fnetd(cos q)
Fnetd(cos q) = 12
m(vf2 − vi
2)
Fnet = m
2
(
d
vf
(
2
co
−s
v
qi2
)
) = =
(450
(
k
2
g
)
)
(
(
7
1
.0
.2
m
m
)
2/s2)
Fnet = 39 N
(450 kg)[(1.1 m/s)2 − (0 m/s)2]
(2)(7.0 m)(cos 0°)
5. vi = 2.40 × 102 km/h
vf = 0 km/h
anet = 30.8 m/s2
m = 1.30 × 104 kg
q = 180°
Wnet = ∆KE = KEf − KEi = 12
mvf2 − vi
2
Wnet = Fnet d (cos q) = manet d(cos q)
manetd(cos q) = 12
m(vf2 − vi
2)
d = 2 a
v
n
f
e
2
t (
−c
v
oi
s
2
q) =
d = = 72.2 m(−5.76 × 104 km2/h2)(103 m/km)2 (1h/3600 s)2
−(2)(30.8 m/s2)
[(0 km/h)2 − (2.40 × 102 km/h)2] (103 m/km)2(1h/3600 s)2
(2)(30.8 m/s2)(cos 180°)
Section Five—Problem Bank V Ch. 5–5
V
6. m = 50.0 kg
vi = 47.00 m/s
vf = 5.00 m/s
Wnet = ∆KE = KEf − KEi = 12
mvf2 − 1
2 mvi
2
Wnet = 12
m(vf2 − vi
2) = 12
(50.0 kg)[(5.00 m/s)2 − (47.00 m/s)2]
Wnet = 12
(50.0 kg)(25.0 m2/s2 − 2209 m2/s2) = 12
(50.0 kg)(−2184 m2/s2)
Wnet = −5.46 × 104 J
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7. m = 2.0 × 106 kg
d = 7.5 m
anet = 7.5 × 10−2 m/s2
KEi = 0 J
Wnet = ∆KE = KEf − KEi = KEf
Wnet = Fnet d (cos q) = manetd(cos q)
KEf = manet d (cos q) = (2.0 × 106 kg)(7.5 × 10−2 m/s2)(7.5 m)(cos 0°)
KEf = 1.1 × 106 J
8. Fapplied = 92 N
m = 18 kg
mk = 0.35
d = 7.6 m
g = 9.81 m/s2
q = 0°
KEi = 0 J
Wnet = ∆KE = KEf − KEi = KEf
Wnet = Fnet d (cos q)
Fnet = Fapplied − Fk = Fapplied − mk mg
KEf = (Fapplied − mk mg) d (cos q)
= [92 N − (0.35)(18 kg)(9.81 m/s2)](7.6 m)(cos 0°)
KEf = (92 N − 62 N)(7.6 m) = (3.0 × 101 N)(7.6 m)
KEf = 228 J
9. m = 2.00 × 102 kg
Fwind = 4.00 × 102 N
d = 0.90 km
vi = 0 m/s
q = 0°
Wnet = ∆KE = KEf − KEi = 12
mvf2 − 1
2 mvi
2
Wnet = Fnet d (cos q) = Fwind d(cos q)
12
mvf2 − 1
2 mvi
2 = Fwind d (cos q)
vf = 2Fwindm
d (cosq)+ vi
2 = + (0m/s)2vf = vf = 6.0 × 101 m/s
(2)(4.00 × 102 N)(9.0 × 102 m)
2.00 × 102 kg
(2)(4.00 × 102 N)(0.90 × 103 m)(cos 0°)
2.00 × 102 kg
10. m = 20.0 g
d = 2.5 m
Fforward = 7.3 × 10−2 N
mk = 0.20
vi = 0 m/s
g = 9.81 m/s2
q = 0°
Wnet = ∆KE = KEf − KEi = 12
mvf2 − 1
2 mvi
2
Wnet = Fnetd(cos q) = (Fforward − Fk) d(cos q) = (Fforward − mk mg)d(cos q)
12
mvf2 − 1
2 mvi
2 = (Fforward − mk mg)d(cos q)
vf = vf = vf = = vf = 2.9 m/s
(2)(3.4 × 10−2 N)(2.5 m)
2.00 × 10−2 kg
(2)(7.3 × 10−2 N − 3.9 × 10−2 N)(2.5 m)
20.0 × 10−3 kg
(2)[7.3 × 10−2 N − (0.20)(20.0 × 10−3 kg)(9.81 m/s2)](2.5 m)(cos 0°) + (0 m/s)2
20.0 × 10−3 kg
2[(Fforward − mk mg) d (cos q) + vi2]
m
Holt Physics Solution ManualV Ch. 5–6
V
1. x = 5.00 cm
KEcar = 1.09 × 10−4 J
Assuming all of the kinetic energy becomes stored elastic potential energy,
KEcar = PEelastic = 12
kx2
k = 2 PE
xe2lastic =
(
(
5
2
.
)
0
(
0
1.
×09
10
×−120
m
4
)
J2)
k = 8.72 × 106 N/m
Additional Practice 5D
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2. PEelastic = 5.78 × 107 J
x = 102 m for each spring
PEelastic = 2 12
kx2PEelastic = kx2
k = PE
xel
2astic =
5
(
.7
1
8
02
×m
10
)
7
2J
k = 5.56 × 103 N/m
3. m = 0.76 kg
x = 2.3 cm
g = 9.81 m/s2
elastic potential energy stored = decrease in gravitational potential energy
PEelastic = mgx
PEelastic = 12
kx2 = mgx
k = 2 m
x2gx
= 2 m
x
g =
k = 6.5 × 102 N/m
(2)(0.76 kg)(9.81 m/s2)
2.3 × 10−2 m
4. m = 5.0 kg
q = 25.0°
PEg = 2.4 × 102 J
PEg = mgh = mgd(sin q)
d = mg
P
(
E
si
g
n q) =
d = 12 m
2.4 × 102 J(5.0 kg)(9.81 m/s2)(sin 25.0°)
5. k = 1.5 × 104 N/m
PEelastic = 120 J
PEelastic = 12
kx2
x = ± 2PE
kelastic = ± 1.5
(2×) (
1
10
240
NJ)
/m
Spring is compressed, so negative root is selected.
x = −0.13 m = −13 cm
6. m = 1750 kg
PEg = 1.69 × 1010 J
g = 6.44 m/s2
PEg = mgh
h = P
m
E
g
g =
h = 1.50 × 106 m = 1.50 × 103 km
1.69 × 1010 J(1750 kg)(6.44 m/s2)
7. h = 7.0 m
PEg = 6.6 × 104 J
g = 9.81 m/s2
PEg = mgh
m = P
g
E
hg =
(9.81
6.
m
6
/
×s2
1
)
0
(
4
7.
J
0 m)
m = 9.6 × 102 kg
Section Five—Problem Bank V Ch. 5–7
V
8. PEg = 3.36 × 109 J
h = 1.45 km
PEg = mgh
m = P
g
E
h
g =
m = 2.36 × 105 kg
3.36 × 109 J(9.81 m/s2)(1.45 × 103 m)
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9. k = 550 N/m
x = −1.2 cm
PEelastic = 12
kx2 = 12
(550 N/m)(–1.2 × 10−2 m)2 = 4.0 × 10−2 J
1. m = 0.500 g
h = 0.250 km
g = 9.81 m/s2
PEi = KEf
mgh = KEf
KEf = (0.500 × 10−3 kg)(9.81 m/s2)(0.250 × 103 m) = 1.23 J
2. d = 96.0 m
q = 18.4°
m = 70.0 kg
g = 9.81 m/s2
a. PE1 = mgh = mgd(sin q)
PE2 = 0 J
MEi = PE1 + PE2 = mgd(sin q)
MEi = (70.0 kg)(9.81 m/s2)(96.0 m)(sin 18.4°) =
b. PE1 = 0 J
PE2 = mgh = mgd(sin q)
MEf = PE1 + PE2 = mgd(sin q) = MEi
MEf =
c. MEi = MEf
PE1,i + PE2,i + KE1 + KE2 = PE1,f + PE2,f + KE1 + KE2
The kinetic energy of each passenger remains unchanged during the trip once thecars are in motion, so
PE1,i + PE2,i = PE1,f + PE2,f
mgh1,i + 0 J = mgh1,f + PE2,f
PE2,f = mgh1,f − mgh1,f = mgd(sin q) − mgh1,f
PE2,f = (70.0 kg)(9.81 m/s2)(96 m)(sin 18.4°) − (70.0 kg)(9.81 m/s2)(20.0 m)
PE2,f = 2.1 × 104 J − 1.37 × 104 J
PE2,f = 7.0 × 103 J
2.08 × 104 J
2.08 × 104 J
10. h = 5334 m
m = 64.0 kg
g = 9.81 m/s2
PEg = mgh = (64.0 kg)(9.81 m/s2)(5334 m) = 3.35 × 106 J
Additional Practice 5E
v1 = v2 = 1.0 m/s
h1,f = 20.0 m
Holt Physics Solution ManualV Ch. 5–8
V
3. hi = 75.0 m
vi = 1.2 m/s + 3.5 m/s = 4.7 m/s
vf = 0 m/s
g = 9.81 m/s2
m = 20.0 g
PEi + KEi = PEf + KEf
mghi + 12
mvi2 = mghf + 1
2mvf
2
hf =12
m(vi2 − vf
2)+ hi = + himg
hf = + 75.0 m = 1.1 m + 75.0 m = 76.1 m(4.7 m/s)2 − (0 m/s)2
(2)(9.81 m/s2)
vi2 − vf
2
2g
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4. m = 25.0 kg
v = 12.5 m/s
g = 9.81 m/s2
PEi = KEf
mgh = 12
mv2
h = 2
v2
g =
(2
(
)
1
(
2
9
.
.
5
81
m
m
/s
/
)
s
2
2) = 7.96 m
5. m = 50.0 g
vi = 3.00 × 102 m/s
vf = 89.0 m/s
MEi + ∆ME = MEf
MEi = KEi = 12
mvi2
MEf = KEf = 12
mvf2
∆ME = MEf − MEi = 12
m(vf2 − vi
2)
∆ME = 12
(50.0 × 10−3 kg)[(89.0 m/s)2 − (3.00 × 102 m/s)2]
∆ME = 12
(5.00 × 10−2 kg)(7.92 × 103 m2/s2 − 9.00 × 104 m2/s2)
∆ME = 12
(5.00 × 10−2 kg) (−8.21 × 104 m2/s2)
∆ME = −2.05 × 103 J
6. m = 50.0 g
vi = 3.00 × 102 m/s
vf = 89.0 m/s
For upward flight,
PE1,f − KE1,i = ∆ME1
where
∆ME1 = Wnet,1 = Fnet,1 h (cos 180°) = (mg + Fresist) h
For downward flight,
KE2,f − PE2,i = ∆ME2
Where
∆ME2 = Wnet,2 = Fnet,2 h (cos 0°) = (mg − Fresist) h
Solving for h,
∆ME2 − ∆ME1 = (mg − Fresist) h − [−(mg + Fresist)h] = 2 mgh
h = ∆ME
22 −
m
∆g
ME1 =
KE2,f = 12
mvf2
KE1,i = 12
mvi2
PE1,f = PE2,i = mgh
h = = vf
2
4
+g
vi2
− h
h = vf
2
8
+g
vi2
=(89.0 m/s)2 + (3.00 × 102 m/s)2
(8)(9.81 m/s2)
12
mvf2 − mgh − mgh + 1
2 mvi
2
2 mg
(KE2,f − PE2,i) − (PE1,f − KE1,i)2 mg
Section Five—Problem Bank V Ch. 5–9
V
h = = 9
(
.7
8
9
)(
×9.
1
8
0
1
3
m
m
/s
2
2/
)
s2
h = 1.25 × 103 m = 1.25 km
7.92 × 103 m2/s2 + 9.00 × 104 m2/s2
(8)(9.81 m/s2)
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7. m = 50.0 kg
k = 3.4 × 104 N/m
x = 0.65 m
hf = 1.00 m − 0.65 m = 0.35 m
PEg,i = PEelastic,f + PEg,f
mghi = 12
kx2 + mghf
hi = hf + 2
kx
m
2
g = 0.35 m + = 0.35 m + 15 m
hi = 15 m
(3.4 × 104 N/m)(0.65 m)2
(2)(50.0 kg)(9.81 m/s2)
8. h = 3.0 m
g = 9.81 m/s2
PEi = KEf
mgh = 12
mvf2
vf =√
2gh =√
(2)(9.81 m/s2)(3.0 m)
vf = 7.7 m/s
10. mw = 546 kg
h = 5.64 m
g = 9.81 m/s2
mflyer = 273 kg
PEw = KEflyer
mw gh = 12
mflyer vflyer2
vflyer = 2
m
mf
w
ly
e
g
rh =
vflyer = 14.9 m/s
(2)(546 kg)(9.81 m/s2)(5.64 m)
273 kg
9. m = 100.0 g
x = 30.0 cm
k = 1250 N/m
PEelastic = KE
12
kx2 = 12
mv2
v = k
mx2
= v = 33.5 m/s
(1250 N/m)(30.0 × 10−2 m)2
100.0 × 10−3 kg
Additional Practice 5F
1. P = 380.3 kW
W = 4.5 × 106 J∆t =
W
P =
38
4
0
.
.
5
3
××
1
1
0
0
6
3J
W = 12 s
2. P = 331 W
h = 442 m
m = 55 kg
g = 9.81 m/s2
W = mgh
∆t = W
P =
m
P
gh =
(55 kg
3
)(
3
9
1
.8
W
1 m/s2) = 720 s = 12 min
3. F = 334 N
d = 50.0 m
q = 0°
P = 2100 W
W = Fd(cos q)
∆t = W
P =
Fd(c
P
os q) = = 7.95 s
(334 N)(50.0 m)(cos 0°)
2100 W
Holt Physics Solution ManualV Ch. 5–10
V
4. P = 13.0 MW
∆t = 15.0 minW = P ∆t = (13.0 × 106 W)(15.0 min)(60 s/min) = 1.17 × 1010 J
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5. P = 1 hp = 745.7 W
∆t = 0.55 sW = P ∆t = (745.7 W)(0.55 s) = 4.1 × 102 J
6. P = (4)(300.0 kW)
∆t = 25 sW = P ∆t = (4)(300.0 × 103 W)(25 s) = 3.0 × 107 J
7. ∆t = 39 s
P = 158 kWW = P ∆t = (158 × 103 W)(39 s) = 6.2 × 106 J
8. W = 1.4 × 1013 J
∆t = 8.5 minP =
∆W
t = = 2.7 × 1010 W = 27 GW
1.4 × 1013 J(8.5 min)(60 s/min)
9. W = 2.82 × 107 J
∆t = 30.0 minP =
∆W
t = =
P = (1.57 × 104 W)(1 hp/745.7 W) = 21.1 hp
1.57 × 104 W = 15.7 kW2.82 × 107 J
(30.0 min)(60 s/min)
10. W = 3.0 × 106 J
∆t = 5.0 minP =
∆W
t = = 1.0 × 104 W
3.0 × 106 J(5.0 min)(60 s/min)
Section Five—Problem Bank V Ch. 6–1
Chapter 6Momentum and Collisions
V
1. m = 1.46 × 105 kg
p = 9.73 × 105 kg•m/s to thesouth
v = m
p =
v = 6.66 m/s to the south
9.73 × 105 kg•m/s
1.46 × 105 kg
Additional Practice 6A
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2. m = 25 kg
p = 6.8 × 104 kg•m/sv =
m
p = = 2.7 × 103 m/s
v = (2.7 km/s)(3600 s/h) = 9.7 × 103 km/h
v = 2.7 × 103 m/s = 9.7 × 103 km/h
6.8 × 104 kg•m/s
25 kg
3. m = 5.00 × 102 kg
p = 8.22 × 103 kg•m/s to thewest
v = m
p =
v = 16.4 m/s to the west
8.22 × 103 kg•m/s
5.00 × 102 kg
4. mc = 177.4 kg
md = 61.5 kg
p = 4.416 × 103 kg•m/s
v = m
p =
mc +p
md = =
v = 18.48 m/s = (18.48 m/s)(3600 s/h)(1 km/103 m) = 66.5 km/h
v = 18.48 m/s = 66.53 km/h)
4.416 × 103 kg•m/s
238.9 kg
4.416 × 103 kg•m/s177.4 kg + 61.5 kg
5. ∆x = 200.0 m
∆t = 19.32 s
m = 77 kg
vavg = ∆∆
x
t =
2
1
0
9
0
.3
.0
2
m
s = 10.35 m/s
p = mv
pavg = mvavg = (77 kg)(10.35 m/s) = 7.8 × 102 kg•m/s
6. ∆x = 274 m to the north
∆t = 8.65 s
m = 50.0 kg
vavg = ∆∆
x
t =
2
8
7
.6
4
5
m
s = 31.7 m/s to the north
p = mv
pavg = mvavg = (50.0 kg)(31.7 m/s) = 1.58 × 103 kg•m/s to the north
7. m = 7.10 × 105 kg
v = 270 km/h
p = mv = (7.10 × 105 kg)(270 km/h)(103 m/km)(1 h/3600 s)
p = 5.33 × 107 kg•m/s
1. F = 10.0 N to the right
m = 3.0 kg
∆t = 5.0 s
vi = 0 m/s
∆p = mvf − mvi = F∆t
vf = F∆t
m
+ mvi = = 17 m/s
vf = 17 m/s to the right
(10.0 N)(5.05) + (3.0 kg)(0 m/s)
3.0 kg
Holt Physics Solution ManualV Ch. 6–2
V
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8. v = 50.0 km/h
p = 0.278 kg•m/sm =
p
v =
m = 2.00 × 10−2 kg = 20.0 g
0.278 kg•m/s(50.0 km/h)(103 m/km)(1 h/3600 s)
9. vavg = 96 km/h to the southeast
pavg = 4.8 × 104 kg•m/s tothe southeast
m = p
v =
p
va
a
v
v
g
g =
m = 1.8 × 103 kg
4.8 × 104 kg•m/s(96 km/h)(103 m/km)(1 h/3600 s)
10. v = 255 km/s
p = 8.62 × 1036 kg•m/sm =
p
v = = 3.38 × 1031 kg
8.62 × 1036 kg•m/s
255 × 103 m/s
Additional Practice 6B
2. m = 60.0 g
F = −1.5 N
∆t = 0.25 s
vf = 0 m/s
∆p = mvf − mvi = F∆t
vi = mvf
m
− F∆t = =
vi = 6.2 m/s
(1.5 N)(0.25 s)60.0 × 10−3 kg
(60.0 × 10−3 kg)(0 m/s) − (−1.5 N)(0.25 s)
60.0 × 10−3 kg
3. F = 75 N
m = 55 kg
∆t = 7.5 s
vi = 0 m/s
∆p = mvf − mvi = F∆t
vf = F∆t
m
+ mvi =
vf = 1.0 × 101 m/s
(75 N)(7.5 s) + (55 kg)(0 m/s)
55 kg
4. m = 0.195 kg
vi = 0.850 m/s to the right = +0.850 m/s
F = 3.50 N to the left = −3.50 N
∆t = 0.0750 s
∆p = mvf − mvi = F∆t
vf = F∆t
m
+ mvi =
vf = = = −0.49 m/s
vf = 0.49 m/s to the left
−0.096 kg•m/s
0.195 kg
−0.262 kg•m/s + 0.166 kg•m/s
0.195 kg
(−3.50 N)(0.0750 s) + (0.195 kg)(0.850 m/s)
0.195 kg
5. m = 5.00 g
vi = 255 m/s to the right
vf = 0 m/s
∆t = 1.45 s
∆p = mvf − mvi = F∆t
F = mvf
∆−t
mvi = = −0.879 N
F = 0.879 N to the left
(5.00 × 10−3 kg)(0 m/s) − (5.00 × 10−3 kg)(255 m/s)
1.45 s
Section Five—Problem Bank V Ch. 6–3
V
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6. m = 1.1 × 103 kg
vf = 9.7 m/s to the east
vi = 0 m/s
∆t = 19 s
F = ∆∆
p
t =
mvf
∆−t
mvi
F = = 560 N
F = 560 N to the east
(1.1 × 103 kg)(9.7 m/s) − (1.1 × 103 kg)(0 m/s)
19 s
7. m = 3.00 × 103 kg
vi = 0 m/s
vf = 8.9 m/s to the right
∆t = 5.5 s
F = ∆∆
p
t =
mvf
∆−t
mvi
F = =
F = 4.9 × 103 N to the right
(3.00 × 103 kg)(8.9 m/s)
5.5 s
(3.00 × 103 kg)(8.9 m/s) − (3.00 × 103 kg)(0 m/s)
5.5 s
8. m = 0.17 kg
∆v = −9.0 m/s
g = 9.81 m/s2
mk = 0.050
10. vf = 15.8 km/s
vi = 0 km/s
F = 12.0 N
m = 0.20 g
∆t = ∆F
p =
mvf −F
mvi
∆t =
∆t =
∆t = 0.26 s
(0.20 × 10−3 kg)(15.8 × 103 m/s)
12.0 N
(0.20 × 10−3 kg)(15.8 × 103 m/s) − (0.20 × 10−3 kg)(0 m/s)
12.0 N
1. vi = 382 km/h to the right
vf = 0 km/h
mc = 705 kg
md = 65 kg
∆t = 12.0 s
F = ∆∆
p
t =
F =
F = = − 6.81 × 103 N
F =
∆x = 12
(vi + vf)∆t = 12
(382 km/h + 0 km/h)(103 m/km)(1 h/3600 s)(12.0 s)
∆x = 637 m to the right
6.81 × 103 N to the left
−(7.70 × 102 kg)(382 km/h)(103 m/km)(1 h/3600 s)
12.0 s
[(705 kg + 65 kg)(0 km/h) − (705 kg + 65 kg)(382 km/h)](103 m//km)(1 h/3600 s)
12.0 s
(mc + md)vf − (mc + md)vi∆t
F∆t = ∆p = m∆v
F = Fk = −mgmk
∆t = −m
m
∆gm
v
k =
−∆gm
v
k =
−(9.81
−9
m
.0
/s
m2)
/
(
s
0.050)
∆t = 18 s
9. m = 12.0 kg
Fapplied = 15.0 N
q = 20.0°
Ffriction = 11.0 N
vi = 0 m/s
vf = 4.50 m/s
F = Fapplied(cos q) − Ffriction
∆t = ∆F
p = =
∆t = =
∆t = 1.7 s
5.40 kg•m/s
3.1 N
5.40 kg•m/s − 0 kg•m/s
14.1 N − 11.0 N
(12.0 kg)(4.50 m/s) − (12.0 kg)(0 m/s)
(15.0 N)(cos 20.0°) − 11.0 N
mvf − mviFapplied(cos q) − Ffriction
Additional Practice 6C
Holt Physics Solution ManualV Ch. 6–4
V
2. vi = 7.82 × 103 m/s
vf = 0 m/s
m = 42 g
∆t = 1.0 × 10−6 s
F = = mvf
∆−t
mvi
F =
F =
F =
∆x = 12
(vi + vf)∆t = 12
(7.82 × 103 m/s + 0 m/s)(1.0 × 10−6 s)
∆x = 3.9 × 10−3 m = 3.9 mm
−3.3 × 108 N
−(42 × 10−3 kg)(7.82 × 103 m/s)
1.0 × 10−6 s
(42 × 10−3 kg)(0 m/s) − (42 × 10−3 kg)(7.82 × 103 m/s)
1.0 × 10−6 s
∆p∆t
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3. m = 63 kg
vi = 7.0 m/s to the right
vf = 0 m/s
∆t = 14.0 s
F = ∆∆
p
t =
mvf
∆−t
mvi
F = = = −32 N
F =
∆x = 12
(vi + vf) ∆t = 12
(7.0 m/s + 0 m/s)(14.0 s)
∆x = 49 m to the right
32 N to the left
−(63 kg)(7.0 m/s)
14.0 s
(63 kg)(0 m/s) − (63 kg)(7.0 m/s)
14.0 s
4. m = 455 kg
∆t = 12.2 s
mk = 0.071
g = 9.81 m/s2
vf = 0 m/s
∆p = F∆t
F = Fk = −mgmk
∆p = −mgmk∆t = −(455 kg)(9.81 m/s2)(0.071)(12.2 s) = −3.9 × 103 kg•ms
∆p =
vi = mvf
m
− ∆p =
vi = = 8.6 m/s
∆x = 12
(vi + vf)∆t = 12
(8.6 m/s + 0 m/s)(12.2 s)
∆x = 52 m
3.9 × 103 kg•ms
455 kg
(455 kg)(0 m/s) − (−3.9 × 103 kg•ms)
455 kg
3.9 × 103 kg•ms opposite the polar bear’s motion
5. m = 75.0 g
∆t = 1.2 s
g = 9.81 m/s2
vf = 0 m/s
∆p = F∆t
F = −mg
∆p = −mg∆t = −(75.0 × 10−3 kg)(9.81 m/s2)(1.2 s)
∆p = −0.88 kg•m/s
∆p =
vi = mvf
m
− ∆p =
vi = 7
0
5
.
.
8
0
8
×k
1
g
0
•−m
3/
k
s
g = 12 m/s upward
∆x = 12
(vi + vf)∆t = 12
(12 m/s + 0 m/s)(1.2 s)
∆x = 7.2 m upward
(75.0 × 10−3 kg)(0 m/s) − (−0.88 kg•m/s)
75.0 × 10−3 kg
0.88 kg•m/s downward
Section Five—Problem Bank V Ch. 6–5
V
6. m = 4400 kg
F = 2200 N to the left = −2200 N
∆t = 8.0 s
vi = 6.5 m/s to the right = +6.5 m/s
∆p = F∆t = (−2200 N)(8.0 s) = −1.8 × 104 kg•m/s
∆p =
vf = ∆p +
m
mvi =
vf = = = 2.5 m/s to the right
∆x = 12
(vi + vf)∆t = 12
(6.5 m/s + 2.5 m/s)(8.0 s) = 12
(9.0 m/s)(8.0 s)
∆x = 36 m to the right
1.1 × 104 kg•m/s
4400 kg
−1.8 × 104 kg•m/s + 2.9 × 104 kg•m/s
4400 kg
−1.8 × 104 kg•m/s + (4400 kg)(6.5 m/s)
4400 kg
1.8 × 104 kg•m/s to the left
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7. F = 25.0 N
∆t = 7.00 s
m = 14.0 kg
vi = 0 m/s
∆p = F∆t = (25.0 N)(7.00 s) =
vf = ∆p +
m
mvi = = 12.5 m/s
∆x = 12
(vi + vf)∆t = 12
(0 m/s + 12.5 m/s)(7.00 s)
∆x = 43.8 m
175 kg•m/s + (14.0 kg)(0 m/s)
14.0 kg
175 kg•m/s
8. m = 2.30 × 103 kg
vi = 22.2 m/s
vf = 0 m/s
F = −1.26 × 104 N
∆t = ∆F
p =
mvf −F
mvi
∆t = =
∆t =
∆x = 12
(vi + vf)∆t = 12
(22.2 m/s + 0 m/s)(4.06 s)
∆x = 45.1 m
4.06 s
−5.11 × 104 kg•m/s
−1.26 × 104 N
(2.30 × 103 kg)(0 m/s) − (2.30 × 103 kg)(22.2 m/s)
−1.26 × 104 N
9. m = 1.35 × 104 kg
vi = 66.1 m/s to the west = −66.1 m/s
vf = 0 m/s
F = 4.00 × 105 N to the east= +4.00 × 105 N
∆t = ∆F
p =
mvf −F
mvi
∆t = =
∆t =
∆x = 12
(vi + vf)∆t = 12
(−66.1 m/s + 0 m/s)(2.23 s) = −73.7 m
∆x = 73.7 m to the west
2.23 s
8.92 × 105 kg•m/s
4.00 × 105 N
(1.35 × 104 kg)(0 m/s) − (1.35 × 104 kg)(−66.1 m/s)
4.00 × 105 N
10. vi = 14.5 m/s
vf = 0 m/s
m = 1.50 × 103 kg
mk = 0.065
a = −1.305 m/s2
∆t = ∆F
p =
mvf −F
mvi
F = Fk = mamk
∆t = v
a
f
m−
k
vi
∆t =
∆t =
∆x = 12
(vi + vf)∆t = 12
(14.5 m/s + 0 m/s)(1.7 × 102 s)
∆x = 1.2 × 103 m = 1.2 km
1.7 × 102 s
0 m/s − 14.5 m/s(− 1.305 m/s2)(0.065)
Holt Physics Solution ManualV Ch. 6–6
V
1. m1 = 68 kg
m2 = 68 kg
v2, i = 0 m/s
v1, f = 0.85 m/s to the west = −0.85 m/s
v2, f = 0.85 m/s to the west = −0.85 m/s
m1v1, i + m2v2, i = m1v1, f + m2v2, f
v1, i = =
v1, i = −0.85 m/s + (−0.85 m/s) = −1.7 m/s
v1, i = 1.7 m/s to the west
(68 kg)(−0.85 m/s) + (68 kg)(−0.85 m/s) − (68 kg)(0 m/s)
68 kg
m1v1, f + m2v2, f − m2v2, imi
Additional Practice 6D
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2. mi = 1.36 × 104 kg
m2 = 8.4 × 103 kg
v2, i = 0 m/s
v1, f = v2, f = 1.3 m/s
m1v1, i + m2v2, i = m1 v1, f + m2v2, f
v1, i =
v1, i =
v1, i = =
v1, i = 2.1 m/s
2.9 × 104 kg•m/s
1.36 × 104 kg
1.8 × 104 kg•m/s + 1.1 × 104 kg•m/s
1.36 × 104 kg
(1.36 × 104 kg)(1.3 m/s) + (8.4 × 103 kg)(1.3 m/s) − (8.4 × 103 kg)(0 m/s)
1.36 × 104 kg
m1v1, f + m2v2, f − m2v2, im1
3. v1, f = 2.2 m/s backwards= −2.2 m/s
v2, f = 5.5 m/s forward= +5.5 m/s
m1 = 38 kg
m2 = 68 kg
m1v1, i + m2v2, i = m1v1, f + m2v2, f
v1, i = v2, i, so
v1, i = m1v
m1,
1
f ++
m
m2
2
v2, f =
v1, i = = = 2.7 m/s
v1, i = 2.7 m/s forward
290 kg•m/s
106 kg
−84 kg•m/s + 370 kg•m/s
106 kg
(38 kg)(−2.2 m/s) + (68 kg)(5.5 m/s)
38 kg + 68 kg
4. m1 = 38 kg
v1, i = 1.6 m/s to the north
m2 = 142 kg
v1, f = 0.32 m/s to the north
v2, f = 0.32 m/s to the north
m1v1, i + m2v2, i = m1v1, f + m2v2, f
v2, i =
v2, i =
v2, i = = = −2.8 × 10−2 m/s
v2, i = 2.8 × 10−2 m/s to the south
−4.0 kg•m/s
142 kg
12 kg•m/s + 45 kg•m/s − 61 kg•m/s
142 kg
(38 kg)(0.32 m/s) + (142 kg)(0.32 m/s) − (38 kg)(1.6 m/s)
142 kg
m1v1, f + m2v2, f − m1v1, im2
5. m1 = 50.0 g
v1, i = 0 m/s
v1, f = 400.0 m/s forward
m2 = 3.00 kg
v2, i = 0 m/s
Because the initial velocities for both rifle and projectile are zero, the momentum con-servation equation takes the following form:
m1v1, f + m2v2, f = 0
v2, f = = = −6.67 m/s
v2, f = 6.67 m/s backward
−(50.0 × 10−3 kg)(400.0 m/s)
3.00 kg
−m1v1, fm2
Section Five—Problem Bank V Ch. 6–7
V
6. mi = 1292 kg
vi = 88.0 km/h to the east
mf = 1255 kg
mivi = mfvf
vf = m
miv
f
i =
vf = 90.6 km/h to the east
(1292 kg)(88.0 km/h)
1255 kg
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7. m = 5.0 × 1014 kg
vi = 74.0 km forward
mi = m2 = 12
m
v1, f = 105 km/s at 15.0°above forward
v2, f is at an angle of −30.0°to the forward direction
mvi = m1v1, f + m2v2, f
To solve for velocity in two dimensions, the momentum conservation equation mustbe written as two equations, one for both the x and y directions.
In the x-direction:
mvi = m1v1, f (cos q1) + m2v2, f (cos q2)
vi = 12
v1, f (cos q1) + 12
v2, f (cos q2)
v2, f =
v2, f =
v2, f = =
v2, f = 54 km/s
In the y-direction (check):
0 = m1v1, f (sin q1) + m2v2, f (sin q2)
v1, f (sin q1) = −v2, f (sin q2)
(105 km/s)(sin 15.0°) = −(54 km/s)[sin(−30.0°)]27.2 km/s = 27 km/s
The slight difference arises from differences in the number of significant figures andfrom rounding.
v2,f = 54 km/s at 30.0° below the initial forward direction
47 km/scos(−30.0°)
148 km/s − 101 km/s
cos(−30.0°)
(2)(74.0 km/s) − (105 km/s)(cos 15.0°)
cos(−30.0°)
2vi − v1, f (cos q1)
cos q2
8. v1, i = 0 cm/s
v1, f = 1.2 cm/s forward = +1.2 cm/s
v2, i = 0 cm/s
v2, f = 0.40 cm/s backward = −0.40 cm/s
m1 = 2.5 g
m1v1, f + m2v2, f = 0
m2 = =
m2 = 7.5 g
−(2.5 g)(1.2 cm/s)
−0.40 cm/s
−m1v1, fv2, f
Holt Physics Solution ManualV Ch. 6–8
V
9. vi = 0 cm/s
m1 = 25.0 g
m2 = 25.0 g
v1 = 7.0 cm/s to the south= 7.0 cm/s at −90° fromeast
v2 = 7.0 cm/s to the west
= 7.0 cm/s at 180° from east
v3 = 3.3 m/s at 45° north ofeast
m1v1, f + m2v2, f + m3v3, f = 0
In the x-direction:
m1v1, f (cos q1) + m2v2, f (cos q2) + m3v3, f (cos q3) = 0
m3 =
m3 =
m3 = =
In the y-direction (check):
m1v1, f (sin q1) + m2v2, f (sin q2) + m3v3, f (sin q3) = 0
(25.0 g)(7.0 cm/s)[sin(−90°)] + (25.0 g)(7.0 cm/s)(sin 180°) + (75 g)(3.3 cm/s)(sin 45°) = 0
−180 g•cm/s + 0 g•cm/s + 180•g•cm/s = 0
75 g(25.0 g)(7.0 cm/s)(3.3 cm/s)(cos 45°)
− (25.0 g)(7.0 cm/s)(cos −90°) + (25.0 g)(7.0 cm/s)(cos 180°)
−(3.3 cm/s)(cos 45°)
m1v1, f (cos q1) + m2v2, f (cos q2)
−v3, f (cos q3)
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10. v1, i = 0 m/s
v2, i = 5.4 m/s to the north
v1, f = 1.5 m/s to the north
v2, f = 1.5 m/s to the north
m1 = 63 kg
m1v1, i + m2v2, i = m1v1, f + m2v2, f
m2 = = =
m2 = 24 kg
(63 kg)(1.5 m/s)
3.9 m/s
(63 kg)(1.5 m/s) − (63 kg)(0 m/s)
5.4 m/s − 1.5 m/s
m1v1, f − m1v1, iv2, i − v2, f
Additional Practice 6E
1. m1 = 1550 kg
m2 = 770 kg
v2, i = 0 m/s
vf = 9.44 m/s forward
v1, i =
v1, i = =
v1, i = 14.0 m/s forward
(2320 kg)(9.44 m/s)
1550 kg
(1550 kg + 770 kg)(9.44 m/s) − (770 kg)(0 m/s)
1550 kg
(m1 + m2)vf − m2v2, im1
2. m1 = 0.17 kg
m2 = 0.75 kg
v2, i = 0.50 m/s to the left = −0.50 m/s
vf = 4.2 m/s to the right = +4.2 m/s
v1, i =
v1, i =
v1, i =
v1, i = =
v1, i = 25 m/s to the right
4.3 kg•m/s
0.17 kg
3.9 kg•m/s + 0.38 kg•m/s
0.17 kg
(0.92 kg)(4.2 m/s) + 0.38 kg•m/s
0.17 kg
(0.17 kg + 0.75 kg)(4.2 m/s) − (0.75 kg)(−0.50 m/s)
0.17 kg
(m1 + m2)vf − m2v2, im1
Section Five—Problem Bank V Ch. 6–9
V
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3. m1 = 45 g
m2 = 75 g
v2, i = 0 m/s
h = 8.0 cm
g = 9.81 m/s2
Use the conservation of mechanical energy to calculate vf .
KE = PE
12
(m1 + m2)vf2 = (m1 + m2)gh
vf =√
2gh =√
(2)(9.81 m/s2)(8.0 × 10−2m)
vf = 1.3 m/s
v1, i = =
v1, i = =
The height that the first ball must have can be determined by using the conservationof mechanical energy.
PE = KE
m1gh1 = 12
m1v1, i2
h1 = v
21,
gi2
= = 0.62 m = 62 cm(3.5 m/s)
(2)(9.81 m/s2)
3.5 m/s(1.20 × 102 g)(1.3 m/s)
45 g
(45 g + 75 g)(1.3 m/s) − (75 g)(0 m/s)
45 g
(m1 + m2)vf − m2v2, im1
4. m1 = m2 = m3 = 5.00 × 102 kg
vf = 3.67 m/s
v3, i = 3.00 m/s
v2, i = 3.50 m/s
m1v1, i + m2v2, i + m3v3, i = (m1 + m2 + m3)vf
v1, i =
v1, i =
−
v1, i =
v1, i =
v1, i = = 4.50 m/s2250 kg•m/s5.00 × 102 kg
5.50 × 103 kg•m/s − 3250 kg•m/s
5.00 × 102 kg
(15.00 × 102 kg)(3.67 m/s) − 1750 kg•m/s − 1.50 × 103 kg•m/s
5.00 × 102 kg
(5.00 × 102 kg)(3.50 m/s) − (5.00 × 102 kg)(3.00 m/s)
5.00 × 102 kg
(5.00 × 102 kg + 5.00 × 102 kg + 5.00 × 102 kg)(3.67 m/s)
5.00 × 102 kg
(m1 + m2 + m3)vf − m2v2, i − m3v3, im1
5. m1 = 8500 kg
v1, i = 4.5 m/s to the right= +4.5 m/s
m2 = 9800 kg
v2, i = 3.9 m/s to the left= −3.9 m/s
vf =
vf = =
vf = 0 m/s
3.8 × 104 kg•m/s − 3.8 × 104 kg•m/s
1.83 × 104 kg
(8500 kg)(4.5 m/s) + (9800 kg)(−3.9 m/s)
8500 kg + 9800 kg
m1v1, i + m2v2, im1 + m2
Holt Physics Solution ManualV Ch. 6–10
V
6. m1 = 1400 kg
v1, i = 45 km/h to the north
m2 = 2500 kg
v2, i = 33 km/h to the east
vf =
The component of vf in the x-direction is given by
vf, x = = =
vf, x = 21 km/h
The component of vf in the y-direction is given by
vf, y = = =
vf, y = 16 km/h
vf =√
vf, x2+ vf,y2 =
√(21km/h)2 + (16 km/h)2
vf =√
440km2/h2+ 260 km2/h2 =√
7.0× 102 km2/h2
vf = 26 km/h
q = tan−1 v
v
f
f
,
,
x
y = tan−1126
1
k
k
m
m
/
/
h
h = 37°
vf = 26 km/h at 37° north of east
(1400 kg)(45 km/h)
3900 kg
(1400 kg)(45 km/h)1400 kg + 2500 kg
m1v1, im1 + m2
(2500 kg)(33 km/h)
3900 kg
(2500 kg)(33 km/h)1400 kg + 2500 kg
m2v2, im1 + m2
m1v1, i + m2v2, im1 + m2
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7. m1 = 50.0 g
v1, i = 0.80 m/s to the west = −0.80 m/s
m2 = 60.0 g
v2, i = 2.50 m/s to the north= +2.50 m/s
m3 = 100.0 g
v3, i = 0.20 m/s to the east = +0.20 m/s
m4 = 40.0 g
v4, i = 0.50 m/s to the south= −0.50 m/s
The component of final velocity in the x-direction is given by
vf, x = =
vf, x = =
vf, x = −8.0 × 10−2 m/s
The component of final velocity in the y-direction is given by
vf, y = =
vf, y = =
vf, y = 0.520 m/s
vf =√
vf, x2+ vf,y2 =
√(−8.0× 10−2m/s)2 + (0.520m/s)2
vf =√
6.4× 10−3m2/s2 + 0.270 m2/s2 =√
0.276m2/s2
vf = 0.525 m/s
q = tan−1v
v
f
f
,
,
x
y = tan−1−8.
0
0
.5
×2
1
0
0
m−2
/s
m/s = −81°
q = 81° north of west, or 9° west of north
vf = 0.53 m/s at 9° west of north
1.30 × 102 g•m/s
250.0 g
1.50 × 102 g•m/s − 2.0 × 101 g•m/s
250.0 g
(60.0 g)(2.50 m/s) + (40.0 g)(−0.50 m/s)
50.0 g + 60.0 g + 100.0 g + 40.0 g
m2v2, i + m4v4, im1 + m2 + m3 + m4
−2.0 × 101 g•m/s
250.0 g
−4.0 × 101 g•m/s + 2.0 × 101 g•m/s
250.0 g
(50.0 g)(−0.80 m/s) + (100.0 g)(0.20 m/s)
50.0 g + 60.0 g + 100.0 g + 40.0 g
m1v1, i + m3v3, im1 + m2 + m3 + m4
Section Five—Problem Bank V Ch. 6–11
V
8. ms = 25.0 kg
mc = 42.0 kg
v1, i = 3.50 m/s
v2, i = 0 m/s
vf = 2.90 m/s
m1 = mass of child and sled = ms + mc = 25.0 kg + 42.0 kg = 67.0 kg
m1v1, i + m2v2, i = (m1 + m2)vf
m2 = =
m2 = =
m2 = 13.8 kg
40.0 kg•m/s
2.90 m/s234 kg•m/s − 194 kg•m/s
2.90 m/s
(67.0 kg)(3.50 m/s) − (67.0 kg)(2.90 m/s)
2.90 m/s − 0 m/s
m1v1, i − m1vfvf − v2, i
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9. v1, i = 5.0 m/s to the right= +5.0 m/s
v2, i = 7.00 m/s to the left= −7.00 m/s
vf = 6.25 m/s to the left= −6.25 m/s
m2 = 150.0 kg
m1 = =
m1 = =
m1 = 9.8 kg
−110 kg•m/s
−11.2 m/s
−1050 kg•m/s + 938 kg•m/s
−11.2 m/s
(150.0 kg)(−7.00 m/s) − (150.0 kg)(−6.25 m/s)
−6.25 m/s − 5.0 m/s
m2v2, i − m2vfvf − v1, i
10. v1, i = 8.0 × 103 m/s to theright = +8.0 × 103 m/s
v2, i = 8.0 × 103 m/s to theleft = −8.0 × 103 m/s
vf = (0.900)v1, i
m2 = (52 000)(2.5 g) = 1.3 × 105 g
m1 = =
m1 = =
m1 = 2.4 × 103 kg
1.9 × 106 kg•m/s
8.0 × 102 m/s
9.4 × 105 kg•m/s + 1.0 × 106 kg•m/s
(8.0 × 103 m/s)(0.100)
(1.3 × 102 kg)(0.900)(8.0 × 103 m/s) − (1.3 × 102 kg)(−8.0 × 103 m/s)
(8.0 × 103 m/s)(1 − 0.900)
m2vf − m2v2, iv1, i − vf
1. m1 = 55 g
v1, i = 1.5 m/s
m2 = 55 g
v2, i = 0 m/s
vf = = =
vf = 0.75 m/s
percent decrease of KE = × 100 = × 100 = − 1 × 100
KEi = 12
m1v1, i2 + 1
2m2v2, i = 1
2(55 × 10−3 kg)(1.5 m/s)2 + 1
2(55 × 10−3 kg)(0 m/s)2
KEi = 6.2 × 10−2 J + 0 J = 6.2 × 10−2 J
KEf = 12
(m1 + m2)vf2 = 1
2(55 g + 55 g)(10−3 kg/g)(0.75 m/s)3
KEf = 3.1 × 10−2 J
percent decrease of KE = − 1 × 100 = (0.50 − 1) × 100 = (−0.50) × 100
percent decrease of KE = −5.0 × 101 percent
3.1 × 10−2 J6.2 × 10−2 J
KEfKEi
KEf − KEiKEi
∆KEKEi
(55 g)(1.5 m/s)
1.10 × 102 g
(55 g)(1.5 m/s) + (55 g)(0 m/s)
55 g + 55 g
m1v1, i + m2v2, im1 + m2
Additional Practice 6F
Holt Physics Solution ManualV Ch. 6–12
V
2. m1 = 4.5 kg
v1, i = 0 m/s
m2 = 1.3 kg
vf = 0.83 m/s
v2, i = =
v2, i = = 3.7 m/s
KEi = 12
m1v1, i2 + 1
2m2v2, i
2 = 12
(4.5 kg)(0 m/s)2 + 12
(1.3 kg)(3.7 m/s)2
KEi = 0 J + 8.9 J = 8.9 J
KEf = 12
(m1 + m2)vf2 = 1
2(4.5 kg + 1.3 kg)(0.83 m/s)2 = 1
2(5.8 kg)(0.83 m/s)2
KEf = 2.0 J
∆KE = KEf − KEi = 2.0 J − 8.9 J = −6.9 J
(5.8 kg)(0.83 m/s)
1.3 kg
(4.5 kg + 1.3 kg)(0.83 m/s) − (4.5 kg)(0 m/s)
1.3 kg
(m1 + m2)vf − m1v1, im2
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3. m1 = 1.50 × 1013 kg
v1, i = 250 m/s
m2 = 6.5 × 1012 kg
v2, i = 420 m/s
vf = =
vf = = = 3.0 × 102 m/s
KEi = 12
m1v1, i2 + 1
2m2v2, i
2 = 12
(1.50 × 1013 kg)(250 m/s)2 + 12
(6.5 × 1012 kg) (420 m/s)2
KEi = 4.7 × 1017 J + 5.7 × 1017 J = 10.4 × 1017 J = 1.04 × 1018
KEf = 12
(m1 + m2)vf2 = 1
2(1.50 × 1013 kg + 6.5 × 1012 kg)(3.0 × 102 m/s)2
KEf = 12
(2.15 × 1013 kg)(3.0 × 102 m/s)2
KEf = 9.7 × 1017 J
∆KE = KEf − KEi = 9.7 × 1017 J − 1.04 × 1018 J = −7.0 × 1016 J
6.5 × 1015 kg•m/s
2.15 × 1013 kg
3.8 × 1015 kg•m/s + 2.7 × 1015 kg•m/s
2.15 × 1013 kg
(1.50 × 1013 kg)(250 m/s) + (6.5 × 1012 kg)(420 m/s)
1.50 × 1013 kg + 6.5 × 1012 kg
m1v1, i + m2v2, im1 + m2
4. m1 = 0.650 kg
v1, i = 15.0 m/s to the right = +15.0 m/s
m2 = 0.950 kg
v2, i = 13.5 m/s to the left = −13.5 m/s
vf = =
vf = = = −1.91 m/s
vf = 1.91 m/s to the left
KEi = 12
m1v1, i2 + 1
2m2v2, i
2 = 12
(0.650 kg)(15.0 m/s)2 + 12
(0.950 kg)(−13.5 m/s)2
KEi = 73.1 J + 86.6 J = 159.7 J
KEf = 12
(m1 + m2)vf2 = 1
2(0.650 kg + 0.950 kg)(1.91 m/s)2 = 1
2(1.600 kg)(1.91 m/s)2
KEf = 2.92 J
∆KE = KEf − KEi = 2.92 J − 159.7 J = −1.57 × 102 J
−3.0 kg•m/s
1.600 kg
9.75 kg•m/s − 12.8 kg•m/s
1.600 kg
(0.650 kg)(15.0 m/s) + (0.950 kg)(−13.5 m/s)
0.650 kg + 0.950 kg
m1v1, i + m2v2, im1 + m2
Section Five—Problem Bank V Ch. 6–13
V
5. m1 = 75.0 kg
v1, i = 1.80 m/s downstream= 1.80 m/s
m2 = (8)(0.30 kg) = 2.4 kg
v2, i = 1.3 m/s upstream = −1.3 m/s
vf = =
vf = = = 1.71 m/s
KEi = 12
m1v1, i2 + 1
2m2v2, i
2 = 12
(75.0 kg)(1.80 m/s)2 + 12
(2.4 kg)(−1.3 m/s)2
KEi = 122 J + 2.0 J = 124 J
KEf = 12
(m1 + m2)vf2 = 1
2(75.0 kg + 2.4 kg)(1.71 m/s)2 = 1
2(77.4 kg)(1.71 m/s)2 = 113 J
∆KE = KEf − KEi = 113 J − 124 J = −11 J
132 kg•m/s
77.4 kg
135 kg•m/s − 3.1 kg•m/s
77.4 kg
(75.0 kg)(1.80 m/s) + (2.4 hg)(−1.3 m/s)
75.0 kg + 2.4 kg
m1v1, i + m2v2, im1 + m2
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6. m1 = 8500 kg
v1, i = 4.5 m/s
m2 = 9800 kg
v2, i = −3.9 m/s
vf = 0 m/s
KEi = 12
m1v1, i2 + 1
2 m2v2, i
2 = 12
(8500 kg)(4.5 m/s)2 + 12
(9800 kg)(−3.9 m/s)2
KEi = 8.6 × 104 J + 7.5 × 104 J = 16.1 × 104 J = 1.61 × 105 J
KEf = 12
(m1 + m2)vf2 = 1
2(8500 kg + 9800 kg)(0 m/s)2 = 0 J
∆KE = KEf − KEi = 0 J − 1.61 × 105 J = −1.61 × 105 J
7. m1 = 45 g
v1, i = 3.5 m/s
m2 = 75 g
v2, i = 0 m/s
vf = 1.3 m/s
KEi = 12
m1v1, i2 + 1
2m2v2, i
2 = 12
(45 × 10−3 kg)(3.5 m/s)2 + 12
(75 × 10−3 kg)(0 m/s)2
KEi = 0.28 J + 0 J = 0.28 J
KEf = 12
(m1 + m2)vf2 = 1
2(45 g + 75 g)(10−3 kg/g)(1.3 m/s)2 = 1
2(1.20 × 10−3 kg)(1.3 m/s)2
KEf = 1.0 × 10−3 J
∆KE = KEf − KEi = 1.0 × 10−3 J − 0.28 J =
The decrease in kinetic energy is almost total.
−0.28 J
8. m1 = 2.4 × 103 kg
v1, i = 8.0 × 103 m/s
m2 = 1.3 × 102 kg
v2, i = −8.0 × 103 m/s
vf = (0.900)(8.0 × 103 m/s)
KEi = 12
m1v1, i2 + 1
2m2v2, i
2
KEi = 12
(2.4 × 103 kg)(8.0 × 103 m/s)2 + 12
(1.3 × 102 kg)(−8.0 × 103 m/s)2
KEi = 7.7 × 1010 J + 4.2 × 109 J = 8.1 × 1010 J
KEf = 12
(m1 + m2)vf2 = 1
2(2.4 × 103 kg + 1.3 × 102 kg)[(0.900)(8.0 × 103 m/s)]2
KEf = 12
(2.5 × 103 kg)(7.2 × 103 m/s)2
KEf = 6.5 × 1010 J
∆KE = KEf − KEi = 6.5 × 1010 J − 8.1 × 1010 J = −1.6 × 1010 J
9. m1 = m2 = m3 = 5.00 × 102 kg
v1, i = 4.50 m/s
v2, i = 3.50 m/s
v3, i = 3.00 m/s
vf = 3.67 m/s
KEi = 12
m1v1, i2 + 1
2m2v2, i
2 + 12
m3v3, i2
KEi = 12
(5.00 × 102 kg)(4.50 m/s)2 + 12
(5.00 × 102 kg)(3.50 m/s)2
+ 12
(5.00 × 102 kg)(3.00 m/s)2
KEi = 5.06 × 103 J + 3.06 × 103 J + 2.25 × 103 J = 10.37 × 103 J = 10.37 kJ
KEf = 12
(m1 + m2 + m3)vf2 = 1
2(5.00 × 102 kg + 5.00 × 102 kg + 5.00 × 102 kg)
(3.67 m/s)2 = 12
(1.500 × 103 kg)(3.67 m/s)2
KEf = 1.01 × 104 J = 10.1 kJ
∆KE = KEf − KEi = 10.1 kJ − 10.37 kJ = −300 J
Holt Physics Solution ManualV Ch. 6–14
V
10. m1 = 50.0 × 10−3 kg
v1, i = −0.80 m/s
m2 = 60.0 × 10−3 kg
v2, i = 2.50 m/s
m3 = 0.1000 kg
v3, i = 0.20 m/s
m4 = 40.0 × 10−3 kg
v4, i = −0.50 m/s
vf = 0.53 m/s
KEi = 12
m1v1, i2 + 1
2m2v2, i
2 + 12
m3v3, i2 + 1
2m4v4, i
2
KEi = 12
(50.0 × 10−3 kg)(−0.80 m/s)2 + 12
(60.0 × 10−3 kg)(2.50 m/s)2
+ 12
(0.1000 kg)(0.20 m/s)2 + 12
(40.0 × 10−3 kg)(−0.50 m/s)2
KEi = 1.6 × 10−2 J + 18.8 × 10−2 J + 0.20 × 10−2 J + 0.50 × 10−2 J
KEi = 0.211 J
KEf = 12
(m1 +m2 + m3 + m4)vf2
KEf = 12
(50.0 × 10−3 kg + 60.0 × 10−3 kg + 0.1000 kg + 40.0 × 10−3 kg)(0.53 m/s)2
KEf = 12
(0.2500 kg)(0.53 m/s)2 = 3.5 × 10−2 J
∆KE = KEf − KEi = 3.5 × 10−2 J − 0.211 J = −0.18 J
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1. v1, i = 6.00 m/s to the right = +6.00 m/s
v2, i = 0 m/s
v1, f = 4.90 m/s to the left = −4.90 m/s
v2, f = 1.09 m/s to the right = +1.09 m/s
m2 = 1.25 kg
Momentum conservation
m1v1, i + m2v2, i = m1v1, f + m2v2, f
m1 = = =
m1 =
Conservation of kinetic energy (check)
12
m1v1, i2 + 1
2m2v2, i
2 = 12
m1v1, f2 + 1
2m2v2, f
2
12
(0.125 kg)(6.00 m/s)2 + 12
(1.25 kg)(0 m/s)2 = 12
(0.125 kg)(−4.90 m/s)2
+ 12
(1.25 kg)(1.09 m/s)2
2.25 J + 0 J = 1.50 J + 0.74 J
2.25 J = 2.24 J
The slight difference arises from rounding.
0.125 kg
1.36 kg•m/s10.90 m/s
(1.25 kg)(1.09 m/s) − (1.25 kg)(0 m/s)
6.00 m/s − (−4.90 m/s)
m2v2, f − m2v2, iv1, i − v1, f
2. m1 = 2.0 kg
v1, i = 8.0 m/s
v2, i = 0 m/s
v1, f = 2.0 m/s
m1v1, i + m2v2, i = m1v1, f + m2v2, f
12
m1v1, i2 + 1
2m2v2, i
2 = 12
m1v1, f2 + 1
2m2v2, f
2
m2 =
12
m1v1, i2 + 1
2 v2, i
2 = 12
m1v1, f2 + 1
2 v2, f
2
v1, i2(v2, i - v2, f) + (v1, f - v1, i )v2, i
2 = v1, f2(v2, i - v2, f) + (v1, f - v1, i )v2, f
2
(v1, i2v2, i + v1, f v2, i
2 - v1, iv2, i2 - v1, f
2v2, i + v2, f (v1, f2 - v1, i
2) = v2, f2(v1, f - v1, i )
Because v2, i = 0, the above equation simplifies to
v1, f2 − v1, i
2 = v2, f (v1, f − v1, i)
v2, f = v1, f + v1, i = 2.0 m/s + 8.0 m/s = 10.0 m/s
m2 = = =
m2 = 1.2 kg
−12 kg•m/s−10.0 m/s
4.0 kg•m/s − 16 kg•m/s
−10.0 m/s
(2.0 kg)(2.0 m/s) − (2.0 m/s)(8.0 m/s)
0 m/s − 10.0 m/s
m1v1, f − m1v1, iv2, i − v2, f
m1v1, f − m1v1, iv2, i − v2, f
m1v1, f − m1v1, iv2, i − v2, f
Additional Practice 6G
Section Five—Problem Bank V Ch. 6–15
V
3. m1 = m2 = 45 g
v2, i = 0 m/s
v1, f = 0 m/s
v2, f = 3.0 m/s
Momentum conservation
m1v1, i + m2v2, i = m1v1, f + m2v2, f
v1, i = v1, f + v2, f − v2, i = 0 m/s + 3.0 m/s − 0 m/s
v1, i =
Conservation of kinetic energy (check)
12
m1v1, i2 + 1
2m2v2, i
2 = 12
m1v1, f2 + 1
2m2v2, f
2
v1, i2 + v2, i
2 = v1, f2 + v2, f
2
(3.0 m/s)2 + (0 m/s)2 = (0 m/s)2 + (3.0 m/s)2
9.0 m2/s2 = 9.0 m2/s2
3.0 m/s
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4. m1 = 3.0 × 107 kg
m2 = 2.5 × 107 kg
v2, i = 4.0 km/h to the north= +4.0 km/h
v1, f = 3.1 km/h to the north= +3.1 km/h
v2, f = 6.9 km/h to the south= −6.9 km/h
Momentum conservation
m1v1, i + m2v2, i = m1v1, f + m2v2, f
v1, i =
v1, i =
v1, i = = −6.0 km/h
v1, i =
Conservation of kinetic energy (check)
12
m1v1, i2 + 1
2m2v2, i
2 = 12
m1v1, f2 + 1
2m2v2, f
2
12
(3.0 × 107 kg)[(−6.0 × 103 m/h)(1 h/3600 s)]2 + 12
(2.5 × 107 kg)[(4.0 × 103 m/h)(1 h/3600 s)]2
= 12
(3.0 × 107 kg)[(3.1 × 103 m/h)(1 h/3600 s)]2 + 12
(2.7 × 107 kg)[(−6.9 × 103 m/h)(1 h/3600 s)]2
4.2 × 107 J + 1.5 × 107 J = 1.1 × 107 J + 4.6 × 107 J
5.7 × 107 J = 5.7 × 107 J
6.0 km/h to the south
−1.8 × 108 kg•km/h
3.0 × 107 kg
9.3 × 107 kg•km/h − 1.7 × 108 kg•km/h − 1.0 × 108 kg•km/h
3.0 × 107 kg
(3.0 × 107 kg)(3.1 km/h) + (2.5 × 107 kg)(−6.9 km/h) − (2.5 × 107 kg)(4.0 km/h)
3.0 × 107 kg
m1v1, f + m2v2, f − m2v2, im1
Holt Physics Solution ManualV Ch. 6–16
V
5. m1 = m2
v1, i = 3.0 m/s to the north
v1, f = 4.0 m/s to the west
v2, f = 3.0 m/s to the north
v1, i = 3.0 m/s
q1, i = 90° counterclockwisefrom east
v1, f = 4.0 m/s
q1, f = 180° counterclock-wise from east
v2, f = 3.0 m/s
q2, f = 90° counterclockwisefrom east
Momentum conservation
In the x-direction:
m1 v1, i(cos q1, i) + m2 v2, i (cos q2, i) = m1v1, f (cos q1, f ) + m2 v2, f (cos q2, f )
v2, i (cos q2, i) = v1, f (cos q1, f ) + v2, f (cos q2, f ) − v1, i (cos q1, i ) = (4.0 m/s)(cos 180°)
+ (3.0 m/s)(cos 90°) − (3.0 m/s)(cos 90°)
v2, i (cos q2, i) = −4.0 m/s + 0 m/s + 0 m/s = −4.0 m/s
In the y-direction:
m1v1, i (sin q1, i) + m2v2, i(sin q2, i) = m1v1, f (sin q1, f ) + m2v2, f (sin q2, f )
v2, i (sin q2, i ) = v1, f (sin q1, f ) + v2, f (sin q2, f ) − v1, i (sin q1, i ) = (4.0 m/s)(sin 180°)
+ (3.0 m/s)(sin 90°) − (3.0 m/s)(sin 90°)
v2, i (sin q2, i) = 0 m/s + 3.0 m/s − 3.0 m/s = 0 m/s
This equation indicates that
sin q2, i = 0, or q2, i = 0° or 180°
v2, i (cos q2, i) = −4.0 m/s
v2, i = 4.0 m/s
q2, i = cos−1(−1.0) = 180°
v2, i =
Conservation of kinetic energy (check)
12
m1v1, i2 + 1
2m2v2, i
2 = 12
m1v1, f2 + 1
2m2v2, f
2
v1, i2 + v2, i
2 = v1, f2 + v2, f
2
(3.0 m/s)2 + (4.0 m/s)2 = (4.0 m/s)2 + (3.0 m/s)2
9.0 m2/s2 + 16 m2/s2 = 16 m2/s2 + 9.0 m2/s2
25 m2/s2 = 25 m2/s2
4.0 m/s to the west
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Section Five—Problem Bank V Ch. 6–17
V
6. m1 = 0.75 kg
m2 = 0.50 kg
m3 = 0.50 kg
v2, i = 0 m/s
v3, i = 0 m/s
v1, f = 0.80 m/s to the east(at 0°)
v2, f = 3.4 m/s at 45° northof east (at 45°)
v3, f = 3.4 m/s at 45° southof east (at −45°)
Momentum conservation
In the x-direction:
m1v1, i (cos q1, i) + m2v2, i (cos q2, i) + m3v3, i (cos q3, i) = m1v1, f (cos q1, f) + m2v2, f (cos q2, f) + m3v3, f (cos q3, f)
m1v1, i (cos q1, i) = m1v1, f (cos q1, f ) + m2v2, f (cos q2, f ) + m3v3, f (cos q3, f ) − m2v2, i(cos q2, i) − m3v3, i(cos q3, i)
m1v1, i (cos q1, i) = (0.75 kg)(0.80 m/s)(cos 0°) + (0.50 kg)(3.4 m/s)(cos 45°) + (0.50 kg)(3.4 m/s)[cos (−45°)] − (0.50 kg)(0 m/s) − (0.50 kg)(0 m/s)
m1v1, i (cos q1, i) = 0.60 kg•m/s + 1.2 kg•m/s + 1.2 kg•m/s − 0 kg•m/s − 0 kg•m/s m1v1, i (cos q1, i) = 3.0 kg•m/s
v1, i (cos q1, i ) = = 4.0 m/s
In the y-direction:
m1v1, i (sin q1, i ) + m2v2, i (sin q2, i ) + m3v3, i (sin q3, i ) = m1v1, f (sin q1, f ) + m2v2, f (sin q2, f ) + m3v3, f (sin q3, f )
Because v2, i, v3, i, and sin q1, f equal 0, m1v1, i (sin q1, i ) = m2v2, f (sin q2, f ) + m3v3, f (sin q3, f ) = (0.50 kg)(3.4 m/s)(sin 45°) + (0.50 kg)(3.4 m/s)[sin(−45°)] = 1.2 kg•m/s − 1.2 kg•m/s = 0 kg•m/s
This result indicates that sin q1, i = 0, or q1, i = 0° or 180°
v1, i (cos q1, i) = 4.0 m/s
v1, i = 4.0 m/s
q1, i = cos−1(1.0) = 0°
v1, i =
Conservation of kinetic energy (check)
12
m1v1, i2 + 1
2m2v2, i
2 + 12
m3v3, i2 = 1
2m1v1, f
2 + 12
m2v2, f2 + 1
2m3v3, f
2
12
(0.75 kg)(4.0 m/s)2 + 12
(0.50 kg)(0 m/s)2 + 12
(0.50 kg)(0 m/s)2
= 12
(0.75 kg)(0.80 m/s)2 + 12
(0.50 kg)(3.4 m/s)2 + 12
(0.50 kg)(3.4 m/s)2
6.0 J + 0 J + 0 J = 0.24 J + 2.9 J + 2.9 J
6.0 J = 6.0 J
4.0 m/s to the east
3.0 kg•m/s
0.75 kg
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Holt Physics Solution ManualV Ch. 6–18
V
7. v2, i = 2.000 m/s upward
v2, f = 1.980 m/s upward
∆yi = −20.4 m
a = −g = −9.81 m/s2
m1 = 0.150 kg
m2 = 325.0 kg
hi = 20.4 m
Velocity of ball dropped from rest is
v1, i = ±√
2a∆yi = ±√
(2)(−9.81 m/s2)(−20.4 m) = ± 20.0m/s
v1, i = −20.0 m/s = 20.0 m/s downward
Momentum conservation
m1v1, i + m2v2, i = m1v1, f + m2v2, f
v1, f =
v1, f =
v1, f = = = 23.3 m/s
v1, f =
∆yf = = = 26.9 m
h = ∆yf − hi = 27 m − 20.4 m =
Conservation of kinetic energy (check)
12
m1v1, i2 + 1
2m2v2, i
2 = 12
m1v1, f2 + 1
2m2v2, f
2
12
(0.150 kg)(−20.0 m/s)2 + 12
(325.0 kg)(2.000 m/s)2 = 12
(0.150 kg)(23 m/s)2
+ 12
(325.0 kg)(1.980 m/s)2
30.0 J + 650.0 J = 4.0 × 101 J + 637.1 J
680.0 J = 677 J
The slight difference arises from rounding.
7 m above the shaft
(23.3 m/s)2
(2)(9.81 m/s2)
v1, f2
2g
23.3 m/s upward
3.50 kg•m/s
0.150 kg
−3.00 kg•m/s + 650.0 kg•m/s − 643.5 kg•m/s
0.150 kg
(0.150 kg)(−20.0 m/s) + (325.0 kg)(2.000 m/s)−(325.0 kg)(1.980 m/s)
0.150 kg
m1v1, i + m2v2, i − m2v2, fm1
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8. v2, i = 2.000 m/s downward= −2.000 m/s
v2, f = 2.017 m/s downward= −2.017 m/s
v1, i = 20.0 m/s downward= −20.0 m/s
g = 9.81 m/s2
m1 = 0.150 kg
m2 = 325.0 kg
hi = 20.4 m
Momentum conservation
m1v1, i + m2v2, i = m1v1, f + m2v2, f
v1, f =
v1, f =
v1, f = = = 17 m/s
v1, f =
∆yf = = = 14.2 m
h = ∆yf − hi = 14.2 m − 20.4 m = −6.20 m
h = 6.20 m below the top of the shaft
(17 m/s)2
(2)(9.81 m/s2)
v1, f2
2g
16.7 m/s upward
2.5 kg•m/s0.150 kg
−3.00 kg•m/s − 650.0 kg•m/s + 655.5 kg•m/s
0.150 kg
(0.150 kg)(−20.0 m/s) + (325.0 kg)(−2.000 m/s)−(325.0 kg)(−2.017 m/s)
0.150 kg
m1v1, i + m2v2, i − m2v2, fm1
Section Five—Problem Bank V Ch. 6–19
V
Conservation of kinetic energy (check)
12
m1v1, i2 + 1
2m2v2, i
2 = 12
mi v1, f2 + 1
2m2v2, f
2
12
(0.150 kg)(−20.0 m/s)2 + 12
(325.0 kg)(−2.000 m/s)2 = 12
(0.150)(17 m/s)2
+ 12
(325.0 kg)(−2.017 m/s)2
30.0 J + 650.0 J = 22 J + 661.1 J
680.0 J = 683 J
The slight difference arises from rounding.
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9. m1 = 0.500 kg
h = 40.0 cm
g = 9.81 m/s2
m2 = 2.50 kg
v2, i = 0 m/s
The velocity of steel ball at point of collision can be determined through conserva-tion of mechanical energy.
PEi = KEf
m1gh = 12
m1v1, i2
v1, i =√
2gh =√
(2)(9.81 m/s2)(0.400 m) = 2.80 m/s
m1v1, i + m2v2, i = m1v1, f + m2v2, f
v2, f =
12
m1v1, i2 + 1
2m2v2, i
2 = 12
m1v1, f2 + 1
2m2v2, f
2
vi, f2 = v1, i
2 + − 2
Because v2, i = 0 m/s, the equation simplifies to the following:
v1, f2 = v1, i
2 − + -
1 + v1, f2 − v1, f + − 1(v1, i)
2 = 0
1 + v1, f2 − v1, f + − 1(2.80 m/s)2 = 0
(1.20)v1, f2 − (1.12 m/s)v1, f − 6.27 m2/s2 = 0
Solving for v1, f by using the quadratic equation,
v1, f =
v1, f = =
v1, f =1.12 m/s ± 5.59 m/s
2.40
1.12 m/s ±√
31.3 m2/s22.40
1.12 m/s ±√
1.25 m2/s2 + 30.1m2/s22.40
1.12 m/s ±√
(−1.12 m/s)2 − (4)(1.20)(−6.27 m2/s2)(2)(1.20)
0.500 kg2.50 kg
(2)(0.500 kg)(2.80 m/s)
2.50 kg
0.500 kg2.50 kg
m1m2
2m1v1, im2
m1m2
m1v1, f2
m2
2m1v1, iv1, fm2
m1v1, i2
m2
m1v1, i + m2v2, i − m1v1, fm2
m2m1
m2v2, i2
m1
m1v1, i + m2v2, i − m1v1, fm2
10. m1 = 7.00 kg
v1, i = 2.00 m/s to the east(at 0°)
m2 = 7.00 kg
v1, i = 0 m/s
v1, f = 1.73 m/s at 30.0°north of east
Momentum conservation
In the x-direction:
m1v1, i (cos q1, i) + m2v2, i (cos q2, i) = m1v1, f (cos q1, f ) + m2v2, f (cos q2, f )
v2, f (cos q2, f) = v1, i (cos q1, i) + v2, i (cos q2, i) − vi, f (cos q1, f)
v2, f (cos q2, f) = (2.00 m/s)(cos 0°) + 0 m/s − (1.73 m/s)(cos 30.0°)
v2, f = 2.00 m/s − 1.50 m/s = 0.50 m/s
In the y-direction:
m1v1, i (sin q1, i) + m2v2, i (sin q2, i) = m1v1, f (sin q1, i) m2v2, f (sin q2, f)
v2, f (sin q2, f ) = v1, i (sin q1, i) + v2, i (sin q2, i) − v2, f (sin q2, f)
v2, f (sin q2, f) = (2.00 m/s)(sin 0°) + 0 m/s − (1.73 m/s)(sin 30.0°) = −0.865 m/s
=
tan q2, f = −1.7
q2, f = tan−1(−1.7) = (−6.0 × 101)°
v2, f = = 1.0 m/s
v2, f =
Conservation of kinetic energy (check)
12
m1v1, i2 + 1
2m2v2, i
2 = 12
miv1, f2 + 1
2m2v2, f
2
12
(7.00 kg)(2.00 m/s)2 + 12
(7.00 kg)(0 m/s)2 = 12
(7.00 kg)(1.73 m/s)2 + 12
(7.00 kg)(1.0m/s)2
14.0 J + 0 J = 10.5 J + 3.5 J
14.0 J = 14.0 J
1.0 m/s at (6.0 × 101)° south of east
0.50 m/scos(−6.0 × 101)°
− 0.865 m/s
0.50 m/s
v2, f (sin q2, f)v2, f (cos q2, f)
Holt Physics Solution ManualV Ch. 6–20
V
The ball’s speed must have a smaller magnitude after collision than before. The positiveroot gives a final, forward speed that is close to the ball’s initial speed. Therefore the neg-ative root gives a more realistic result.
v1, f = −4.
2
4
.
7
40
m/s = −1.86 m/s
v2, f =
v2, f =
v2, f = = = 0.932 m/s
v1, f =
v2, f = 0.932 m/s forward
1.86 m/s backwards
2.33 kg•m/s
2.50 kg
1.40 kg•m/s + 0 kg•m/s + 0.930 kg•m/s
2.50 kg
(0.500 kg)(2.80 m/s) + (2.50 kg)(0 m/s) − (0.500 kg)(−1.86 m/s)
2.50 kg
m1v1, i + m2v2, i − m1v1, fm2
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Section Five—Problem Bank V Ch. 7–1
Chapter 7Rotational Motion and the Law of Gravity
V
1. ∆s = +24.0 m
r = 3.50 m∆θ =
∆r
s =
2
3
4
.5
.0
0
m
m = 6.86 rad
Additional Practice 7A
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2. r = 5.55 m
∆s = +31.3 m∆θ =
∆r
s =
3
5
1
.5
.3
5
m
m = 5.64 rad
3. area = πr2 = 2730 km2
∆s = −545 kmr =
arπea =
2730πkm2
= 29.5 km
∆θ = ∆r
s =
−2
5
9
4
.5
5
k
k
m
m = −18.5 rad
4. ∆s = 4.3 × 1011 m
∆θ = 0.39 radr =
∆∆
θs
= 4.3
0.
×39
10
ra
11
d
m = 1.1 × 1012 m
5. ∆s = 35.0 km
∆θ = 1.75 radr =
∆∆
θs
= 1
3
.
5
7
.
5
0
r
k
a
m
d = 20.0 km
6. ∆s = −36.6 µm
∆θ = −6
π rad
r = ∆∆
θs
= = 69.9 µm−36.6 µm
−6
π rad
7. r = 10.0 m
∆θ = +5.7 rad∆s = r∆θ = (10.0 m)(5.7 rad) = 57 m
8. r = 1.08 × 108 km
∆θ = + π3
rad∆s = r∆θ = (1.08 × 108 km)
π3
rad = 1.13 × 108 km
9. r = 4.48 × 109 km
∆θ = + π3
rad∆s = r∆θ = (4.48 × 109 km)
π3
rad = 4.69 × 109 km
10. r = 28.1 m
∆θ = –7.50 rad∆s = r∆θ = (28.1 m)(–7.50 rad)= 2.11 × 102 m
4. ωavg = 0.75 rad/s
∆θ = 3.3 rad∆t =
ω∆
a
θ
vg =
0.
3
7
.
5
3
r
r
a
a
d
d
/s = 4.4 s
Holt Physics Solution ManualV Ch. 7–2
V
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2. ∆θ = +2π rad
∆t = 4.56 minωavg =
∆∆
θt
= = 2.30 × 10−2 rad/s2π rad
(4.56 min)(60 s/min)
3. v = 280 m/s
∆x = 2.0 m
∆θ = +0.54 rad
∆t = ∆v
x
ωavg = ∆∆
θt
= = v
∆∆x
θ = = 12 rev/s
(280 m/s)(0.54 rad)(1 rev/2π rad)
2.0 m
∆θ
∆v
x
5. ωavg = 8.6 × 10−3 rad/s
∆θ = (6)(2π rad)∆t =
ω∆
a
θ
vg =
8.6
(6
×)(
1
2
0
π−3
ra
r
d
a
)
d/s = 4.4 × 103 s = 1.2 h
6. ωavg = 2.75 rad/s
∆θ = (3)(2π rad)∆t =
ω∆
a
θ
vg =
(
2
3
.
)
7
(
5
2πra
r
d
a
/
d
s
) = 6.85 s
7. ∆t = 4.2 h
ωavg = +2π rad/day∆θ = ωavg∆t = (2π rad/day)(1 day/24 h)(4.2 h) = 1.1 rad
8. ωavg = 1 rev/212 × 106 year
∆t = 4.50 × 109 year
∆θ = ωavg∆t = (1 rev/212 × 106 year)(4.50 × 109 year)(2π rad/rev) = 133 rad
9. ωavg = −1 rev/243 day
∆tE = 365.25 day
∆tV = 224.7 day
ωavg = (−1 rev/243 day)(2π rad/rev) =
∆θ = ωavg∆tE = (−2.59 × 10−2 rad/day)(365.25 day) =
∆θ = ωavg∆tV = (−2.59 × 10−2 rad/day)(224.7 day) = −5.82 rad
−9.46 rad
−2.59 × 10−2 rad/day
10. ∆x = 6.0 m
v = 5.0 m/s
ωavg = 6.00 rad/s
∆t = ∆v
x
∆θ = ωavg∆t = ωavg∆v
x =
(6.00 r
5
a
.0
d/
m
s)
/
(
s
6.0 m) = 7.2 rad
1. αavg = 1.5 rad/s2
ω1 = 3.0 rad/s
∆t = 4.0 s
ω2 = ω1 + αavg∆t = 3.0 rad/s + (1.5 rad/s2)(4.0 s) = 3.0 rad/s + 6.0 rad/s
ω2 = 9.0 rad/s
Additional Practice 7C
1. ∆θ = −106 rad
∆t = 7.5 sωavg =
∆∆
θt
= −1
7
0
.
6
5
r
s
ad = −14.1 rad/s
Additional Practice 7B
Section Five—Problem Bank V Ch. 7–3
V
Givens Solutions
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2. ω1 = 9.5 rad/s
αavg = −5.4 × 10−3 rad/s2
∆t = 22 min
ω2 = ω1 + αavg∆t = 9.5 rad/s + (−5.4 × 10−3 rad/s2)(22 min)(60 s/min)
ω2 = 9.5 rad/s − 7.1 rad/s = 2.4 rad/s
3. αavg = 32 rad/s2
∆t = 1.5 s
ω1 = 0 rad/s
ω2 = ω1 + αavg∆t = 0 rad/s + (32 rad/s2)(1.5 s) = 48 rad/s
4. ω2 = 76 rad/s
ω1 = 0 rad/s
αavg = 9.5 rad/s2
6. ω1 = 5.14 × 10−2 rad/s
ω2 = 3.09 × 10−2 rad/s
αavg = −1.75 × 10−3 rad/s2
∆t = ω2
α−
av
ω
g
1 = =
∆t = 11.7 s
−2.05 × 10−2 rad/s−1.75 × 10−3 rad/s2
3.09 × 10−2 rad/s − 5.14 × 10−2 rad/s
−1.75 × 10−3 rad/s2
9. ω1 = 2.07 rad/s
ω2 = 1.30 rad/s
∆t = 2.2 s
αavg = ω2
∆−t
ω1 = = −0.7
2
7
.2
r
s
ad/s = −0.35 rad/s21.30 rad/s − 2.07 rad/s
2.2 s
∆t = ω2
α−
av
ω
g
1 = 76 r
9
ad
.5
/s
ra
−d
0
/s
r2ad/s
= 8.0 s
5. αavg = 3.91 rad/s2
ω2 = 7.70 rad/s
ω1 = 2.50 rad/s
∆t = ω2
α−
av
ω
g
1 = = 1.33 s7.70 rad/s − 2.50 rad/s
3.91 rad/s2
7. ω1 = 4.0 rad/s
ω2 = 5.0 rad/s
∆t = 7.5 s
αavg = ω2
∆−t
ω1 = 5.0 rad/
7
s
.
−5
4
s
.0 rad/s =
1.0
7.
r
5
ad
s
/s = 0.13 rad/s2
8. ω1 = 7.14 rad/s
ω2 = 2.38 rad/s
∆t = 9.00 s
αavg = ω2
∆−t
ω1 = = 9
−.
4
0
.
0
76
s = −0.529 rad/s22.38 rad/s − 7.14 rad/s
9.0 s
10. ω1 = 2π rad/23.66 h
ω2 = 2π rad/24.00 h
∆t = 70.0 × 106 year
αavg = ω2
∆−t
ω1 =
αavg =
αavg = = −6.2 × 10−15 rad/h2−3.8 × 10−3 rad/h(70.0 × 106 year)(365.25 day/year)(24 h/day)
0.2618 rad/h − 0.2656 rad/h(70.0 × 106 year)(365.25 day/year)(24 h/day)
(2 π rad/24.00 h) − (2π rad/23.66 h)(70.0 × 106 year)(365.25 day/year)(24 h/day)
Holt Physics Solution ManualV Ch. 7–4
V
1. ωi = 0 rad/s
ωf = 3.33 rad/s
α = 0.183 rad/s2
∆t = ωf
α− ωi =
3.33
0
r
.1
a
8
d
3
/s
r
−ad
0
/s
r2ad/s
= 18.2 s
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2. ωi = 0 rad/s
α = 0.13 rad/s2
∆θ = 1.6 rad
Because ωi = 0 rad/s
∆θ = 12
α∆t2
∆t = 2∆αθ =
(
0
2.
)
1(
3
1 .
r
6adra
/s
d2)
= 5.0 s
3. ωi = 5.2 rad/s
ωf = 20.9 rad/s
∆θ = 216 rad
∆t = ω
2
i
∆+
θωf
= = = 16.6 s(2)(216 rad)
26.1 rad/s
(2)(216 rad)5.2 rad/s + 20.9 rad/s
4. ωi = 0.111 rad/s
ωf = 0.178 rad/s
α = 1.1 × 10−2 rad/s2
∆θ = ωf
2
2
−a
ωi2
=
∆θ = = = 0.87 rad1.94 × 10−2 rad2/s2
2.2 × 10−2 rad/s2
3.17 × 10−2 rad2/s2 − 1.23 × 10−2 rad2/s2
2.2 × 10−2 rad/s2
(0.178 rad/s)2 − (0.111 rad/s)2
(2)(1.1 × 10−2 rad/s2)
Additional Practice 7D
5. ωi = 78.0 rev/min
∆t = 30.0 s
α = −0.272 rad/s2
∆θ = ωi∆t + 12
α∆t2
∆θ = (78.0 rev/min)(2π rad/rev)(1 min/60 s)(30.0 s) + 12
(−0.272 rad/s2)(30.0 s)2
∆θ = 245 rad − 122 rad = 123 rad
∆θ = (123 rad)(1 rev/2π rad) = 19.6 rev = 123 rad = 19.6 rev
6. ωi = 298 rad/s
α = −44.0 rad/s2
∆θ = 276 rad
ωf2 = ωi
2 + 2α∆θ
ωf =√
ωi2 + 2α∆θ =
√(298 rad/s)2 + (2)(−44.0 rad/s2)(276 rad)
ωf =√
8.88 × 104 rad2/s2 − 2.43× 104 rad2/s2 =√
6.45 × 104 rad2/s2 f = 254 rad/s
7. ωi = 0 rad/s
∆t = 13.0 s
∆θ = 10.0 rev
ωf = 2
∆∆
t
θ − ωi = − 0 rad/s f = 9.67 rad/s
(2)(10.0 rev)(2π rad/rev)
13.0 s
8. ωi = 1200 rev/min
ωf = 3600 rev/min
∆t = 12 s
α = ωf
∆−t
ωi =
α = = 21 rad/s2(2400 rev/min)(2π rad/rev)(1 min/60 s)
12 s
(3600 rev/min − 1200 rev/min)(2π rad/rev)(1 min/60 s)
12 s
Section Five—Problem Bank V Ch. 7–5
V
9. ∆θ = 158 rad
ωi = 0 rad/s
ωf = 70.0 rad/s
α = ωf
2
2∆−
θωi
2
= = 15.5 rad/s2(70.0 rad/s)2 − (0 rad/s)2
(2)(158 rad)
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10. ωi = 3.29 rad/s
∆t = 2.50 s
∆θ = 12.3 rad
α = 2(∆θ
∆−t2ωi∆t) =
α = = (2)
(
(
2
4
.5
.1
s)
r2ad)
α = 1.3 rad/s2
(2)(12.3 rad − 8.2 rad)
(2.5 s)2
(2)[(12.3 rad) − (3.29 rad/s)(2.5 s)]
(2.5 s)2
1. ω = 2.07 × 10−3 rad/s
r = 1.5 km
vt = rω = (1.5 × 103 m)(2.07 × 10−3 rad/s) = 3.1 m/s
2. ω = 188.5 rad/s
r = 3.73 cmvt = rω = (3.73 × 10−2 m)(188.5 rad/s) = 7.03 m/s
4. r = 0.30 m
vt = 4.5 m/sω =
v
rt =
4
0
.
.
5
30
m
m
/s = 15 rad/s
Additional Practice 7E
3. r = 15.2 m
ω = 6.28 rad/svt = rω = (15.2 m)(6.28 rad/s) = 95.5 m/s
5. r = 2.00 m
vt = 94.2 m/sω =
v
rt =
9
2
4
.
.
0
2
0
m
m
/s = 47.1 rad/s
6. vt = 0.63 m/s
r = 1.5 mω =
v
rt =
0.
1
6
.
3
5
m
m
/s = 0.42 rad/s
7. ω = 3.14 × 10−2 rad/s
vt = 0.45 m/sr =
v
wt =
3.14
0
×.4
1
5
0
m−2
/
r
s
ad/s = 14 m
8. ω = 10.0 rad/s
vt = 4.60 m/sr =
ωvt =
1
4
0
.
.
6
0
0
r
m
ad
/
/
s
s = 0.460 m = 46.0 cm
9. ω = 11 rad/s
vt = 4.0 cm/sr =
ωvt =
4
1
.
1
0
r
c
a
m
d/
/
s
s = 0.36 cm = 3.6 mm
10. vt = 1.5 m/s
ω = 0.33 rad/sr =
ωvt =
0.
1
3
.
3
5
r
m
ad
/s
/s = 4.5 m
Holt Physics Solution ManualV Ch. 7–6
V
1. r = 6.0 cm
α = 35.2 rad/s2at = rα = (6.0 10−2 m)(35.2 rad/s2) = 2.1 m/s2
Additional Practice 7F
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2. α = 105 rad/s2
r = 1.75 cm
at = rα = (1.75 cm)(105 rad/s2) = 184 cm/s2 = 1.84 m/s2
3. r = 5.87 m
α = 1.40 × 10−2 rad/s2at = rα = (5.87 m)(1.40 × 10−2 rad/s2) = 8.22 × 10−2 m/s2
4. ∆ω = 1.23 × 10−2 rad/s
∆t = 10.0 s
at = 7.50 × 10−2 m/s2
α = ∆∆ωt =
1.23 ×1
1
0
0
.0
−2
s
rad/s = 1.23 × 10−3 rad/s2
r = a
αt =
1
7
.2
.5
3
0
××
1
1
0
0−
−
3
2
r
m
ad
/
/
s
s
2
2 = 61.0 m
5. α = 42 rad/s2
at = 64 m/s2r =
a
rt =
4
6
2
4
r
m
ad
/
/
s
s
2
2 = 1.5 m
6. α = 6.25 × 10−2 rad/s2
at = 0.75 m/s2r =
a
rt =
6.25
0
×.7
1
5
0
m−2
/
r
s2
ad/s2 = 12 m
7. at = 0.157 m/s2
r = 0.90 mα =
a
rt =
0.1
0
5
.9
7
0
m
m
/s2
= 0.17 rad/s2
8. r = 1.75 m
at = 0.83 m/s2α =
a
rt =
0.
1
8
.
3
75
m
m
/s2
= 0.47 rad/s2
9. r = 0.50 m
∆v = 5.0 m/s
∆t = 8.5 s
at = ∆∆
v
t =
5.
8
0
.5
m
s
/s = 0.59 m/s2
α = a
rt =
0.
0
5
.
9
50
m
m
/s2
= 1.2 rad/s2
10. r = 16 cm
at = 0.59 m/s2α =
a
rt =
16
0.
×59
1
m
0−/2s2
m = 3.7 rad/s2
1. r = 3.81 m
vt = 124 m/sac =
v
rt2
= (1
3
2
.
4
81
m
m
/s)2
= 4.04 × 103 m/s2
Additional Practice 7G
Section Five—Problem Bank V Ch. 7–7
V
2. r = 6.50 cm
ω = 30.0 rad/sac = rω2 = (6.50 × 10−2 m)(30.0 rad/s)2 = 58.5 m/s2
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3. r = 11 m
vt = 1.92 × 10−2 m/sac =
v
rt2
= (1.92 ×
1
1
1
0
m
−2 m/s)2
= 3.4 × 10−5 m/s2
4. r = 8.9 m
ac = (20.0) g
g = 9.81 m/s2
vt =√
rac =√
(8.9 m)(20.0)(9.81 m/s2) = 42 m/s
5. r = 4.2 m
ac = 2.13 m/s2vt =
√rac =
√(4.2 m)(2.13 m/s2) = 3.0 m/s
6. ac = g = 9.81 m/s2
r = 150 m
vt =√
rac =√
(150 m)(9.81 m/s2) = 38 m/s
7. vt = 75.0 m/s
ac = 22.0 m/s2r =
v
at
c
2
= (7
2
5
2
.
.
0
0
m
m
/
/
s
s
)2
2
= 256 m
8. ω = 3.5 rad/s
ac = 2.0 m/s2r =
ωac
2 =
(3
2
.5
.0
r
m
ad
/
/
s
s
2
)2 = 0.16 m = 16 cm
9. vt = 0.35 m/s
ac = 0.29 m/s2r =
v
at
c
2
= (0
0
.
.
3
2
5
9
m
m
/
/
s
s
)2
2
= 0.42 m = 42 cm
10. ac = 9.81 m/s2
vt = 15.7 m/sr =
v
at
c
2
= (1
9
5
.8
.7
1
m
m
/
/
s
s
)2
2
= 25.1 m
1. m = 40.0 kg
ω = 0.50 rad/s
r = 18.0 m
Fc = mrω2 = (40.0 kg)(18.0 m)(0.50 rad/s)2 = 180 N
Additional Practice 7H
2. r = 0.25 m
vt = 5.6 m/s
m = 0.20 kg
Fc = mv
rt2
= (0.20 kg)(5
0
.6
.2
m
5 m
/s)2
= 25 N
Holt Physics Solution ManualV Ch. 7–8
V
3. vt = 48.0 km/h
r = 35.0 m
mk = 0.500
m = 1250 kg
g = 9.81 m/s2
Fc = mv
rt2
= (1250 kg)
Fc =
Ff = µkFn = µkmg = (0.500)(1250 kg)(9.81 m/s2)
Ff =
The available frictional force is not large enough to maintain the automobile’scircular motion.
6130 N
6350 N
[(48.0 km/h)(103 m/km)(1 h/3600 s)]2
35.0 m
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4. m = 1250 kg
r = 35.0 m
θ = 9.50°
g = 9.81 m/s2
µk = 0.500
F = Ff + mg(sin θ) = µkFn + mg(sinθ) = µkmg(cos θ) + mg(sin θ)
F = (0.500)(1250 kg)(9.81 m/s2)(cos 9.50°) + (1250 kg)(9.81 m/s2)(sin 9.50°)
F = 6.05 × 103 N + 2.02 × 103 N
F =
Fc = F = 8.07 × 103 N
vt = F
mcr = vt = 15.0 m/s = 54.0 km/h
(8.07 × 103 N)(35.0 m)
1250 kg
8.07 × 103 N
5. m = 2.05 × 108 kg
r = 7378 km
Fc = 3.00 × 109 N
vt = F
mcr = vt = 1.04 × 104 m/s = 10.4 km/s
(3.00 × 109 N)(7378 × 103 m)
2.05 × 108 kg
6. m = 55 kg
ω = 2.0 rad/s
Fc = 135 N
r = m
F
ωc
2 = (55 kg)
1
(
3
2
5
.0
N
rad/s)2 = 0.61 m = 61 cm
7. m = 7.55 × 1013 kg
vt = 0.173 km/s
Fc = 505 N
r = m
F
v
c
t2
= = 4.47 × 1015 m(7.55 × 1013 kg)(0.173 × 103 m/s)2
505 N
8. ω = 36.7 rad/s
r = 0.10 m
Fc = 670 N
m = r
F
ωc2 = = 5.0 kg
670 N(0.10 m)(36.7 rad/s)2
9. r = 35.0 cm
vt = 2.21 m/s
Fc = 0.158 N
m = F
vt
c2r
= = 1.13 × 10−2 kg = 11.3 g(0.158 N)(35.0 × 10−2 m)
(2.21 m/s)2
10. Fc = 8.00 × 102 N
r = 0.40 m
vt = 6.0 m/s
m = F
vt
c2r
= = 8.9 kg(8.00 × 102 N)(0.40 m)
(6.0 m/s)2
Section Five—Problem Bank V Ch. 7–9
V
Givens SolutionsC
opyr
ight
©by
Hol
t, R
ineh
art a
nd W
inst
on.A
ll rig
hts
rese
rved
.
1. m1 = 2.04 × 104 kg
m2 = 1.81 × 105 kg
r = 1.5 m
G = 6.673 × 10−11 N•m2
kg2
Fg = G m
r1m
22 = 6.673 × 10−11 = 0.11 N
(2.04 × 104 kg)(1.81 × 105 kg)
(1.5 m)2
N•m2
kg2
2. m1 = 1.4 × 1021 kg
m2 = 5.98 × 1024 kg
r = 3.84 × 108 m
G = 6.67 × 10−11 N•m2
kg2
Fg = G m
r1m
22 = 6.673 × 10−11 = 3.8 × 1018 N
(1.4 × 1021 kg)(5.98 × 1024 kg)
(3.84 × 108 m)2N•m2
kg2
3. m1 = 0.500 kg
m2 = 2.50 × 1012 kg
r = 10.0 km
G = 6.673 × 10−11 N•m2
kg2
Fg = G m
r1m
22 = 6.673 × 10−11 = 8.34 × 10−7 N
(0.500 kg)(2.50 × 1012 kg)
(10.0 × 103 m)2N•m2
kg2
Additional Practice 7I
4. Fg = 2.77 × 10−3 N
r = 2.50 × 10−2 m
m1 = 157 kg
G = 6.673 × 10−11 N•m2
kg2
m2 = = = 165 kg(2.77 × 10−3 N)(2.50 × 10−2 m)2
6.673 × 10−11 (157 kg)
Fgr2
Gm1
5. Fg = 1.636 × 1022 N
m1 = 1.90 × 1027 kg
r = 1.071 × 106 km
G = 6.673 × 10−11 N•m2
kg2
m2 = = = 1.48 × 1023 kg(1.636 × 1022 N)(1.071 × 109 m)2
6.673 × 10−11 (1.90 × 1027 kg)
Fgr2
Gm1
6. Fg = 1.17 × 1018 N
m1 = 1.99 × 1030 kg
r = 4.12 × 1011 m
G = 6.673 × 10−11 N•m2
kg2
m2 = = = 1.50 × 1021 kg(1.17 × 1018 N)(4.12 × 1011 m)2
6.673 × 10−11 (1.99 × 1030 kg)
Fgr2
Gm1
7. m1 = m2 = 9.95 × 1041 kg
Fg = 1.83 × 1029 N
G = 6.673 × 10−11 N•m2
kg2
r = Gm
F1
g
m2 = = 1.90 × 1022 m6.673 × 10−11
N
k
•
g
m2
2
(9.95 × 1041 kg)2
1.83 × 1029 N
N•m2
kg2
N•m2
kg2
N•m2
kg2
Holt Physics Solution ManualV Ch. 7–10
V
8. m1 = 1.00 kg
m2 = 1.99 × 1030 kg
Fg = 274 N
G = 6.673 × 10−11 N•m2
kg2
r = Gm
F1
g
m2 = = 6.96 × 108 m
6.673 × 10−11 N
k
•
g
m2
2
(1.00 kg)(1.99 × 1030 kg)
274 N
Givens Solutions
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9. m1 = 1.00 kg
m2 = 3.98 × 1031 kg
Fg = 2.19 × 10−3 N
G = 6.673 × 10−11 N•m2
kg2
r = Gm
F1
g
m2 = = 1.10 × 1012 m
6.673 × 10−11 N
k
•
g
m2
2
(1.00 kg)(3.98 × 1031 kg)
2.19 × 10−3 N
10. Fg = 125 N
m1 = 4.5 × 1013 kg
m2 = 1.2 × 1014 kg
G = 6.673 × 10−11 N•m2
kg2
r = Gm
F1
g
m2 = = 5.4 × 107 m
6.673 × 10−11 N
k
•
g
m2
2
(4.5 × 1013 kg)(1.2 × 1014 kg)
125 N
Section Five—Problem Bank V Ch. 8–1
Chapter 8Rotational Equilibrium and Dynamics
V
1. d = 1.60 m
t = 4.00 × 102 N • m
q = 80.0°
F = d(si
tn q) =
F = 254 N
4.00 × 102 N • m(1.60 m)(sin 80.0°)
Additional Practice 8A
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2. tnet = 14.0 N • m
d ′ = 0.200 m
q ′ = 80.0°
t = 4.00 × 102 N • m
tnet = t − t ′
t ′ = Fgd ′(sin q ′) = t − tnet
Fg = d
t′(
−sin
tn
qet
′) =
Fg = = 1.96 × 103 N386 N • m
(0.200 m)(sin 80.0°)
4.00 × 102 N • m − 14. N • m
(0.200 m)(sin 80.0°)
3. d = 2.44 m
t = 50.0 N • m
q = 90°
F = =
F = 20.5 N
50.0 N • m(2.44 m)(sin 90°)
td(sin q)
4. t = 1.4 N • m
d = 0.40 m
q = 60.0°
F = d(si
tn q) =
F =
tmax is produced when q = 90°, or
tmax = Fd = (4.0 N)(0.40 m) = 1.6 N • m
4.0 N
1.4 N • m(0.40 m)(sin 60.0°)
5. Fmax = 2.27 × 105 N • m
r = 0.660 m
d = 12
r
tmax = Fmaxd = Fm
2axr
tmax = = 7.49 × 104 N • m(2.27 × 105 N • m)(0.660 m)
2
6. m = 1.6 kg
l = 43 cm
x = 15 cm
q = 90°
g = 9.81 m/s2
t = F d(sin q) = mg (l − x)(sin q)
t = (1.6 kg)(9.81 m/s2)(0.43 m − 0.15 m)(sin 90°) = (1.6 kg)(9.81 m/s2)(0.28 m)
t = 4.4 N • m
7. t = 0.46 N • m
F = 0.53 N
q = 90°
d = F(si
tn q) =
d = 0.87 m
0.46 N • m(0.53 N)(sin 90°)
9. m = 28 kg
g = 9.81 m/s2
q = 89°
t = 1.84 × 104 N • m
d = F(si
tn q) =
mg(s
tin q) =
d = 67 m
1.84 × 104 N • m(28 kg)(9.81 m/s2)(sin 89°)
Holt Physics Solution ManualV Ch. 8–2
V
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8. t = 8.25 × 103 N • m
F = 587 N
q = 65.0°
d = F(si
tn q) =
(5
8
8
.2
7
5
N
×)
1
(s
0
i
3
n
N
65
•
.0
m
°)
d = 15.5 m
10. Fb = 1.200 × 103 N
qb = 90.0°
m = 60.0 kg
g = 9.81 m/s2
qg = 87.7°
tnet = −2985 N • m
tnet = tg − tb = Fgd(sin qg) − Fbd(sin qb)
d = =
d =
d = 588
−−2
1
9
.
8
2
5
00
N
ו
1
m
03 N
d = −29
−8
6
5
12
N
N
• m = 4.88 m
−2985 N • m(60.0 kg)(9.81 m/s2)(sin 87.7°) − (1.200 × 103 N)(sin 90.0°)
tnetmg(sin qg) − Fb(sin qb)
tnetFg(sin qg) − Fb(sin qb)
1. m1 = 2.3 kg
m2 = 0.40 kg
l = 1.00 m
g = 9.81 m/s2
Apply the second condition of equilibrium.
tnet = t1 − t2 = 0
t1 = Fg,1d1 = m1gx
t2 = Fg,2d2 = m2g(l − x)
m1gx = m2g(l − x)
(m1 + m2)x = m2l
x = m1
m
+2l
m2 =
(0
2
.
.
4
3
0
k
k
g
g
+)(
0
1
.
.
4
0
0
0
k
m
g
) =
(0.40 k
2
g
.7
)(
k
1
g
.00 m)
x = 0.15 m from the ostrich egg
2. mms = 139 g
dms = 49.7 cm
mw = 50.0 g
dw = 10.0 cm
g = 9.81 m/s2
Apply the second condition of equilibrium.
tnet = tms − tw = 0
Let x be the distance from the fulcrum to the zero mark.
tms = mmsg(dms − x)
tw = mwg(x − dw)
mmsg(dms − x) = mwg(x − dw)
mmsdms + mwdw = (mw + mms)x
x = mm
msd
m
m
s
s
++
m
m
w
wdw =
x = =
x = 39.2 cm from the zero mark
7.41 × 103 g • cm
189 g
6.91 × 103 g • cm + 5.00 × 102 g • cm
189 g
(139)(49.7 cm) + (50.0 g)(10.0 cm)
139 g + 50.0 g
Additional Practice 8B
Section Five—Problem Bank V Ch. 8–3
V
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3. mA = 4.64 × 107 kg
dA = 3.00 × 102 m
Fcs = 3.22 × 107 N
Fs = 7.55 × 108 N
g = 9.81 m/s2
Apply the first condition of equilibrium to solve for the weight of the cantilever arm, mcg.
Fs − mAg − mCg − Fcs = 0
mCg = Fs − mAg − Fcs = 7.55 × 108 N − (4.64 × 107 kg)(9.81 m/s2) − 3.22 × 107 N
mCg = 7.55 × 108 N − 4.55 × 108 N − 3.22 × 107 N = 2.68 × 108 N
To solve for dC, apply the second condition of equilibrium using the support for the can-tilever as the pivot point.
mAgd
2A − mCg
d
2C − FcsdC = 0
m2Cg + FcsdC =
mA
2
gdA
dC = m
mA
C
gd
gA + 2Fcs =
dC = =
dC = 411 m
(4.64 × 107 kg)(9.81 m/s2)(3.00 × 102 m)
3.32 × 108 N
(4.64 × 107 kg)(9.81 m/s2)(3.00 × 102 m)
2.68 × 108 N + 6.44 × 107 N
(4.64 × 107 kg)(9.81 m/s2)(3.00 × 102 m)
2.68 × 108 N + (2)(3.22 × 107 N)
4. mf = 70.0 kg
g = 9.81 m/s2
q = 10.0°
t = 7.08 × 103 N • m
Fu = 3.14 × 103 N
Apply the first condition of equilibrium to find the weight of the ladder, ml g.
Fu − mfg − mlg = 0
mlg = Fu − mf g
mlg = 3.14 × 103 N − (70.0 kg)(9.81 m/s2) 3.14 × 103 N − 687 N = 2.45 × 103 N
To solve for the length of the ladder (d), apply the second condition of an equilib-rium, using the base of the ladder as the pivot point.
t − mlgd
2(sin q) − mf gd(sin q) = 0
m
2l g + mf g(sin q)d = t
d = =
d = = = 21.3 m(2)(7.08 × 103 N • m)(3.82 × 103 N)(sin 10.0°)
(2)(7.08 × 103 N • m)(2.45 × 103 N + 1.37 × 103 N)(sin 10.0°)
(2)(7.08 × 103 N • m)[2.45 × 103 N + (2)(70.0 kg)(9.81 m/s2)](sin 10.0°)
2t(mlg + 2 mf g)(sin q)
5. dc = 32.0 m
q = 60.0°
FT,x = 1.233 × 104 N
FT,y = 1.233 × 104 N
g = 9.81 m/s2
Apply the first condition of equilibrium in the x direction.
Fx = Rx,base − FI,x = 0
Rx,base = R(cos q) = FT,x
R = c
F
oT
s,x
q
Apply the first condition of equilibrium in the y direction; solve for the flagpole’s weight.
Fy = Ry,base − mg − FT,y = 0
Ry,base = R(sin q) − FT,y = mg
cFoT
s,x
q(sin q) − FT,y = mg
mg = FT,x(tan q) − FT,y = (1.233 × 104 N)(tan 60.0°) − 1.233 × 104 N
mg = (1.233 × 104 N)[(tan 60.0°) − 1.00] = (1.233 × 104 N)(1.73 − 1.00)
mg = (1.233 × 104 N)(0.73) = 9.0 × 103 N
Holt Physics Solution ManualV Ch. 8–4
V
To solve for the length of the flagpole (l ), apply the second condition of equilibrium,using the base of the flagpole as the pivot point.
FT,ydc(sin q) − FT,ydc(cos q) − mg (cos q) = 0
l =
l =
l = =
l = 64 m
1.45 × 105 N • m12
(9.0 × 103 N)(cos 60.0°)
3.42 × 105 N • m − 1.97 × 105 N • m
12
(9.0 × 103 N)(cos 60.0°)
(1.233 × 104 N)(32.0 m)(sin 60.0°) − (1.233 × 104 N)(32.0 m)(cos 60.0°)
12
(9.0 × 103 N)(cos 60.0°)
FT,xdc(sin q) − FT,ydc(cos q)
12
mg(cos q)
l2
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6. mE = 5.98 × 1024 kg
g = 9.81 m/s2
dE = 1.00 m
dapplied = 3.8 × 1016 m
Apply the first condition of equilibrium.
Ffulcrum − Fapplied − mEg = c
Ffulcrum = Fapplied + mEg
To solve for Fapplied, apply the second condition of equilibrium, using the fulcrum asthe pivot point.
Fapplieddapplied − mEgdE = 0
Fapplied = d
m
ap
E
p
g
l
d
ie
E
d =
Fapplied =
Substitute the value for Fapplied into the first-condition equation and solve forFfulcrum.
Ffulcrum = 1.5 × 109 N + (5.98 × 1024 kg)(9.81 m/s2) = 1.5 × 109 N + 5.87 × 1025 N
Ffulcrum = 5.87 × 1025 N
1.5 × 109 N
(5.98 × 1024 kg)(9.81 m/s2)(1.00 m)
3.8 × 1016 m
7. d = 2.00 m
q = 30.0°
t = 1.47 × 103 N • m
Apply the first condition of equilibrium in the x and y directions.
Fx = Frod − FT(cos q) = 0
Frod = FT(cos q)
Fy = FT(sin θ) − Fg = 0
Fg = FT(sin q)
Apply the second condition of equilibrium, using the end of the rod anchored in thewall as the pivot point.
t − Fgd = 0
Fg = d
t =
Fg =
Substitute the value for Fy into the second first-condition equation to solve for FT.Substitute the value of FT into the first first-condition equation to solve for Frod.
FT = si
F
ng
q =
si
7
n
3
3
5
0
N
.0° = 1470 N
Frod = FT(cos q) = (1470 N)(cos 30.0°)
Frod = 1270 N
735 N
1.47 × 103 N • m
2.00 m
Section Five—Problem Bank V Ch. 8–5
V
8. Fg = 7.10 × 102 N
ms = 0.75
dx = 1.22 m
dy = 1.00 m
Apply the first condition of equilibrium to determine the largest force that will notovercome static friction.
Fapplied − Fs = 0
Fapplied = Fs = msFn = msFg
Fapplied = (0.75)(7.10 × 102 N) = 532 N
Apply the second condition of equilibrium to determine the largest force that will notlift the door from the rail. Choose one of the wheels for the axis of rotation.
Fapplieddy − Fgd
2x = 0
Fapplied = F
2g
d
d
y
x =
Fapplied = 433 N
The largest force that will not upset either equilibrium condition is the smaller of thetwo forces.
Fapplied = 433 N
(7.10 × 102 N)(1.22 m)
(2)(1.00 m)
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9. m = 307 kg
h = 2.44 m
q = 70.0°
d = 1.22 m
g = 9.81 m/s2
Apply the first condition of equilibrium in the x direction.
Fx = Fapplied(sin q) − R(cos q) = 0
R = Fapplied(tan q)
To solve for Fapplied, either use the first condition of equilibrium in the y direction orthe second condition of equilibrium.
Fy = Fapplied(cos q) + R(sin q) − mg = 0
Fapplied(cos q) + [Fapplied(tan q)](sin q) = mg
Fapplied = =
Fapplied = =
Fapplied =
Alternatively,
Fappliedd − mgh
2(cos q) = 0
Fapplied = mgh(
2
c
d
os q) =
Fapplied = 1030 N
Substitute the value of Fapplied into the first-condition equation in the x direction tosolve for R.
R = Fapplied(tan q) = (1030 N)(tan 70.0°) = 2830 N
(307 kg)(9.81 m/s2)(2.44 m)(cos 70.0°)
(2)(1.22 m)
1030 N
(307 kg)(9.81 m/s2)
2.92
(307 kg)(9.81 m/s2)
0.342 + 2.58
(307 kg)(9.81 m/s2)(cos 70.0°) + (tan 70.0°)(sin 70.0°0)
mg(cos q) + (tan q)(sin q)
Holt Physics Solution ManualV Ch. 8–6
V
10. Fg = 1.96 × 103 N
d1 = 0.250 m
d2 = 1.50 m
Apply the first condition of equilibrium to the x and y directions.
Fx = F − Rx = 0
Rx = F
Fy = Ry − Fg = 0
Ry = Fg = 1.96 × 103 N
To solve for F, apply the second condition of equilibrium, using the hinge as the pivotpoint.
Fgd1 − Fd2 = 0
F = F
dgd
2
1 =
F =
Substitute the value for F in the first-condition equation for the x direction, thensolve for R using the Pythagorean theorem.
Rx = 327 N
R =√
Rx2 + Ry
2 =√
(327 N)2 + (1.96 × 103 N)2R =
√1.07 × 105 N2+ 3.84× 106 N2 =
√3.95 × 106 N2
R = 1.99 × 103 N
327 N
(1.96 × 103 N)(0.250 m)
1.50 m
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1. t = 2.98 N • m
wi = 0 rad/s
wf = 55 rad/s
Dt = 0.75 s
R = 20.0 cm
I = at
= =
I =
For a hoop spinning on an axis along its diameter,
I = 12
MR2
M = R
2I2 =
M = 2.0 kg
(2)(0.041 kg • m2)
(0.200 m)2
0.041 kg • m2
2.98 N • m
55 rad
0
/
.
s
7
−5
0
s
rad/s
t
wf
∆−twi
2. t = 1.7 N • m
a = 5.5 rad/s2I =
at
= = 0.3 kg • m21.7 N • m5.5 rad/s2
Additional Practice 8C
3. t = 0.750 N • m
a = 499 rad/s2I =
at
= = 1.50 × 10−3 kg • m30.750 N • m499 rad/s2
4. M = 7.91 × 103 kg
R = 1.83 m
a = 6.13 rad/s2
I = 25
MR2 = 25
(7.91 × 103 kg)(1.83 m)2 = 1.06 × 104 kg • m2
t = Ia = (1.06 × 104 kg • m2)(6.13 rad/s2) = 6.50 × 104 N • m
Section Five—Problem Bank V Ch. 8–7
V
5. a = −6.53 × 10−27 rad/s2
M = 5.98 × 1024 kg
R = 6.37 × 106 m
I = 0.331 MR2 = (0.331)(5.98 × 1024 kg)(6.37 × 106 m)2 = 8.03 × 1037 kg • m2
t = Ia = (8.03 × 1037 kg • m2)(−6.53 × 10−22 rad/s2) = −5.24 × 1016 N • m
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6. a = 1.05 rad/s2
I = 8.14 × 104 kg • m2
t = Ia = (8.14 × 104 kg • m2)(1.05 rad/s2) = 8.55 × 104 N • m
8. t = 1.01 N • m
I = 3.85 × 10−5 kg • m2a =
tI
= = 2.62 × 104 rad/s21.01 N • m3.85 × 10−5 kg • m2
10. M = 15 kg
R = 0.25 m
wi = 9.5 rad/s
wf = 0 rad/s
t = −0.80 N • m
I = 12
MR2 = 12
(15 kg)(0.25 m)2 = 0.47 kg • m2
a = tI
= =
∆t = wf
a− wi =
0 ra
−d
1
/s
.7
−r
9
ad
.5
/s
r2ad/s
= −−1
9
.
.
7
5
r
r
a
a
d
d
/
/
s
s2
∆t = 5.6 s
−1.7 rad/s2−0.80 N • m0.47 kg • m2
Additional Practice 8D
7. t = 108 N • m
I = 5.40 kg • m2a =
tI
= = 20.0 rad/s2108 N • m5.40 kg • m2
9. m = 0.15 kg
r = 0.35 m
t = 1.5 N • m
Dt = 0.26 s
a = tI
= m
tr2
a = =
w = a∆t = (82 rad/s2)(0.26 s) = 21 rad/s
82 rad/s21.5 N • m(0.15 kg)(0.35 m)2
1. r = 0.120 m
m = 22.0 g
wi = 50.00 rad/s
wf = 50.24 rad/s
Li = Lf
Iiwi = Ifwf
(I + 25 mr2)wi = Iwf
I(wf − wi) = 25 mr2 wf
I = 25
wm
f −r2
ww
i
f =
I = 1.7 kg • m2
(25)(22.0 × 10−3 kg)(0.120 m)2 (50.24 rad/s)
50.24 rad/s − 50.00 rad/s
2. M = 755 kg
li = 1.75 m × 2
= 3.50 m
wi = 1.25 rad/s
wf = 1.70 × 10−2 rad/s
Li = Lf
Iiwi = If wf
1
1
2Mli
2wi = Ifwf
If = =
If = 5.67 × 104 kg • m2
(755 kg)(3.50 m)2 (1.25 rad/s)
(12)(1.70 × 10−2 rad/s)
Mli2wi
12wf
Holt Physics Solution ManualV Ch. 8–8
V
3. ri = 3.00 m
rf = 0.20 m
m = 55.0 kg
wi = 2.00 rad/s
wf = 2.35 rad/s
Li = Lf
Iiwi = If wf
(I + 4 mri2)wi = (I + 4 mrf
2)wf
I(wf − wi) = 4 m(ri2wi − rf
2wf)
I = 4 m(r
wi2
f
w−i −
wr
i
f2wf) =
I = =
I = 1.1 × 104 kg • m2
(4)(55.0 kg)(17.9 m2/s)
0.35 rad/s
(4)(55.0 kg)(18.0 m2/s − 0.094 m2/s)
0.35 rad/s
(4)(55.0 kg)[(3.00 m)2 (2.00 rad/s) − (0.20 m)2 (2.35 rad/s)]
2.35 rad/s − 2.00 rad/s
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4. wi = 2.1 rad/s
ri = 1.2 m
rf = 0.50 m
Li = Lf
Iiwi = Ifwf
mri2wi = mrf
2wf
wf = ri
r
2
f
w2
i = (1.2
(
m
0
)
.5
2
0
(2
m
.1
)
r2ad/s)
wf = 12 rad/s
5. v1 = 43.5 km/s
r1 = 7.00 × 107 km
r2 = 1.49 × 108 km
L1 = L2
I1w1 = I2w2
mr12w1 = mr2
2w2
r1v1 = r2v2
v2 = =
v2 = 20.4 km/s
(7.00 × 107 km)(43.5 km/s)
1.49 × 108 km
r1v1r2
6. wi = 57.7 rad/s
Mw = 3.81 kg
Rw = 0.350 m
Itot = 2.09 kg • m2
Li = Lf
Iwwi = Itotwf
MwRw2wi = Itotwf
wf = Mw
I
R
to
w
t
2wi =
wf = 12.9 rad/s
(3.81 kg)(0.350 m)2 (57.7 rad/s)
2.09 kg • m2
7. wi = 1.50 rad/s
wf = 2.04 × 10−2 rad/s
M = 7.55 kg
li = 3.50 m
Li = Lf
Iiwi = If wf
1
1
2Mli
2wi = 1
1
2Mlf
2wf
lf = = lf = 30.0 m
(3.50 m)2 (1.50 rad/s)
2.04 × 10−2 rad/s
l i2wiwf
Section Five—Problem Bank V Ch. 8–9
V
8. v1 = 3.68 km/s
r1 = 7.35 × 109 km
v2 = 6.14 km/s
L1 = L2
I1w1 = I2w2
mr12w1 = mr2
2w2
r1vi = r2v2
r2 = r1
v
v
2
1 =
r2 = 4.41 × 109 km
(7.35 × 109 km)(3.68 km/s)
6.14 km/s
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9. v1 = 3403 m/s
r1 = 3593 km
h1 = 2.00 × 102 km
v2 = 3603 m/s
Li = Lf
I1w1 = I2w2
mr12w1 = mr2
2w2
r1v1 = r2v2
r2 = r1
v
v
2
1 =
r2 =
r = r2 − h2 = r1 − h1
h2 = r2 − r1 + h1 = 3394 km − 3593 km + 2.00 × 102 km
h2 = 1 km
3394 km
(3593 km)(3403 m/s)
3603 m/s
10. m1 = m2 = 55.0 kg
ri = 5.0
2
0 m = 2.50 m
vi = 5.00 m/s
vf = 15.0 m/s
Li = Lf
Iiwi = Ifwf
(m1ri2 + m2ri
2)wi = (mirf2 + m2rf
2)wf
ri2wi = rf
2wf
rivi = rf vf
rf = r
viv
f
i = (2.50
1
m
5.
)
0
(5
m
.0
/
0
s
m/s)
rf = 0.833 m
distance between skaters = 2rf2 = 1.67 m
1. k = 1.05 × 104 N/m
x = 4.0 × 10−2 m
KErot = 2.8 J
MEi = MEf
PEelastic = KEtrans + KErot
12
kx2 = KEtrans + KErot
KEtrans = 12
kx2 − KErot = 12
(1.05 × 104 N/m)(4.0 × 10−2 m)2 − 2.8 J
KEtrans = 8.4 J − 2.8 J − 5.6 J
Additional Practice 8E
Holt Physics Solution ManualV Ch. 8–10
V
2. vi = 2.2 m/s
m = 55 g
MEi = MEf
KEi = KEtrans − KErot
12
mvi2 = 1
2mvf
2 + 12
Iwf2 = 1
2mvf
2 + 12
23
mr2v
rf
2
12
mvi2 = 1
2mvf
2+23
= 12
mvf25
3 = 5
3KEtrans
KEtrans = 1
3
0mvi
2
KErot = KEi − KEtrans = 12
mvi2 −
1
3
0mvi
2 = 1
2
0mvi
2 = 15
mvi2
KErot = 15
(55 × 10−3 kg)(2.2 m/s)2 = 5.3 × 10−2 J
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3. h = 46.0 m
m = 25.0 kg
g = 9.81 m/s2
MEi = Ef
PEg = KEtrans + KErot
mgh = 12
mvf2 + 1
2Iwf
2 = 12
mvf2 + 1
2(mr2)
v
rf
2
= 12
mvf2(1 + 1) = mvf
2
KEtrans = 12
mgh = 12
(25.0 kg)(9.81 m/s2)(46.0 m)
KEtrans = 5.64 × 103 J
4. k = 1.05 × 104 N/m
x = 4.0 cm
wf = 43.5 rad/s
MEi = MEf
PEelastic = KEtrans + KErot
12
kx2 = 12
mvf2 + 1
2Iwf
2 = 12
mr2v
rf
2
+ 12
Iwf2 = 1
2Iwf
2 + 12
Iwf2 = Iwf
2
I = 2
k
wx
f
2
2 =
I = 4.4 × 10−3 kg • m2
(1.05 × 104 N/m)(4.0 × 10−2 m)2
(20943.5 rad/s)2
5. KEi = 45 J
wf = 27 rad/s
r = 0.11 m
MEi = MEf
KEi = KEtrans + KErot = 12
mvf2 + 1
2Iwf
2
KEi = 12
mr2v
rf
2
+ 12
Iwf2 = 1
25
2Iwf
2 + 12
Iwf2 = 1
2Iwf
2(52
+ 1) = 74
Iwf2
I = 4
7
K
wE
f2i =
(7)
(
(
4
2
)
7
(4
r
5
ad
J
/
)
s)2 =
m = = = 7.2 kg(5)(3.5 × 10−2 kg • m2)
(2)(0.11 m)2
52
I
r2
3.5 × 10−2 kg • m2
Section Five—Problem Bank V Ch. 8–11
V
6. h = 0.60 m
g = 9.81 m/s2
MEi = MEf
KEtrans + KErot = PEg
12
mvi2 + 1
2Iwi
2 = mgh
12
mvi2 + 1
22
5mr2
v
ri
2
= 12
mvi 1 + 25
= 1
7
0mvi
2 = mgh
vi = = vi = 2.9 m/s
(10)(9.81 m/s2+)(0.60 m)
7
10gh
7
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7. k = 150 N/m
x = 6.0 cm
m = 67 g
MEi = Ef
PEelastic = KEtrans + KErot
12
kx2 = 12
mvf2 + 1
2Iwf
2 = 12
mvf2 + 1
22
5mr2
v
rf
2
= mvf21
2 + 1
5 =
1
7
0mvf
2
vf = 5
7
k m
x2
= vf = 2.4 m/s
(5)(150 N/m)(6.0 × 10−2 m)2
(7)(67 × 10−3 kg)
8. vi = 2.4 m/s
q = 3.5°
g = 9.81 m/s2
MEi = MEf
KEtrans + KErot = PEg
12
mvi2 + 1
2Iwf
2 = mgh
12
mvi2 + 1
22
5mr2
v
ri
2
= mgd(sin q)
mvi21
2 + 1
5 =
1
7
0mvi
2 = mgd(sin q)
d = 10g
7
(
v
sii
n
2
q) =
d =
The ball has more than enough kinetic energy to reach the back of the pinball machine.
6.7 m
(7)(2.4 m/s)2
(10)(9.81 m/s2)(sin 3.5°)
9. vi = 4.6 m/s
g = 9.81 m/s2
MEi = MEf
KEtrans + KErot = PEg
12
mvi2 + 1
2Iwi
2 = mgh
12
mvi2 + 1
22
5mr2
v
ri
2
= mvi21
2 + 1
5 =
1
7
0mvi
2 = mgh
h = 7
1
v
0i
g
2
= (1
(7
0
)
)
(
(
4
9
.
.
6
81
m
m
/s
/
)
s
2
2)
h = 1.5 m
Holt Physics Solution ManualV Ch. 8–12
V
10. r = 15 cm
wf = 102 rad/s
g = 9.81 m/s2
MEi = MEf
PEg = KEtrans + KErot
mgh = 12
mvf2 + 1
2Iwf
2
mgh = 12
(mr2)v
rf
2
+ 12
Iwf2 = 1
23
2Iwf
2 + 12
Iwf2 = 1
2Iwf
232
+ 1 = 54
Iwf2
mgh = gh = 54
Iwf2
h = 5r
6
2wg
f2
=
h = 2.0 × 101 m
(5)(0.15 m)2(102 rad/s)2
(6)(9.81 m/s2)
32
Ir2
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Section Five—Problem Bank V Ch. 9–1
Chapter 9Fluid Mechanics
V
1. rgasoline = 675 kg/m3
Vs = 1.00 m3
g = 9.81 m/s2
FB = Fg rga
rso
s
line = m
rs
s
g rgasoline = Vsg rgasoline
FB = (1.00 m3)(9.81 m/s2)(675 kg/m3) = 6.62 × 103 N
Additional Practice 9A
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2. rr = 2.053 × 104kg/m3
Vr = (10.0 cm)3
g = 9.81 m/s2
apparent weight = 192 N
FB = Fg − apparent weight
FB = mrg − apparent weight = rrVrg − apparent weight
FB = (2.053 × 104 kg/m3)(10.0 cm)3(10−2 m/cm)3(9.81 m/s2) − 192 N = 201 N − 192 N
FB = 9 N
3. mh = 1.47 × 106 kg
Ah = 2.50 × 103 m2
rsw = 1.025 × 103 kg/m3
g = 9.81 m/s2
FB = Fg = mhg
FB = (1.47 × 106 kg)(9.81 m/s2) =
volume of hull submerged = Vsw = m
rs
s
w
w = rm
sw
h
h = V
As
h
w = Ah
m
rh
sw
h = = 0.574 m1.47 × 106 kg
(2.50 × 103 m2)(1.025 × 103 kg/m3)
1.44 × 107 N
4. msh = 1.47 × 106 kg
rsteel = 7.86 × 103 kg/m3
rgold = 1.93 × 104 kg/m3
Ah = 2.50 × 103 m2
rsw = 1.025 × 103 kg/m3
g = 9.81 m/s2
FB = Fg = mghg = rgoldVhg = rgold rm
st
s
e
h
el g
FB = =
h = V
As
h
w = A
m
hrgh
sw =
Ah
F
rB
swg
h = = 1.41 m3.54 × 107 N
(2.50 × 103 m2)(1.025 × 103 kg/m3)(9.81 m/s2)
3.54 × 107 N(1.93 × 104 kg/m3)(1.47 × 106 kg)(9.81 m/s2)
7.86 × 103 kg/m3
5. V = 166 cm3
apparent weight = 35.0 N
rw = 1.00 × 103 kg/m3
g = 9.81 m/s2
Fg = FB + apparent weight
rosmiumVg = rwVg + apparent weight
rosmium = rw + appare
V
nt
g
weight
rosmium = 1.00 × 103 kg/m3+
rosmium = 1.00 × 103 kg/m3+ 2.15 × 104 kg/m3
rosmium = 2.25 × 104 kg/m3
35.0 N(166 cm3)(10−6 m3/cm3)(9.81 m/s2)
Holt Physics Solution ManualV Ch. 9–2
V
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6. V = 2.5 × 10−3 m3
apparent weight = 7.4 N
rw = 1.0 × 103 kg/m3
g = 9.81 m/s2
Fg = FB + apparent weight
rebonyVg = rwVg + apparent weight
rebony = rw + appare
V
nt
g
weight
rebony = 1.0 × 103 kg/m3+
= 1.0 × 103 kg/m3+ 3.0 × 102 kg/m3
rebony = 1.3 × 103 kg/m3
7.4 N(2.5 × 10−3 m3)(9.81 m/s2)
7. Vsg = 7.52 cm3
msg = 45.0 g
Vlg = 7.38 × 10−6 m3
g = 9.81 m/s2
rsg = m
Vs
s
g
g =
r of solid gallium =
Fg = FB
msgg = rlgVlgg
rlg = m
Vl
s
g
g = 7
4
.
5
3
.
8
0
××
1
1
0
0−
−
6
3
m
kg3
r of liquid gallium = 6.10 × 103 kg/m3
5.91 × 103 kg/m3
45.0 × 10−3 kg(7.62 cm3)(10−6 m3/cm3)
8. rplatinum = 21.5 g/cm3
rw = 1.00 g/cm3
apparent weight = 40.2 N
g = 9.81 m/s2
Fg = FB + apparent weight
mg = rwVg + apparent weight = rwrpla
m
tinumg + apparent weight
mg 1 − rpla
r
t
w
inum = apparent weight
m = =
m = =
m = 4.30 kg
40.2 N(9.81 m/s2)(0.953)
40.2 N(9.81 m/s2)(1 − 0.047)
40.2 N
(9.81 m/s2)1 − 1
2
.
1
0
.
0
5
g
g
/
/
c
c
m
m
3
3apparent weight
g 1 − rpla
r
t
w
inum
9. rlithium = 534 kg/m3
rgasoline = 675 kg/m3
Vgasoline = 5.93 × 10−4 m3
Fg = FB
mlithiumg = rgasolineVgasolineg
mlithium = rgasolineVgasoline = (675 kg/m3)(5.93 × 10−4 m3)
mlithium =
Vlithium = m
rl
l
i
i
t
t
h
h
i
i
u
u
m
m = 5
0
3
.
4
40
k
0
g/
k
m
g3 = 7.49 × 10−4 m3
0.400 kg
Section Five—Problem Bank V Ch. 9–3
V
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10. h1 = 5.00 cm
h2 = 4.30 cm
A = 3.4 m2
rw = 1.0 × 103kg/m3
Before the mass is submerged,
Fg = FB
(mp + mb + mm)g = rwV1g = rwAh1g
After the mass is submerged,
(mp + mb)g = rwV2g = rwAh2g
Substituting the second equation into the first,
rwAh2g + mmg = rwAh1g
mm = rwA(h1 − h2)
mm = (1.0 × 103 kg/m3)(3.4 m2)(5.00 cm − 4.3 cm)(10−2 m/cm)
mm = (1.0 × 103kg/m3)(3.4 m2)(0.70 × 10−2 m)
mm = 24 kg
2. P = 1.01 × 105 Pa
F = 2.86 × 108 N
A = P
F =
1
2
.
.
0
8
1
6
××
1
1
0
05
8
P
N
a =
A = 4pr2
r = 4
A
p = 2.83 ×4
1
p03
m2
= 15.0 m
2.83 × 103 m2
Additional Practice 9B
1. P = 1.50 × 106 Pa
F = 1.22 × 104 N
A = P
F =
1
1
.
.
5
2
0
2
××
1
1
0
06
4
P
N
a = 8.13 × 10−3 m2
3. F = 5.0 N
P = 9.6 × 103 PaA =
P
F =
9.6
5
×.0
10
N3 Pa
= 5.2 × 10−4 m2
4. m = 1.40 × 103 kg
h = 0.076 m
rice = 917 kg/m3
P1 = P2
A
F1
1 =
A
F2
2
m
A1
g =
m
Aic
2
eg = m
Vic
i
e
c
h
e
g = ricehg
A1 = r
m
iceh = = 2.0 × 101 m21.40 × 103 kg
(917 kg/m3)(0.076 m)
Holt Physics Solution ManualV Ch. 9–4
V
5. A = 1.54 m2
P1 = (1 + 1.00 × 10−2)P2
P2 = 1.013 × 105 Pa
g = 9.81 m/s2
F = PnetA = (P1 − P2)A = (1 + 1.00 × 10−2 − 1)P2A = (1.00 × 10−2)P2A
F = (1.00 × 10−2)(1.013 × 105 Pa)(1.54 m2) =
m = F
g =
1.
9
5
.
6
81
×m
10
/
3
s2N
= 159 kg
1.56 × 103 N
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6. r = 6.0 km
P = 1.2 × 1016 Pa
F = PA = P(4pr2) = (1.2 × 1016 Pa)(4p)(6.0 × 103 m)2
F = 5.4 × 1024 N
7. F1 = 4.45 × 104 N
h1 = 448 m
h2 = 8.00 m
P1 = P2
A
F1
1 =
A
F2
2
A
F1
1
h
h1
1 =
F1
V
h1 = A
F2
2
h
h2
2 =
F2
V
h2
F2 = F
h1h
2
1 =
F2 = 2.49 × 106 N
(4.45 × 104 N)(448 m)
8.00 m
8. h = 760 mm
r = 13.6 × 103 kg/m3
g = 9.81 m/s2
P = A
F =
m
A
g =
m
A
g
h
h =
m
V
gh = rgh
P = (13.6 × 103 kg/m3)(9.81 m/s2)(760 × 10−3 m) = 1.0 × 105 Pa
9. m = 2.4 × 1013 kg
A = 3.14 km2
g = 9.81 m/s2
P = A
F =
m
A
g =
P = 7.5 × 107 Pa
(2.4 × 1013 kg)(9.81 m/s2)(3.14 km2)(106 m2/km2)
10. F = 4.4 × 103 N
A = 2.9 × 10−2 m2
Po = 1.0 × 105 Pa
P = A
F =
2
4
.9
.4
××1
1
0
0−2
3
m
N2 =
Pgauge = P − Po = 1.5 × 105 Pa − 1.0 × 105 Pa = 5 × 104 Pa
1.5 × 105 Pa
1. P = 6.9 × 104 Pa
r = 0.55 kg/m3
g = 9.81 m/s2
Po = 1.01 × 105 Pa
P = Po + pgh
h = P
r−
g
Po = =
h = −5.9 × 103 m = 5.9 km above sea level
−3.2 × 104 Pa(0.55 kg/m3)(9.81 m/s2)
6.9 × 104 Pa − 1.01 × 105 Pa
(0.55 kg/m3)(9.81 m/s2)
2. P − Po = 1.47 × 106 Pa
r = 1.00 × 103 kg/m3
g = 9.81 m/s2
P = Po + rgh
h = P
r−
g
Po =
h = 1.50 × 102 m
1.47 × 106 Pa(1.00 × 103 kg/m3)(9.81 m/s2)
Additional Practice 9C
Section Five—Problem Bank V Ch. 9–5
V
3. Po = 9.0 × 104 Pa
P = 5.00 × 106 Pa
r = 1.0 × 103 kg/m3
g = 9.81 m/s2
P = Po + rgh
h = P
r−
g
Po = =
h = 5.0 × 102 m
4.91 × 106 Pa(1.0 × 103 kg/m3)(9.81 m/s2)
5.00 × 106 Pa − 9.0 × 108 Pa(1.0 × 103 kg/m3)(9.81 m/s2)
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4. P = 4.03 × 105 Pa
Po = 1.01 × 105 Pa
r = 1.025 × 103 kg/m3
g = 9.81 m/s2
P = Po − rgh
h = P
r−
g
Po = =
h = 30.0 m
3.02 × 105 Pa(1.025 × 103 kg/m3 (9.81 m/s2)
4.03 × 105 Pa − 1.01 × 105 Pa(1.025 × 103 kg/m3)(9.81 m/s2)
5. h = 9.1 m
r = 1.0 × 103 kg/m3
g = 9.81 m/s2
Pgauge = P − Po = rgh
Pgauge = (1.0 × 103 kg/m3)(9.81 m/s2)(9.1 m) = 8.9 × 104 Pa
6. h = 86 m
r = 1.29 kg/m3
Po = 1.01 × 105 Pa
g = 9.81 m/s2
P = Po + rgh = 1.01 × 105 Pa + (1.29 kg/m3)(9.81 m/s2)(86 m) = 1.01 × 105 Pa + 1.1 × 103 Pa
P = 1.02 × 105 Pa
7. r = 13.6 × 103 kg/m3
Po = 1.01 × 105 Pa
h = 1.50 m
g = 9.81 m/s2
P = Po + rgh = 1.01 × 105 Pa + (13.6 × 103 kg/m3)(9.81 m/s2)(1.50 m)
= 1.01 × 105 Pa + 2.00 × 105 Pa
P = 3.01 × 105 Pa
8. Po = 9.10 × 106 Pa
h = −1.00 km
P = 8.60 × 106 Pa
gv = 8.87 m/s2
P = Po + rgVh
r = P
g
−
Vh
Po = =
r = 56 kg/m3
−5.0 × 105 Pa(8.87 m/s2)(−1.00 × 103 m)
8.60 × 106 Pa − 9.10 × 106 Pa(8.87 m/s2)(−1.00 × 103 m)
9. Po = 1.01 × 105 Pa
P = 2.23 × 105 Pa
h = 3.99 m
g = 9.81 m/s2
P = Po + rgh
r = P
g
−h
Po = =
r = 3.12 × 103 kg/m3
1.22 × 105 Pa(9.81 m/s2)(3.99 m)
2.23 × 105 Pa − 1.01 × 105 Pa
(9.81 m/s2)(3.99 m)
10. Po = 1.01 × 105 Pa
P = 1.29 × 105 Pa
h = 3.99 m
g = 9.81 m/s2
P = Po + rgh
r = P
g
−h
Po = =
r = 7.2 × 102 kg/m3
2.8 × 104 Pa(9.81 m/s2)(3.99 m)
1.29 × 105 Pa −1.01 × 105 Pa
(9.81 m/s2)(3.99 m)
Holt Physics Solution ManualV Ch. 9–6
V
1. ∆P = P1 − P2 = 1.5 × 103 Pa
v2 = 17.0 m/s
r = 1.00 × 103 kg/m3
P1 + 12
rv12 + rgh1 = P2 + 1
2rv2
2 + rgh2
Assume v1 ≈ 0 m/s, so the above equation simplifies to
∆P = rg(h2 − h1) + 12
rv22
h2 − h1 = ∆r
P
g −
v
22
g
2
= − (2
(
)
1
(
7
9
.
.
0
81
m
m
/s
/
)
s
2
2)
h = h2 − h1 = 0.15 m − 14.7 m = −14.6 m
h = h2 − h1 = −14.6 m = 14.6 m below the surface
1.5 × 103 Pa(1.00 × 103 kg/m3)(9.81 m/s2)
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2. ∆x = 120 m
∆y = − 69 m
g = 9.81 m/s2
To calculate v2, use the equations for a horizontally-launched projectile.
v2 = vx = ∆∆
x
t
∆y = vy,i∆t − 12
g∆t2
vy,i = 0 m/s, so
∆y = − 12
g∆t2
∆t = −2
g
∆y
v2 =
P1 + 12
rv12 + rgh1 = P2 + 1
2rv2
2 + rgh2
Because P1 = P2, and assuming v1 ≈ 0 m/s, Bernoulli’s equation simplifies to
rg(hi − h2) = 12
rv22 = 1
2r
2
h1 − h2 = = −∆4
x
∆
2
y
h = h1 − h2 = (−4
(
)
1
(
2
−0
6
m
9.0
)2
m) = 52 m
∆x2
2g−2
g
∆t
∆x
−2
g
∆y
∆x
−2
g
∆y
3. v1 = 2.00 m/s
v2 = 7.93 m/s
g = 9.81 m/s2
P1 + 12
rv12 + rgh1 = P2 + 1
2rv2
2 +rgh2
P1 = P2, so
rg(h1 − h2) = 12
r(v22 + v1
2)
h1 − h2 = v2
2
2
−g
v12
=
h1 − h2 = = (2
5
)(
8
9
.9
.8
m
1 m
2/s
/
2
s2)
h = h1 − h2 = 3.00 m
62.9 m2/s2 − 4.00 m2/s2
(2)(9.81 m/s2)
(7.93 m/s)2 − (2.00 m/s)2
(2)(9.81 m/s2)
Additional Practice 9D
Section Five—Problem Bank V Ch. 9–7
V
4. v1 = 24.45 m/s
v2 = 0.55 m/s
g = 9.81 m/s2
P1 + 12
rv12 + rgh1 = P2 + 1
2rv2
2 + pgh2
P1 = P2, so
rg(h2 − h1) = 12
r(v12 − v2
2)
h2 − h1 = v1
2
2
−g
v22
=
h2 − h1 = = (2
5
)
9
(
7
9
.
.
5
81
m
m
2/
/
s
s
2
2)
h = h2 − h1 = 30.5 m
597.8 m2/s2 − 0.30 m2/s2
(2)(9.81 m/s2)
(24.45 m/s)2 − (0.55 m/s)2
(2)(9.81 m/s2)
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5. v1 = 2.50 m/s
h1 − h2 = 3.00 m
g = 9.81 m/s2
P1 + 12
rv12 + rgh1 = r2 + 1
2rv2
2 + rgh2
P1 = P2, so
12
rv12 + rg(h1 − h2) = 1
2rv2
2
v2 =√
v12+ 2g(h1− h2) =√
(2.50m/s)2 + (2)(9.81m/s2)(3.00m)
v2 =√
6.25 m2/s2 + 58.9m2/s2 =√
65.2 m2/s2
v2 = 8.07 m/s
6. v1 = 0.90 m/s
∆P = P2 − P1 = 311 Pa
P = 1.025 × 103 kg/m3
P1 = 12
rv12 + rgh1 = P2 + 1
2rv2
2 + rgh2
Because h1 = h2, so the above equation simplifies to
12
rv12 − ∆P = 1
2rv2
2
v2 = v12− 2∆
PP = (0.90m/s)2 −
1.02
(2
5)
×(3
1
10
13 P
ka
g
)/m3v2 =
√0.81 m2/s2 − 0.607 m2s2 =
√0.20 m2/s2
v2 = 0.45 m/s
7. rair = 1.3 kg/m3
rmercury = 1.36 × 104 kg/m3
h2 − h1 = 3.5 cm
g = 9.81 m/s2
For the static mercury columns, v1 = v2.
P1 + 12
rmercuryv12 + rmercurygh1 = P2 + 1
2rmercuryv2
2 + rmercurygh2
P1 − P2 = rmercuryg(h2 − h1)
For the flowing air, v1 = 0 m/s, because the tube is fixed to the wing, and h1 = h2.
P1 + 12
rairv12 + rairgh1 = P2 + 1
2rairv2
2 + rairgh2
P1 − P2 = 12
rairv22
Substituting the first equation into the second,
rmercury g (h2 − h1) = 12
rairv22
v2 = 2rmercur
ryg
ai
(r
h2− h1) =
v2 = 85 m/s
(2)(1.36 × 104 kg/m3)(9.81 m/s2)(3.5 × 10−2 m)
1.3 kg/m3
Holt Physics Solution ManualV Ch. 9–8
V
8. r = 950 kg/m3
v1 = 10.00 m/s
v2 = 9.90 m/s
P1 + 12
rv12 + rgh1 = P2 + 1
2rv2
2 + rgh2
h1 = h2, so
P2 − P1 = 12
r(v12 − v2
2) = 12
(950 kg/m3)[(10.00 m/s)2 − (9.90 m/s)2]
P2 − P1 = 12
(950 kg/m3)(100.0 m2/s2 − 98.0 m2/s2) = 12
(950 kg/m3)(2.0 m2/s2)
∆P = P2 − P1 = 950 Pa
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9. A1 = 7.8 × 10−3 m2
v1 = 2.0 m/s
A2 = 3.1 × 10−4 m2
r = 1.00 × 103 kg/m3
h2 − h1 = 10.5 m
g = 9.81 m/s2
From the continuity equation, an expression for v2 can be derived.
A1v1 = A2v2
v2 = A
A1
2v1
P1 + 12
rv12 + rgh1 = P2 + 1
2rv2
2 + rgh2
P1 − P2 = 12
r(v22 − v1
2) + rg(h2 − h1) = 12
rv12
A
A1
2
2
2 − 1 + rg(h2 − h1)
P1 − P2 = 12
(1.00 × 103 kg/m3)(2.0 m/s)2((7
3
.
.
8
1
××
1
1
0
0
−
−
3
4m
m
2
2)
)
2
2 − 1+ (1.00 × 103 kg/m3)(9.81 m/s2)(10.5 m)
P1 − P2 = 12
(1.00 × 103 kg/m3)(2.0 m/s)2(6.3 × 102 − 1) + 1.03 × 105 Pa
P1 − P2 = 12
(1.00 × 103 kg/m3)(2.0 m/s)2 (6.3 × 102) + 1.03 × 105 Pa
P1 − P2 = 1.3 × 106 Pa + 1.03 × 105 Pa
∆P = P1 − P2 = 1.4 × 106 Pa
10. A1 = 9.2 × 10−2 m2
v1 = 8.3 m/s
A2 = 4.6 × 10−2 m2
r = 1.0 × 103 kg/m3
From the continuity equation, an expression for v2 can be derived.
A1v1 = A2v2
v2 = A
A1
2v1
P1 + 12
rv12 + rgh1 = P2 + 1
2rv2
2 + rgh2
h1 = h2, so
P1 − P2 = 12
r(v22 − v1
2) = 12
rv12
A
A
2
12
2 − 1P1 − P2 = 1
2(1.0 × 103 kg/m3)(8.3 m/s)((9
4
.
.
2
6
××
1
1
0
0
−
−
2
2m
m
2
2)
)
2
2 − 1P1 − P2 = 1
2(1.0 × 103 kg/m3)(8.3 m/s)2(4 − 1) = 1
2(1.0 × 103 kg/m3)(8.3 m/s)2(3)
∆P = P1 − P2 = 1.0 × 105 Pa
1. T1 = 184
T2 = 331 K
V1 = 3.70 m3
At constant pressure:
V
T1
1 = T
V
2
2
V2 = V
T1T
1
2 = (3.70 m
18
3
4
)(
K
331 K) = 6.66 m3
Additional Practice 9E
Section Five—Problem Bank V Ch. 9–9
V
2. V1 = 0.455 m3
P1 = 9.1 × 106 Pa
P2 = 5.0 × 104 Pa
At constant temperature:
P1V1 = P2V2
V2 = P
P1V
2
1 = = 83 m3(9.1 × 106 Pa)(0.455 m3)
5.0 × 104 Pa
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3. V1 = 65.4 m3
P1 = 1.03 × 105 Pa
P2 = 5.84 × 104 Pa
At constant temperature:
P1V1 = P2V2
V2 = P
P1V
2
1 = = 115 m3(1.03 × 105 Pa)(65.4 m3)
5.84 × 104 Pa
4. T1 = 295 K
P1 = 1.01 × 105 Pa
T2 = 225 K
P2 = 5.10 × 104 Pa
N = 5.55 × 1022 particles
kB = 1.38 × 10−23 J/K
PV = NkBT
V1 = Nk
PB
1
T1 =
V1 =
V2 = Nk
PB
2
T2 =
V2 = 3.38 × 10−3 m3
(5.55 × 1022)(1.38 × 10−23 J/K)(225 K)
5.10 × 104 Pa
2.24 × 10−3 m3
(5.55 × 1022)(1.38 × 10−23 J/K)(295 K)
1.01 × 105 Pa
5. P1 = 7.5 × 104 Pa
T1 = 250 K
P2 = 2.0 × 106 Pa
At constant volume:
T
P1
1 =
T
P
2
2
T2 = P
P2T
1
1 = = 6.7 × 103 K(2.0 × 106 Pa)(250 K)
7.5 × 104 Pa
Holt Physics Solution ManualV Ch. 9–10
V
6. V1 = 1.00 m3
T1 = 295 K
V2 = 65.4 m3
At constant pressure:
V
T1
1 = T
V
2
2
T2 = V
V2T
1
1 = (65.4
1
m
.0
3
0
)
m
(2395 K)
= 1.93 × 104 K
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7. N = 2.1 × 1057 particles
P = 2.1 × 1016 Pa
V = 2.1 × 1025 m3
KB = 1.38 × 10−23 J/K
PV = NkBI
T = N
P
k
V
B = = 1.5 × 107 K
(2.1 × 1016 Pa)(2.1 × 1025 m3)(2.1 × 1057)(1.38 × 10−23 J/K)
8. T1 = 295 K
P1 = 2.50 × 105 Pa
T2 = 506 K
At constant volume:
T
P1
1 =
T
P
2
2
P2 = P
T1T
1
2 = = 4.29 × 105 Pa(2.50 × 105 Pa)(506 K)
295 K
9. T = 1.0 × 102 K
V = 3.3 × 1043 m2
r = 10.0 atoms/cm3
kB = 1.38 × 10−23 J/K
N = rV
PV = NkBT = rVkBT
P = rkBT = (10.0 atoms/cm3)(106 cm3/m3)(1.38 × 10−23 J/K)(1.0 × 102 K)
P = 1.38 × 10−14 Pa
10. P1 = 1.42 × 105 Pa
V1 = 1.00 m3
V2 = 1.83 m3
At constant temperature:
P1V1 = P2V2
P2 = P
V1V
2
1 = = 7.76 × 104 Pa(1.42 × 105 Pa)(1.00 m3)
1.83 m3
Section Five—Problem Bank V Ch. 10–1
Chapter 10Heat
V
1. T1 = 463 K
T2 = 93 K
TC,1 = (T − 273)°C = (463 − 273)°C =
TF,1 = 9
5 TC,1 + 32 =
9
5 (1.90 × 102)°F + 32°F = 342°F + 32°F =
TC,2 = (T − 273)°C = (93 − 273)°C =
TF,2 = 9
5 TC,2 + 32 =
9
5 (−1.80 × 102)°F + 32°F = −324°F + 32°F = −292°F
−180 × 102 °C
374°F
1.90 × 102 °C
Additional Practice 10A
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2. T = 330.0 KTC = (T − 273)°C = (330.2 − 273.2)°C =
TF = 9
5 TC + 32 =
9
5 (56.8)°F + 32°F = 102°F + 32°F = 134°F
56.8°C
3. Ti = 237 K
Tf = 283 K
∆TC = (Tf − 273)°C − (Ti − 273)°C = Tf − Ti
∆TC = (283 − 237)°C =
∆TF = 9
5TC, f + 32°F −
99
5TC, i + 32°F =
9
5 ∆TC°F
∆TF = 9
5(46)°F = 83°F
46° C
4. TF,i = − 5 °F
TF,f = + 37°FTC,i =
5
9(TF, i − 32)°C =
5
9(−5 − 32)°C =
5
9(− 37)°C = −21°C
TC,f = 5
9(TF, f − 32)°C =
5
9(37 − 32)°C =
5
9(5)°C = 3°C
∆T = (TC,f + 273 K) − (TC,i + 273 K) = TC,f − TC,i
∆T = [3 − (−21)] K = 24 K
6. TC,1 = 47°C
TC, 2 = 42°CT1 = (TC,1 + 273)K = (47 + 273) K =
T2 = (TC,2 + 273)K = (42 + 273) K = 315 K
3.20 × 102 K
5. TF = 78°FT = TC + 273 =
5
9(TF − 32) + 273
T = 5
9 (78 − 32) K + 273 K =
5
9 (46) K + 273 K = 26 K + 273 K
T = 299 K
9. TF = 188.6°FTC =
5
9(TF − 32.0)°C =
5
9(188.6 − 32.0)°C =
5
9(156.6)°C
TC = 87.00°C
Holt Physics Solution ManualV Ch. 10–2
V
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8. TF = 2192°FTC =
5
9(TF − 32)°C =
5
9(2192 − 32)°C =
5
9(2.160 × 103)°C
TC = 1.200 × 103°C
10. T = 2.70 K TC = (T − 273)°C = (2.70 − 273)°C = −270°C
1. m = 5.25 g
vi = 3.27 m/s
∆Upenny = 12
∆U
k = 2.03 J/1.00° C
∆PE + ∆KE + ∆U = 0
The change in potential energy from before to after impact is zero, as is KEf.
∆PE + ∆KE + ∆U = 0 + KEf − KEi + ∆V = − KEi + ∆U = 0
∆U = KEi = 12
mvi2
∆Upenny = 12
∆U = 12
(12
mvi2) =
m
4
vi2
∆T = ∆Up
kenny =
∆T = = 6.91 × 10−3 °C(5.25 × 10−3 kg)(3.27 m/s)2
(4)(2.03 J/1.00°C)
mvi2
4k
2. h = 9.5 m
g = 9.81 m/s2
∆Uacorn = (0.85)∆U
k/m = 12
1
0
.0
0
°J
C
/kg
∆PE + ∆KE + ∆U = 0
The change in kinetic energy from before the acorn is dropped to after it has landedis zero, as is PEf.
∆PE + ∆KE + ∆U = PEf − PEi + 0 + ∆U = −PEi + ∆U = 0
∆U = PEi = mgh
∆Uacorn = (0.85) ∆U = (0.85)mgh
∆T = ∆U
kacorn =
(0.85
k
)mgh =
(0
(
.
k
8
/
5
m
)g
)
h
∆T = = 6.6 × 10−2 °C(0.85)(9.81 m/s2)(9.5 m)
12
1
0
.
0
0°J
C
/kg
Additional Practice 10B
7. TF = −67°FTC =
5
9(TF − 32) + 273K
T = 5
9(− 67 − 32) + 273K =
55
9(− 97) + 273K = (− 55 + 273)K
T = 218 K
Section Five—Problem Bank V Ch. 10–3
V
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4. vi = 47.5 m/s
h = 151 m
m = 7.32 g
∆Utwig = (0.100)∆U
k = 8.5 J/1.0°C
g = 9.81 m/s2
∆PE + ∆KE + ∆U = 0
∆U = −∆PE − ∆KE = PEi − PEf + KEi − KEf
Both PEf and KEf equal zero.
∆U = PEi + KEi = mgh + 12
mvi2
∆Utwig = (0.100)(mgh + 12
mvi2)
∆T = ∆U
ktwig =
k
∆T =
∆T = = (0.1
8
0
.
0
5
)
J
(
/
1
°9
C
.1 J) = 0.22°C
(0.100)(10.8 J + 8.26 J)
8.5 J/°C
(0.100)[(7.32 × 10–3 kg)(9.81 m/s2)(151 m) + 12
(7.32 × 10–3 kg)(47.5 m/s)2]
8.5 J/1.0°C
3. PEi = 6.2 J
∆Ucone = (0.100)∆U
k = 180 J/1.0°C
∆PE + ∆KE + ∆U = 0
The change in kinetic energy from before the ice cream cone is dropped to after it haslanded is zero, as is PEf.
∆PE + ∆KE + ∆U = PEf − PEi + 0 + ∆U = −PEi + ∆U = 0
∆U = PEi
∆Ucone = (0.100)∆U = (0.100)PEi
∆T = ∆U
kcone =
(0.10
k
0)PEi
∆T = (0
1
.
8
1
0
00
J/
)
1
(
.
6
0
.
°2
C
J) = 3.4 × 10−3 °C
5. h = 561.7 m
∆U = 105 J
g = 9.81 m/s2
∆PE = ∆KE + ∆U = 0
When the stone lands, its kinetic energy is transferred to the internal energy of thestone and the ground. Therefore, overall, ∆KE = 0 J
∆PE = PEf − PEe = 0 − mgh = − ∆U
m = ∆g
U
h = = 1.91 × 10−2 kg = 19.1 g
105 J(9.81 m/s2)(561.7 m)
6. h = 5.5 m
g = 9.81 m/s2
∆U = 2.77 × 103 J
∆PE + ∆KE + ∆U = 0
When the prey lands, all of the kinetic energy is transferred to the ground. Therefore∆KE = 0 J
∆PE = PEf − PEi = 0 − mgh = −∆U
m = ∆g
U
h = = 51 kg
2.77 × 103 J(9.81 m/s2)(5.5 m)
7. vf = 0 m/s
vi = 13.4 m/s
∆U = 5836 J
∆PE + ∆KE + ∆U = 0
The bicyclist remains on the bicycle, which does not change elevation, so ∆PE = 0 J.
∆KE = KEf − KEi = 0 − 12
mvi2 = −∆U
m = 2
v
∆
i2U =
(
(
1
2
3
)(
.4
58
m
36
/s)
J)2 = 65.0 kg
(0.100)(mgh + 12
mvi2)
Holt Physics Solution ManualV Ch. 10–4
V
10. ∆KE = KEf − KEi= −7320 J
∆Uhands = (1 − 0.300)∆U
∆PE + ∆KE + ∆U = 0
The hands don’t change height, so ∆PE = 0 J.
∆U = −∆KE = −(−7320 J) = 7320 J
∆Uhands = (1− 0.300)(7320 J) = (0.700)(7320 J) = 5120 J
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1. mw = 15 g = 0.015 kg
mhp = 15 g = 0.015 kg
∆Tw = 1.0°C
∆Thp = 1.6°C
cp,w = 4186 J/kg • °C
cp,hpmhp∆Thp = cp,wmw∆Tw
cp,hp = cp
m,w
h
m
p∆w
T
∆
h
T
p
w =
cp,hp = 2.6 × 103 J/kg • °C
(4186 J)(kg • °C)(0.015 kg)(1.0°C)
(0.015 kg)(1.6°C)
2. mv = 0.340 kg
mw = 1.00 kg
Tv,i = 21.0°C
Tw,i = 90.0°C
Tf = 73.7°C
cp,w = 4186 J/kg • °C
cp,vmv∆Tv = cp,wmw∆Tw
∆Tw = Tw,i − Tf = 90.0°C − 73.7°C = 16.3°C
∆Tv = Tf − Tv,i = 73.7°C − 21.0°C = 52.7°C
cp,v = cp,
mwm
v∆w
T
∆
v
Tw =
cp,v = 3.81 × 103 J/kg •°C
(4186 J/kg • °C)(1.00 kg)(16.3°C)
(0.340 kg)(52.7°C)
3. ma = 0.250 kg
mw = 1.00 kg
∆Tw = 1.00° C
∆Ta = 17.5° C
cp,w = 4186 J/kg • °C
cp,ama∆Ta = cp,wmw∆Tw
cpa = =
cp,a = 957 J/kg • °C
(4186 J/kg • °C)(1.00 kg)(1.00°C)
(0.250 kg)(17.5° C)
cp,wmw∆Twma∆Ta
4. mi = 3.0 kg
mw = 5.0 kg
∆Tw = 2.25° C
∆Ti = 29.6° C
cp,w = 4186 J/kg • °C
cp,imi∆Ti = cp,wmw∆Tw
cp,i = =
cp,i = 530 J/kg • °C
(4186 J/kg • °C)(5.0 kg)(2.25°C)
(3.0 kg)(29.6°C)
cp,wmw∆Twmi∆Ti
Additional Practice 10C
8. ∆KE = KEf − KEi
= −2.15 × 104 J
∆PE + ∆KE + ∆U = 0
There is no change in the height of the sticks, so ∆PE = 0 J.
∆U = −∆KE = −(−2.15 × 104 J) = 2.15 × 104 J
9. vi = 20.5 m/s
vf = 0 m/s
m = 61.4 kg
∆PE + ∆E + ∆U = 0
The height of the skater does not change, so ∆PE = 0 J.
∆KE = KEf − KEi = 0 − 12
mvi2
∆U = −∆KF = −(− 12
mvi2) = 1
2(61.4 kg)(20.5 m/s)2 = 1.29 × 104 J
Section Five—Problem Bank V Ch. 10–5
V
7. Q = 45 × 106 J
∆Ta = 55° C
cp,a = 1.0 × 103 J/kg • °C
cp,a = ma
Q
∆Ta
ma = cp,a
Q
∆Ta = a = 818 kg
45 × 106 J(1.0 × 103 J/kg • °C)(55°C)
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8. mc = 0.190 kg
Q = 6.62 × 104 J
cp,c = 387 J/kg • °C
cp,c = mc
Q
∆Tc
∆Tc = mc
Q
cp,c = = 9.00 × 102°C
6.62 × 104 J(0.190 kg)(387 J/kg • °C)
9. mt = 0.225 kg
cp,t = 2.2 × 103 J/kg • °C
Q = −3.9 × 104 J
cp,t = mt
Q
∆Tt
∆Tt = m
Q
tcp,t = = −79°C
−3.9 × 104 J(0.225 kg)(2.2 × 103 J/kg • °C)
10. cp,t = 140 J/kg • °C
mt = 0.23 kg
Q = −3.0 × 104 J
cp,t = mt
Q
∆Tt
∆Tt = m
Q
tcp,t = = −930°C
−3.0 × 104 J(0.23 kg)(140 J/kg • °C)
5. Q = −1.09 × 1010 J
cp,w = 4186 J/kg • °C
∆Tw = −5.0°C
cp,w = mw
Q
∆Tw
mw = cp,w
Q
∆Tw = = 5.2 × 105 kg
−1.09 × 1010 J(4186 J/kg • °C)(−5.0°C)
6. cp,b = 121 J/kg • °C
Q = 25 J
∆Tb = 5.0°C
cp,b = mb
Q
∆Tb
mb = cp,b
Q
∆Tb = = 4.1 × 10−2 kg
25 J(121 J/kg • °C)(5.0°C)
Additional Practice 10D
1. Q = 1.10 × 106 J
m = 5.33 kg
Q = mLf
Lf = m
Q =
1.1
5
0
.3
×3
1
k
0
g
6 J = 2.06 × 105 J/kg
2. Q = 9.6 × 105 J
m = 5.33 kg
Lf = m
Q =
9.
5
6
.3
×3
1
k
0
g
5 J = 1.8 × 105 J/kg
3. Q = 3.72 × 105 J
m = 0.65 kg
Lsubl = m
Q =
3.7
0
2
.6
×5
1
k
0
g
5 J = 5.7 × 105 J/kg
4. Q = 8.5 × 104 J
Lv = 2.26 × 106 J/kg
Q = mLv
m = L
Q
v =
2.2
8
6
.5
××1
1
0
06
4
J/
J
kg = 3.8 × 10−2 kg
Holt Physics Solution ManualV Ch. 10–6
V
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6. m = 250 kg
Qtot = 1.380 × 108 J
cp,i = 605 J/kg •°C
Lf = 2.47 × 105 J/kg
Qtot = mcp,i ∆T + mLf
mcp,i ∆T = Qtot − mLf
∆T = m
Q
ct
p
ot
,i −
c
L
p
f
,i
∆T = −
∆T = (910 − 408)°C = 5.0 × 102°C
2.47 × 105 J/kg605 J/kg •°C
1.380 × 108 J(250 kg)(605 J/kg •°C)
Givens Solutions
7. m = 1.45 kg
Tf = 330°C
Qtot = 4.46 × 104 J
cp,i = 120 J/kg •°C
Lf = 2.45 × 104 J/kg
Qtot = mcp,i ∆T + mLf
∆T = m
Q
ct
p
ot
,i −
c
L
p
f
,i = −
∆T = (2.6 × 102 − 2.0 × 102)°C = 60°C
∆T = Tf − Ti = 330°C − Ti = 60°C
Ti = 330°C − 60°C = 270°C
2.45 × 104 J/kg120 J/kg •°C
4.46 × 104 J(1.45 kg)(120 J/kg •°C)
8. m = 0.75 g
Ti = 100.0°C
Qtot = 2.0 × 103 J
cp,w = 4200 J/kg •°C
Lv = 2.26 × 106 J/kg
Qtot = mcp,w ∆T + mLv
∆T = m
Q
ct
p
o
,
t
w −
c
L
p,
v
w = −
∆T = (6.3 × 102 − 5.4 × 102)°C = 90°C
∆T = Ti − Tf = 100.0°C − Tf = 90°C
Tf = 100.0°C − 90°C = 10°C
2.26 × 106 J/kg4200 J/kg •°C
2.0 × 103 J(0.75 × 10−3 kg)(4200 J/kg •°C)
9. m = 0.35 kg
Lf = 8.02 × 104 J/kg
Q = mLf = (0.35 kg)(8.02 × 104 J/kg)
Q = 2.8 × 104 J
10. m = 55.0 g
Ti = 20.0°C
Tf = 357°C
cp,m = 130 J/kg •°C
Lv = 2.95 × 105 J/kg
Qtot = mcp,m ∆T + mLv = mcp,m (Tf − Ti) + mLv
Qtot = (55.0 × 10−3 kg)(130 J/kg •°C)(357°C − 20.0°C)
+ (55.0 × 10−3 kg)(2.95 × 105 J/kg)
Qtot = (55.0 × 10−3 kg)(130 J/kg •°C)(337°C)
+ (55.0 × 10−3 kg)(2.95 × 105 J/kg) = 2.4 × 103 J + 1.62 × 104 J
Qtot = 1.86 × 104 J
5. Q = 2.11 × 106 J
Lv = 8.45 × 105 J/kg
m = L
Q
v =
8.
2
4
.
5
11
××10
150
J
6
/
J
kg = 2.50 kg
Chapter 11Thermodynamics
V
1. W = 3.29 × 106 J
∆V = 2190 m3P =
∆W
V =
3.
2
2
1
9
9
×0
1
m
03
6 J = 1.50 × 103 Pa = 1.50 kPa
Additional Practice 11A
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Section Five—Problem Bank V Ch. 11–1
2. W = 1.06 × 106 J
Vi = 2.80 × 103 m3
Vf = 8.50 × 105 m3
∆V = Vf − Vi = 8.50 × 105 m3 − 2.80 × 103 m3 = 8.47 × 105 m3
P = ∆W
V =
8
1
.4
.0
7
6
××1
1
0
05
6
m
J3 = 1.25 Pa
3. W = 1.3 J
∆V = 5.4 × 10−4 m3P =
∆W
V =
5.4 ×1
1
.3
0−J4 m3 = 2.4 × 103 Pa = 2.4 kPa
4. W = 472.5 J
P = 25.0 kPa = 2.50 × 104 Pa∆V =
W
P =
2.50
47
×2
1
.5
04J
Pa = 1.89 × 10−2 m3
5. W = 393 J
P = 655 kPa = 6.55 × 105 Pa∆V =
W
P =
6.55
3
×93
10
J5 Pa
= 6.00 × 10−4 m3
6. W = 0.20 J
P = 39 Pa∆V =
W
P =
0
3
.
9
20
Pa
J = 5.1 × 10−3 m3
7. W = 3.2 × 102 J
P = 4.0 × 105 Pa∆V =
W
P =
4
3
.0
.2
××1
1
0
05
2
P
J
a = 8.0 × 10−4 m3
8. P = 2.07 × 107 Pa
∆V = 0.227 m3
W = P ∆V = (2.07 × 107 Pa)(0.227 m3) = 4.70 × 106 J
9. Vi = 3.375 × 10−6 m3
Vf = 5.694 × 10−6 m3
P = 101.33 kPa
= 1.0133 × 105 Pa
∆V = Vf − Vi = 5.694 × 10−6 m3 − 3.375 × 10−6 m3 = 2.319 × 10−6 m3
W = P ∆V = (1.0133 × 105 Pa)(2.319 × 10−6 m3) = 0.2350 J
10. Vi = 2.0 × 10−3 m3
Vf = 5.0 × 10−3 m3
P = 0.90 kPa = 900 Pa
∆V = Vf − Vi = 5.0 × 10−3 m3 − 2.0 × 10−3 m3 = 3.0 × 10−3 m3
W = P ∆V = (900 Pa)(3.0 × 10−3 m3) = 2.7 J
Holt Physics Solutions ManualV Ch. 11–2V Ch. 11–2
2. Ui = 39 J
Uf = 163 J
Q = 114 J
∆U = Uf − Ui = Q − W
W = Q − ∆U = Q − (Uf − Ui) = Q − Uf + Ui
W = 114 J − 163 J + 39 J = −10 J
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1. mi = 2.0 kg
mf = 15 kg
Ti = 28°C
Tf = 52°C
d = 2.3 mm
W = Fd = m2gd = (15 kg)(9.81 m/s2)(2.3 × 10−3 m) = 0.34 J
3. Ui = 8093 J
Uf = 2.092 × 104 J
Q = 6932 J
∆U = Uf − Ui = Q − W
W = Q − ∆U = Q − (Uf − Ui) = Q − Uf + Ui
W = 6932 J − 2.092 × 104 J + 8093 J = −5895 J
4. Q = 4.50 × 108 J
W = 3.21 × 108 J
∆U = Q − W = 4.50 × 108 J − 3.21 × 108 J = 1.29 × 108 J
5. Q = 632 kJ
W = 102 kJ∆U = Q − W = 632 kJ − 102 kJ = 5.30 × 102 kJ = 5.30 × 105 J
6. Q = 867 J
W = 623 J∆U = Q − W = 867 J − 623 J = 244 J
7. W = 192 kJ
∆U = 786 kJ
∆U = Q − W
Q = ∆U + W = 786 kJ + 192 kJ = 978 kJ
8. W = 602 kJ
∆U = 1.09 × 105 J
∆U = Q − W
Q = ∆U + W = 602 kJ + 109 kJ = 711 kJ
Additional Practice 11B
9. ∆U = 873 J ∆V = 0, so W =
∆U = Q − W
Q = ∆U + W = 873 J + 0 J = 873 J
0 J
10. ∆U = 986 J ∆V = 0, so W =
∆U = Q − W
Q = ∆U + W = 986 J + 0 J = 986 J
0 J
Section Five—Problem Bank V Ch. 11–3V Ch. 11–3
V
Givens SolutionsC
opyr
ight
©by
Hol
t, R
ineh
art a
nd W
inst
on.A
ll rig
hts
rese
rved
.
1. eff = 0.17
Qh = 5.5 × 109 JWnet = eff Qh = (0.17)(5.5 × 109 J) = 9.4 × 108 J
2. eff = 0.35
Qh = 7.37 × 108 J
4. eff = 0.29
Qh = 693 JWnet = eff Qh = (0.29)(693 J) = 2.0 × 102 J
7. Wnet = 544 J
eff = 0.2225Qh =
W
efn
fet =
0
5
.2
4
2
4
2
J
5 = 2.44 × 103 J
Wnet = eff Qh = (0.35)(7.37 × 108 J) = 2.6 × 108 J
3. eff = 0.15
Qh = 9.36 × 108 JWnet = eff Qh = (0.15)(9.36 × 108 J) = 1.4 × 108 J
Additional Practice 11C
5. eff = 0.11
Wnet = 1150 JQh =
W
efn
fet =
1
0
1
.
5
1
0
1
J = 1.0 × 104 J
6. eff = 0.19
Wnet = 998 JQh =
W
efn
fet =
9
0
9
.1
8
9
J = 5.3 × 103 J
8. Qh = 365 J
Qc = 223 Jeff = 1 −
Q
Q
h
c = 1 − 2
3
2
6
3
5
J
J = 1 − 0.611 = 0.389
9. Qh = 571 J
Qc = 463 Jeff = 1 −
Q
Q
h
c = 1 − 4
5
6
7
3
1
J
J = 1 − 0.811 = 0.189
10. Wnet = 128 J
Qh = 581 Jeff =
W
Qn
h
et = 1
5
2
8
8
1
J
J = 0.220
Section Five—Problem Bank V Ch. 12–1
Chapter 12Vibrations and Waves
V
1. k = 420 N/m
x = 4.3 × 10−2 m
Felastic = −kx = −(420 N/m)(4.3 × 10−2 m) = −18 N
Additional Practice 12A
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2. k = 65 N/m
x = –1.5 × 10−1 mFelastic = −kx = −(65 N/m)(–1.5 × 10−1 m) = 9.8 N
3. k = 49 N/m
x = –1.2 × 10−1 mFelastic = −kx = −(49 N/m)(–1.2 × 10−1 m) = −5.9 N
4. k = 26 N/m
x = –5.0 × 10−2 mFelastic = −kx = −(26 N/m)(–5.0 × 10−2 m) = 1.3 N
5. Fg = –669 N
x = –6.5 × 10−2 m
Fnet = 0 = Felastic + Fg = −kx + Fg
Fg = kx
k = F
x
g = –6.5
–6
×6
1
9
0
N−2 m
= 1.0 × 104 N/m
6. Fg = –550 N
x = –15 m
Fnet = 0 = Felastic + Fg = −kx + Fg
Fg = kx
k = F
x
g = –
–
5
1
5
5
0
m
N = 37 N/m
7. Fg = 620 N
x = 7.2 × 10−2 m
Fnet = 0 = Felastic + Fg = −kx + Fg
Fg = kx
k = F
x
g = 7.2
6
×20
10
N−2 m
= 8.6 × 103 N/m
8. Felastic = 12 N
k = 180 N/m
Felastic = −kx
x = − Fel
kastic = −
18
1
0
2
N
N
/m = –0.067 m = −6.7 cm
9. Felastic = 52 N
k = 490 N/m
Felastic = −kx
x = − Fel
kastic = −
49
5
0
2
N
N
/m = − 0.11 m = −11 cm
Holt Physics Solution ManualV Ch. 12–2
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10. m = 3.0 kg
g = 9.81 m/s2
k = 36 N/m
Fnet = 0 = Felastic + Fg = −kx −mg
mg = −kx
x = − m
k
g = −
(3.0 k
3
g)
6
(
N
9.8
/m
1 m/s2) = –0.82 m = −82 cm
1. L = 3.0 × 10−1 m
T = 1.16 sT = 2p
L
g
T2 = 4p
g
2 L
g = 4p
T
2
2L
= = 8.8 m/s2(4p 2)(3.0 × 10−1 m)
(1.16 s)2
2. L = 0.650 m
T = 2.62 sT = 2p
L
g
T2 = 4p
g
2 L
g = 4p
T
2
2L
= (4p
(
2
2
)
.
(
6
0
2
.6
s
5
)
02
m) = 3.74 m/s2
Additional Practice 12B
3. L = 1.14 m
T = 3.55 sT = 2p
L
g
T2 = 4p
g
2 L
g = 4p
T
2
2L
= (4p
(
2
3
)
.5
(1
5
.1
s)
42m)
= 3.57 m/s2
4. L = 5.00 × 10−1 m
T = 2.99 sT = 2p
L
g
T2 = 4p
g
2 L
g = 4p
T
2
2L
= = 2.21 m/s2(4p 2)(5.00 × 10−1 m)
(2.99 s)2
5. f = 1.0 Hz
g = 9.81 m/s2 T = 2p L
g =
1
f
f
12 =
4pg
2 L
L = 4p
g2f 2 =
(4p9
2.8
)
1
(1
m
.0
/
s
s−
2
1)2 = 0.25 m = 25 cm
Section Five—Problem Bank V Ch. 12–3
V
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olt,
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8. L = 6.200 m
g = 9.819 m/s2 T = 2p L
g = 2p
9.6
8
.1
2
900mm
/s2 =
f = T
1 =
4.9
1
93 s = 0.2003 Hz
4.993 s
9. L = 2.500 m
g = 9.780 m/s2 T = 2p L
g = 2p
9.2
7
.8
5
000mm
/s2 =
f = T
1 =
3.1
1
77 s = 0.3148 Hz
3.177 s
10. L = 3.120 m
g = 9.793 m/s2 T = 2p L
g = 2p
9.3
7
.9
1
320mm
/s2 =
f = T
1 =
3.5
1
46 s = 0.2820 Hz
3.546 s
6. f = 0.50 Hz
g = 9.81 m/s2 T = 2p L
g =
1
f
f
12 =
4pg
2 L
L = 4p
g2f 2 =
(4p92.
)
8
(
1
0.
m
50
/s
s
2
−1)2 = 0.99 m = 99 cm
7. f = 2.5 Hz
g = 9.81 m/s2 T = 2p L
g =
1
f
f
12 =
4pg
2 L
L = 4p
g2f 2 =
(4p9
2.8
)(
1
2
m
.5
/
s
s−
2
1)2 = 4.0 × 10−2 m
1. f = 3.00 × 102 Hz
k = 8.65 × 104 N/m T = 2p m
k =
1
f
f
12 =
4pk
2 m
m = 4p
k2f 2 =
(
8
4
.6
p52)
×(3
1
0
0
0
4 N
s−/1m
)2 = 2.43 × 10−2 kg
Additional Practice 12C
Holt Physics Solution ManualV Ch. 12–4
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5. F = 66 N
T = 2.9 s
g = 9.81 m/s2
f = mg
m = F
g
T = 2p m
k
T 2 = 4p
k
2 m
k = 4p
T
2
2m
= 4
g
pT
2
2F
= (9.8
(
1
4pm
2
/
)
s
(26
)
6
(2
N
.9
)
s)2 = 32 N/m
6. f = 87 N
g = 9.81 m/s2
T = 0.64 s
f = mg
m = F
g =
9.8
8
1
7
m
N
/s2 = 8.9 kg
T = 2pm
k
T 2 = 4p2
k
m
k = 4p
T
2
2m
= 4
g
pT
2
2F
= (9.8
(
1
4pm
2)
/s
(28
)
7
(0
N
.6
)
4 s) = 850 N/m
3. k = 2.03 × 103 N/m
f = 0.79 HzT = 2p
m
k =
1
f
f
12 =
4pk
2 m
m = 4p
k2f 2 =
(
2
4
.
p03
2)
×(0
1
.
0
7
3
9
N
H
/
z
m
)2 = 82 kg
4. F = 32 N
T = 0.42 s
g = 9.81 m/s2
F = mg
m = F
g =
9.8
3
1
2
m
N
/s2 = 3.3 kg
T = 2p m
k
T 2 = 4p
k
2 m
k = 4p
T
2
2m
= 4
g
pT
2
2F
= = 730 N/m4p2 (32 N)
(9.81 m/s2)(0.42 s)2
2. T = 0.079 s
k = 63 N/mT = 2p
m
k
T 2 = 4p
k
2 m
m = 4
kT
p
2
2 = (63 N/m
4
)
p(20.079 s)2
= 9.96 × 10−3 kg
Section Five—Problem Bank V Ch. 12–5
V
4. f = 288 Hz
l = 5.00 mv = fl = (288 Hz)(5.00 m) = 1.44 × 103 m/s
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7. f = 20.0 Hz
v = 331 m/s
v = fl
l = v
f =
3
2
3
0
1
.0
m
H
/
z
s = 16.6 m
7. k = 364 N/m
m = 24 kgT = 2p
m
k = 2p
362
4
4 N
kg
/m = 1.6 s
8. m = 55 kg
k = (36)(458 N/m)T = 2p
m
k = 2p(36)(
5
45
58
kg
N/m) = 0.36 s
Additional Practice 12D
9. m = 8.2 kg
k = 221 N/mT = 2p
m
k = 2p
228
1
.2
Nk
/
gm = 1.2 s
10. m = 3(24 g) = 72 g = 7.2 × 10−2 kg
k = 99 N/m
T = 2pm
k = 2p
7.2
9×9 1
N0
/
−
m
2
kg = 0.17 s
1. l = 2.3 × 104 m
f = 0.065 Hz
v = fl = (0.065 Hz)(2.3 × 104 m) = 1.5 × 103 m/s
2. f = 2.8 × 105 Hz
l = 0.51 cm = 5.1 × 10−3 m
v = fl = (2.8 × 105 Hz)(5.1 × 10−3 m) = 1.4 × 103 m/s
3. l = 6.0 m
T = 2.0 sv = fl =
T
l =
6
2
.
.
0
0
m
s = 3.0 m/s
5. f = 1.6 Hz
l = 2.5 mv = fl = (1.6 Hz)(2.5 m) = 4.0 m/s
6. f = 2.5 × 104 Hz
v = 331 m/s
v = fl
l = v
f =
2.5
33
×1
1
m
04/s
Hz = 1.3 × 10−2 m
8. l = 14 m
v = 7.0 m/s
v = fl
f = lv
= 7
1
.0
4
m
m
/s = 0.50 Hz
Holt Physics Solution ManualV Ch. 12–6
V
9. l = 10.6 m
v = 331 m/s
v = fl
f = lv
= 3
1
3
0
1
.6
m
m
/s = 31.2 Hz
Givens Solutions
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olt,
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10. l = 1.1 m
v = 2.42 × 104 m/s
v = fl
f = lv
= 2.42
1
×.1
10
m
4 m/s = 2.2 × 104 Hz
Chapter 13Sound
V
1. P = 5.88 × 10−5 W
Intensity = 3.9 × 10−6 W/m2Intensity =
4pP
r2
r = (4p)(InP
tensity) =
r = 1.1 m
5.88 × 10−5 W(4p) (3.9 × 10−6 W/m2)
Additional Practice 13A
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2. P = 1.57 × 10−3 W
Intensity = 5.20 × 10−3 W/m2Intensity =
4pP
r2
r = (4p)(InP
tensity) =
r = 0.155 m
1.57 × 10−3 W(4p)(5.20 × 10−3 W/m2)
3. P = 4.80 W
Intensity = 7.2 × 10−2 W/m2Intensity =
4pP
r2
r = (4p)(InP
tensity) = = 2.3 m
4.80 W(4p)(7.2 × 10−2 W/m2)
4. P = 151 kW = 1.51 × 105 W
Intensity = .025 W/m2Intensity =
4pP
r2
r = (4p)(InP
tensity) = (41
p.)
5
(1
.0×25
1 0
W5
/
W
m2) = 693 m
5. P = 402 W
r = 32 mIntensity =
4pP
r2 = (4p
4
)
0
(
2
32
W
m)2 = 3.1 × 10−2 W/m2
6. P = 3.5 W
r = 0.50 mIntensity =
4pP
r2 = (4p)
3
(
.
0
5
.5
W
0 m)2 = 1.11 W/m2
7. P = 2.76 × 10−2 W
r = 5.0 × 10−2 mIntensity =
4pP
r2 = (4p
2
)
.
(
7
5
6
.0
××10
1
−
0
2
−2W
m)2 = 0.88 W/m2
8. Intensity = 9.3 × 10−8 W/m2
r = 0.21 mIntensity =
4pP
r2
P = 4pr2 Intensity = (4p)(0.21 m)2 (9.3 × 10−3 W/m2)
P = 5.2 × 10−3 W
Section Five—Problem Bank V Ch. 13–1
10. Intensity = 1.0 × 104 W/m2
r = 1.0 mIntensity =
4pP
r2
P = 4pr2 Intensity = (4p) (1.0 m)2 (1.0 × 104 W/m2)
P = 1.3 × 105 W
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9. Intensity = 4.5 × 10−4 W/m2
r = 1.5 mIntensity =
4pP
r2
P = 4pr2 Intensity = (4p) (1.5 m)2 (4.5 × 10−4 W/m2)
P = 1.3 × 10−2 W
1. n = 2
f2 = 466.2 Hz
v = 331 m/s
fn = 4
nv
L
L = 4
n
f
v
n =
(
(
4
2
)
)
(
(
4
3
6
3
6
1
.2
m
H
/s
z
)
) = 0.355 m = 35.5 cm
2. n = 3
f3 = 370 Hz
v = 331 m/s
fn = 4
nv
L , n = 1, 3, 5, . . .
L = 4
n
f
v
n =
(
(
3
4
)
)
(
(
3
3
3
7
1
0
m
H
/
z
s
)
) = 0.671 m = 67.1 cm
3. n = 1
f1 = 392.0 Hz
v = 331 m/s
fn = 4
nv
L
L = 4
n
f
v
n =
(
(4
1
)
)
(
(
3
3
9
3
2
1
.0
m
H
/s
z
)
) = 0.211 m = 21.1 cm
4. n = 1
f1 = 370.0 Hz
v = 331 m/s
fn = 2
nv
L
L = 2
n
f
v
n =
(
(
2
1
)
)
(
(
3
3
7
3
0
1
.0
m
H
/s
z
)
) = 0.447 m = 44.7 cm
5. n = 1
L = 35.0 cm = 3.50 × 10−1 m
v = 346 m/s
fn = 2
nv
L
f1 = (2)
(
(
1
3
)
.5
(3
0
4
×6
1
m
0−/s
1)
m) = 494 Hz
6. n = 1
L = 4.20 × 10−1 m
v = 329 m/s
fn = 2
nv
L
f1 = (2)
(
(
1
4
)
.2
(3
0
2
×9
1
m
0−/s
1)
m) = 392 Hz
Additional Practice 13B
Holt Physics Solution ManualV Ch. 13–2
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olt,
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7. n = 1
v = 499 m/s
L = 0.85 m
fn = 2
nv
L
f1 = (
(
1
2
)
)
(
(
4
0
9
.8
9
5
m
m
/s
)
) = 294 Hz
8. n = 1
f1 = 277.2 Hz
L = 0.75 m
fn = 2
nv
L
v = 2
n
Lfn =
v = 420 m/s
(2)(0.75 m)(277.2 Hz)
1
9. n = 7
f1 = 466.2 Hz
L = 1.53 m
fn = 4
nv
L
v = 4
n
Lfn = = 408 m/s(4)(1.53 m)(466.2 Hz)
7
10. n = 1
f1 = 125 Hz
L = 1.32 m
fn = 2
nv
L
v = 2L
n
fn = = 330 m/s(2)(1.32 m)(125 Hz)
1
Section Five—Problem Bank V Ch. 13–3
Chapter 14Light and Reflection
V
1. f = 7.6270 × 108 Hz
l = 3.9296 × 10−1 m
c = fl = (7.6270 × 108 s−1)(3.9296 × 10−1 m) =
The radio wave travels through Earth’s atmosphere.
2.9971 × 108 m/s
Additional Practice 14A
Givens Solutions
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olt,
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ed.
Section Five—Problem Bank V Ch. 14–1
2. f = 1.17306 × 1011 Hz
l = 2.5556 × 10−3 m
c = fl = (1.17306 × 1011 s−1)(2.5556 × 10−3 m) =
The microwave travels through space.
2.9979 × 108 m/s
3. l = 3.2 × 10−9 m f = lc
= 3
3
.0
.2
0
××
1
1
0
0−
8
9m
m
/s = 9.4 × 1016 Hz
4. l = 5.291 770 × 10−11 ma. f =
lc
= =
b. You cannot see atoms because light in the visible part of the spectrum has wavelengthsthat are much larger than atoms.
5.67 × 1018 Hz3.00 × 108 m/s
5.291 770 × 10−11 m
5. UVA: l1 = 3.2 × 10−7 m
l2 = 4.0 × 10−7 m
UVB: l1 = 2.8 × 10−7 m
l2 = 3.2 × 10−7 m
for UVA waves:
f1 = lc
1 =
3
3
.0
.2
0
××
1
1
0
0−
8
7m
m
/s =
f2 = lc
2 =
3
4
.0
.0
0
××
1
1
0
0−
8
7m
m
/s =
for UVB waves:
f1 = lc
1 =
3
2
.0
.8
0
××
1
1
0
0−
8
7m
m
/s =
f2 = lc
2 =
3
3
.0
.2
0
××
1
1
0
0−
8
7m
m
/s = 9.4 × 1014 Hz
1.1 × 1015 Hz
7.5 × 1014 Hz
9.4 × 1014 Hz
6. l = 1.67 × 10−10 m f = lc
= 1
3
.
.
6
0
7
0
××
1
1
0
0−
8
1m0 m
/s = 1.80 × 1018 Hz
7. f = 9.5 × 1014 Hz l = c
f =
3
9
.
.
0
5
0
××
1
1
0
01
8
4m
s−/1s
= 3.2 × 10−7 m = 320 nm
8. f = 2.85 × 109 Hz l = c
f =
3
2
.
.
0
8
0
5
××
1
1
0
0
8
9m
s−/1s
= 0.105 m = 10.5 cm
Holt Physics Solutions ManualV Ch. 14–2
10. f = 2.5 × 1010 Hz a. l = c
f =
3
2
.
.
0
5
0
××
1
1
0
01
8
0m
s−/1s
= 1.2 × 10−2 m =
b. 1.2 cm > 1. 2 mm
The microwave’s wavelength is larger than the holes, so it cannot pass through.This is analogous to a ball being trapped by a net.
c. 400 nm < 1.2 mm
700 nm < 1.2 mm
Yes, visible light can pass through the holes in the microwave oven’s door. This iswhy we can look through the holes and see the food as it cooks.
1.2 cm
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olt,
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9. f1 = 1800 MHz = 1.8 × 109 Hz
f2 = 2000 MHz = 2.0 × 109 Hz
l1 = f
c
1 =
3
1
.0
.8
0
××
1
1
0
09
8
s
m−1
/s = 0.17 m =
l2 = f
c
2 =
3
2
.0
.0
0
××
1
1
0
09
8
s
m−1
/s = 0.15 m = 15 cm
17 cm
1. f = 32.0 cm
h = 75 cm
a. You want to appear to be shaking hands with yourself, so the image must appearto be where your hand is. So
p = q 1
f =
p
1 +
1
q =
p
2
p = 2f = (2)(32.0 cm) =
q = p =
b. M = h
h
′ = −
p
q
h′ = − q
p
h = −
(64.0
6
c
4
m
.0
)(
c
7
m
.5 cm) = −7.5 cm
64.0 cm
64.0 cm
2. f = 9.5 cm
q = 15.5 cm
h = 3.0 cm
a. 1
f =
p
1 +
q
1
p
1 =
1
f −
1
q =
9.5
1
cm −
15.5
1
cm
p
1 =
0
1
.1
c
0
m
5 −
0
1
.0
c
6
m
5 =
0
1
.0
c
4
m
0
p =
b. h′ = − q
p
h = −
(15.5 c
2
m
5
)
c
(
m
3.0 cm) = −1.9 cm
25 cm
Additional Practice 14B
3. f = 17 cm
q = 23 cm
h = 2.7 cm
a. 1
f =
p
1 +
q
1
p
1 =
1
f −
1
q =
17
1
cm −
23
1
cm
p
1 =
0
1
.0
c
5
m
9 −
0
1
.0
c
4
m
35 =
0
1
.0
c
1
m
6
p =
b. h′ = − q
p
h = −
(23 cm
62
)(
c
2
m
.7 cm) = −1.0 cm
62 cm
Section Five—Problem Bank V Ch. 14–3
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olt,
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4. p = 5.0 cm A car’s beam has rays that are parallel, so q = ∞.
1
q =
∞1
= 0 1
f =
p
1 +
1
q =
p
1 + 0 =
p
1
f = p =
R
2 =
p
1 +
1
q =
1
f
R = 2 f = (2)(5.0 cm) = 1.0 × 101 cm
5.0 cm
7. p = 3.00 cm = 3.00 × 102 cm
f = 30.0 cm
h = 15 cm
a. 1
q =
1
f −
p
1 =
30.0
1
cm −
3.00 ×1
102 cm
1
q =
0
1
.0
c
3
m
33 −
0.
1
00
c
3
m
33 =
0
1
.0
c
3
m
00
q =
b. M = h
h
′ = −
p
q
h′ = − q
p
h = −
(3
3
3
.
.
0
3
0
c
×m
1
)(
0
125
c
c
m
m) = 1.7 cm
33.3 cm
5. p = 19 cm
q = 14 cm1
f =
p
1 +
1
q =
19
1
cm +
14
1
cm
1
f =
0
1
.0
c
5
m
3 +
0
1
.0
c
7
m
1 =
1
0.
c
1
m
2
f =
R
2 =
p
1 +
1
q =
1
f
R = 2f = (2)(8.3 cm) = 17 cm
8.3 cm
6. p = 35 cm
q = 42 cm1
f =
p
1 +
1
q =
R
2 =
35
1
cm +
42
1
cm
1
f =
0
1
.0
c
2
m
9 +
0
1
.0
c
2
m
4 =
0
1
.0
c
5
m
3
f =
R
2 =
1
f
R = 2f = (2)(19 cm) = 38 cm
19 cm
8. f = 17.5 cm
p = 15.0 cma.
1
q =
1
f −
p
1 =
17.5
1
cm −
15.0
1
cm
1
q =
0
1
.0
c
5
m
71 −
0
1
.0
c
6
m
67 =
0.
1
00
c
9
m
60
q =
b. M = − p
q = −
−1
1
5
0
.0
4
c
c
m
m = 6.93
−104 cm
Holt Physics Solutions ManualV Ch. 14–4
V
9. f = 60.0 cm
p = 35.0 cm
a. 1
q =
1
f −
p
1 =
60.0
1
cm −
35.0
1
cm =
0
1
.0
c
1
m
67 −
0
1
.0
c
2
m
86 =
−0
1
.0
cm
119
q =
b. M = − p
q = −
−3
8
5
4
.0
.0
c
c
m
m = 2.4
−84.0 cm
Givens Solutions
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olt,
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10. f = 23.0 cm
p = 3.00 cm
p = 33.0 cm
a. 1
q =
1
f −
p
1 =
23.0
1
cm −
3.00
1
cm =
0
1
.0
c
4
m
35 −
0
1
.3
c
3
m
3 =
−1
0.
c
2
m
90
q =
M = − p
q = −
3
3
.
.
4
0
5
0
c
c
m
m =
The image is real, and upright, so you can read the writing.
b. 1
q =
1
f −
p
1 =
23.0
1
cm −
33.0
1
cm =
0
1
.0
c
4
m
35 −
0
1
.0
c
3
m
03 =
0
1
.0
c
1
m
32
q =
M = − p
q = −
7
3
5
3
.
.
8
0
c
c
m
m =
The image is inverted and virtual, so you cannot read the writing (unless you canread upside-down).
−2.30
75.8 cm
1.15
−3.45 cm
1. h′ = 9.0 cm
h = 1.5 m = 150 cm
p = 3 m = 300 cm
a. M = h
h
′ =
1
9
5
.0
0
c
c
m
m =
b. M = − p
q
q = −Mp = −(0.060)(300 cm) =
1
f =
p
1 +
1
q =
3
1
m +
−0.1
1
8 m =
0
1
.3
m
3 −
−1
5
m
.6 =
−1
5
m
.3
f = −0.19 m =
c. R = 2f = 2 (19 cm) = −38 cm
−19 cm
−18 cm
0.060
Additional Practice 14C
Section Five—Problem Bank V Ch. 14–5
V
3. f = −6.3 cm
q = −5.1 cmp
1 =
1
f −
1
q =
−6.3
1
cm −
−5.1
1
cm =
0
1
.1
c
5
m
9 −
0
1
.1
c
9
m
6 =
0
1
.0
c
3
m
7
p =
M = −p
q =
−(−2
5
7
.1
cm
cm) = 0.19
27 cm
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olt,
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4. f = −33 cm
q = −16.1 cmp
1 =
1
f −
1
q =
−33
1
cm −
−16.
1
1 cm =
−1
0.
c
0
m
30 −
0
1
.0
c
6
m
2 =
0
1
.0
c
3
m
2
p =
M = − q
f = −
(−1
3
6
1
.1
cm
cm) = 0.52
31 cm
6. h = 1.75 m
M = 0.11
q = −42 cm = −0.42 m
a. M = h
h
′ = −
p
q
h′ = Mh = (0.11)(1.75 m) =
b. p = − M
q = −
0.
0
4
.
2
11
m = 3.8 m
0.19 m
5. f = −12 cm
q = −9.0 cmp
1 =
1
f −
1
q =
−12
1
cm −
−9.0
1
cm =
0
1
.0
c
8
m
3 +
0
1
.1
c
1
m
1 =
0
1
.0
c
2
m
8
p =
M = − p
q = −
9
3
.
6
0
c
c
m
m = 0.25
36 cm
7. f = −27 cm
p = 43 cm1
q =
1
f −
p
1 =
−27
1
cm −
43
1
cm =
−1
0.
c
0
m
37 −
0
1
.0
c
2
m
3 =
−1
0.
c
0
m
60
q =
M = − p
q = −
−4
1
3
7
c
c
m
m = 0.40
−17 cm
2. q = −5.2 cm
p = 17 cm
h = 3.2 cm
a. 1
f =
p
1 +
q
1 =
17
1
cm +
−5.2
1
cm =
0
1
.0
c
5
m
9 −
1
0.
c
1
m
9 =
−1
0
c
.
m
13
f =
b. 1
f =
R
2 =
p
1 +
q
1
R = 2f = (2)(−7.7 cm) = −15 cm
−7.7 cm
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olt,
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t and
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.All
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s re
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ed.
Holt Physics Solutions ManualV Ch. 14–6
10. M = 0.24
p = 12 cm
M = − p
q
q = −Mp = −(0.24)(12 cm) = −2.9 cm
9. f = −39 cm
p = 16 cm
h = 6.0 cm
a. 1
q =
1
f −
p
1 =
−39
1
cm −
16
1
cm =
−1
0.
c
0
m
26 −
0
1
.0
c
6
m
2 =
−1
0.
c
0
m
88
q =
b. M = h
h
′ = −
p
q
h′ = − q
p
h = −
(−11 c
1
m
6
)
c
(
m
6.0 cm) = 4.1 cm
−11 cm
8. f = −8.2 cm
p = 18 cm1
q =
1
f −
p
1 =
−8.2
1
cm −
18
1
cm =
−1
0.
c
1
m
22 −
0
1
.0
c
5
m
6 =
−1
0
c
.
m
18
q =
M = − p
q = −
−1
5
8
.6
c
c
m
m = 0.31
−5.6 cm
Chapter 15Refraction
V
1. qr = 35°
nr = 1.553
ni = 1.000
qi = sin−1nr(s
n
in
i
qr) = sin−1(1.553
1
)
.0
(s
0
i
0
n 35°) = 63°
Additional Practice 15A
Givens Solutions
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olt,
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s re
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ed.
Section Five—Problem Bank V Ch. 15–1
2. qr = 41°
nr = 1.486
ni = 1.00
qi = sin−1nr(s
n
in
i
qr) = sin−1(1.486
1
)
.0
(s
0
i
0
n 41°) = 77°
3. qr = 33°
nr = 1.555
ni = 1.000
qi = sin−1nr(s
n
in
i
qr) = sin−1(1.555
1
)
.0
(s
0
i
0
n 33°) = 58°
4. c = 3.00 × 108 m/s
v = 2.07 × 108 m/sn =
v
c =
3
2
.
.
0
0
0
7
××
1
1
0
0
8
8m
m
/
/
s
s = 1.45
5. c = 3.00 × 108 m/s
v = 1.97 × 108 m/sn =
v
c =
3
1
.
.
0
9
0
7
××
1
1
0
0
8
8m
m
/
/
s
s = 1.52
6. c = 3.00 × 108 m/s
v = 1.95 × 108 m/sn =
v
c =
3
1
.
.
0
9
0
5
××
1
1
0
0
8
8m
m
/
/
s
s = 1.54
7. qi = 59.2°
nr = 1.61
ni = 1.00
qr = sin−1ni(s
n
in
r
qi) = sin−1(1.00)(
1
s
.
i
6
n
1
59.2°) = 32.2°
8. qi = 35.2°
ni = 1.00
nr,1 = 1.91
nr,2 = 1.66
qr,1 = sin−1ni(s
n
i
r
n
,1
qi) = sin−1(1.00)(
1
s
.
i
9
n
1
35.2°) =
qr,2 = sin−1ni(s
n
i
r
n
,2
qi) = sin−1(1.00)(
1
s
.
i
6
n
6
35.2°) = 20.3°
17.6°
Holt Physics Solutions ManualV Ch. 15–2
10. qi = 22°
ni = 1.000
nr = 1.544
qi = 14°
ni = 1.544
nr = 1.000
1st surface:
qr = sin−1ni(s
n
in
r
qi) = sin−1(1.000
1
)
.5
(s
4
i
4
n 22°) =
2nd surface:
qr = sin−1(1.544
1
)
.0
(s
0
i
0
n 14°) = 22°
14°
V
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olt,
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t and
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ston
.All
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s re
serv
ed.
9. qi = 17°
ni = 1.00
nr = 1.54
qi = 11°
ni = 1.54
nr = 1.00
1st surface:
qr = sin−1ni(s
n
in
r
qi) = sin−1(1.00)
1
(
.
s
5
i
4
n 17°) =
2nd surface:
qr = sin−1(1.54)
1
(
.
s
0
i
0
n 11°) = 17°
11°
1. p = 13 cm
q = 19 cm
h′ = 3.0 cm
1
f =
p
1 +
1
q =
13
1
cm +
19
1
cm =
0
1
.0
c
7
m
7 +
0
1
.0
c
5
m
3 =
1
0.
c
1
m
3
f =
M = h
h
′ = −
p
q
h = − p
q
h′ = −
(13 cm
19
)(
c
3
m
.0 cm) = 2.1 cm
7.7 cm
2. p = 44 cm
q = −14 cm
h = 15 cm
1
f =
p
1 +
1
q =
44
1
cm +
−14
1
cm =
0
1
.0
c
2
m
3 +
0
1
.0
c
7
m
1 =
−1
0
c
.0
m
48
f =
M = h
h
′ = −
p
q
h′ = − q
p
h = −
(−14 c
4
m
4
)
c
(
m
15 cm) = 4.8 cm
−21 cm
3. f = −13.0 cm
M = 5.00M = −
p
q
q = −Mp = −(5.00)p
1
f =
p
1 +
1
q =
p
1 +
−(5.
1
00)p =
−5
−.0
(5
0
.0
+0
1
)p
.00 =
−(
−5
4
.0
.0
0
0
)p =
(5
4
.0
.0
0
0
)p
M = h
h
′ = −
p
q
p = (4
5
.
.
0
0
0
0
)f =
(4.00)(
5
−.0
1
0
3.0 cm) = −10.4 cm
Additional Practice 15B
Section Five—Problem Bank V Ch. 15–3
V
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4. h = 18 cm
h′ = −9.0 cm
f = 6.0 cm
a. M = h
h
′ =
−1
9
8
.0
c
c
m
m =
b. M = − p
q
q = −Mp = −(−0.50)p = (0.50)p
1
f =
p
1 +
1
q =
p
1 +
(0.5
1
0)p =
(0
0
.5
.5
0
0
)p +
(0.5
1
0)p =
(0
1
.5
.5
0
0
)p =
3
p
.0
p = (3.0) f = (3.0)(6.0 cm) =
c. q = −Mp = (0.50)(18 cm) = 9.0 cm
18 cm
−0.50 cm
8. q = −12 cm
f = −44 cmp
1 =
1
f −
1
q =
−44
1
cm −
−12
1
cm = −1
0
c
.0
m
23 − −1
0.
c
0
m
83 =
0
1
.0
c
6
m
0
p = 17 cm
5. p = 4 m
f = 4 m1
f =
p
1 +
1
q, but f = p, so
1
q = 0. That means q =
M = − p
q = −
4
•m =
The rays are parallel, and the light can be seen from very far away.
•
•
6. p = 0.5 m
f = 0.5 m1
f =
p
1 +
1
q
p = f, so 1
q = 0, and q =
M = −p
q =
−0.
•5 m =
The rays are parallel, and the light can be seen from far away.
−•
•
7. f = 3.6 cm
q = 15.2 cmp
1 =
1
f −
1
q =
3.6
1
cm −
15.2
1
cm =
1
0.
c
2
m
8 −
0
1
.0
c
6
m
6 =
1
0.
c
2
m
1
p = 4.8 cm
9. f = 9.0 cm = 0.09 m
q = 18 mp
1 =
1
f −
1
q =
0.09
1
0 m −
18
1
m =
1
1
m
1 −
0
1
.0
m
56 =
1
1
m
1
p = 0.091 m = 9.1 cm
10. f = 5.5 m
q = 5.5 cm = 0.055 mp
1 =
1
f −
1
q =
5.5
1
m −
−0.0
1
55 m =
0
1
.1
m
8 −
1
−1
m
8 =
1
1
m
8
p = 5.5 × 10−2 m = 5.5 cm
Holt Physics Solutions ManualV Ch. 15–4
V
1. qc = 37.8°
nr = 1.00sin qc =
n
nr
i
ni = sin
nr
qc =
sin
1.
3
0
7
0
.8° = 1.63
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2. qc = 39.18°
nr = 1.000sin qc =
n
nr
i
ni = sin
nr
qc =
sin
1.
3
0
9
0
.
0
18 = 1.583
3. qc,1 = 35.3°
qc,2 = 33.1°
nr = 1.00
sin qc = n
nr
i
ni,1 = sin
n
qr
c,1 =
sin
1.
3
0
5
0
.3° =
ni,2 = sin
n
qr
c,2 =
sin
1.
3
0
3
0
.1° = 1.83
1.73
4. ni = 1.670
qc = 62.85°sin qc =
n
nr
i
nr = ni(sin qc) = (1.670)(sin 62.85°) = 1.486
Additional Practice 15C
5. ni = 1.80
qc = 57.0°sin qc =
n
nr
i
nr = ni(sin qc) = (1.80)(sin 57.0°) = 1.51
6. ni = 1.64
qc = 69.9°sin qc =
n
nr
i
nr = ni(sin qc) = (1.64)(sin 69.9°) = 1.54
7. ni- = 1.766
nr = 1.000
sin qc = n
nr
i
qc = sin−1n
nr
i = sin−111.
.
0
7
0
6
0
6 = 34.49°
8. ni = 1.774
nr = 1.000
sin qc = n
nr
i
qc = sin−1n
nr
i = sin−111.
.
0
7
0
7
0
4 = 34.31°
9. ni = 1.61
nr = 1.00sin qc =
n
nr
i
qc = sin−1n
nr
i = sin−1
1
1
.
.
0
6
0
1 = 38.4°
10. ni = 1.576
nr = 1.000sin qc =
n
nr
i
qc = sin−1n
nr
i = sin−111.
.
0
5
0
7
0
6 = 39.38°
Section Five—Problem Bank V Ch. 16–1
Chapter 16Interference and Diffraction
V
1. 14 450 lines/cm
l = 6.250 × 10–7 m
q < 90°
m = 1: q1 = sin–1(ml/d)
q1 = sin–1[(1)(6.250 × 10–7 m) ÷ (1 450 000 lines/m)–1]
q1 = 64.99°
m = 2: q2 = sin–1[(2)(6.250 × 10–7 m) ÷ (1 450 000 lines/m)–1]
q2 = ∞
Therefore, 1 is the highest-order number that can be observed.
Additional Practice 16A
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2. 12 260 lines/cm
l = 5.896 × 10–7 m
q < 90°
m = 1: q1 = sin–1(ml/d)
q1 = sin–1[(1)(5.896 × 10–7 m) ÷ (1 226 000 lines/m)–1]
q1 = 46.29°
m = 2: q2 = sin–1[(2)(5.896 × 10–7 m) ÷ (1 450 000 lines/m)–1]
q2 = ∞
Therefore, 1 is the highest-order number that can be observed.
3. l= 5.461 × 10–7 m
m = 1
q = 75.76°
d = s
m
in
lq
= = 5.634 × 10–7 m
d = 5.634 × 10–5 cm
# lines/cm = (5.634 × 10–5 cm)–1 = 1.775 × 104 lines/cm
(1)(5.461 × 10–7 m)
sin (75.76°)
4. l = 4.471 × 10–7 m
m = 1
q = 40.25°
d = s
m
in
lq
= = 6.920 × 10–7 m
d = 6.920 × 10–7 cm
# lines/cm = (6.920 × 10–7 cm)–1 = 1.445 × 104 lines/cm
(1)(4.471 × 10–7 m)
sin (40.25°)
5. 1950 lines/cm
l = 4.973 × 10–7 m
m = 1: q1 = sin–1(ml /d)
q1 = sin–1[(1)(4.973 × 10–7 m) ÷ (195 000 lines/m)–1]
q1 =
m = 2: q2 = sin–1[(2)(4.973 × 10–7 m) ÷ (195 000 lines/m)–1]
q2 = 11.18°
5.565°
10. 2400 lines/cm
m = 2
q = 26.54 °
l = d(s
m
in q) = = 9.296 × 10–7 m
(240 000 m)–1sin (26.54°)
2
1. l = 5.875 × 10–7 m
m = 2
q = 0.130°
d = s
m
in
lq
= = 5.18 × 10–4 m
d = 0.518 mm
2(5.875 × 10–7 m)
sin (0.130°)
2. l = 6.563 × 10–7 m
m = 4
q = 0.626°
d = s
m
in
lq
= = 2.40 × 10–4 m
d = 0.240 mm
4(6.563 × 10–7 m)
sin (0.626°)
Holt Physics Solution ManualV Ch. 16–2
V
6. 7500 lines/cm
l1 = 5.770 × 10–7 m
l2 = 5.790 × 10–7 m
m = 2
q1 = sin–1(ml1/d)
q1 = sin–1[(2)(5.770 × 10–7 m) ÷ (750 000 lines/m)–1]
q1 = 59.94°
q2 = sin–1(ml2/d)
q2 = sin–1[(2)(5.790 × 10–7 m) ÷ (750 000 lines/m)–1]
q2 = 60.28°
q = q2 – q1 = 59.94° – 60.28° = 0.3400°
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7. 3600 lines/cm
m = 3
q = 76.54°
l = d(s
m
in q) = = 9.000 × 10–7 m
(360 000 m)–1sin (76.54°)
3
8. 9550 lines/cm
m = 2
q = 54.58°
l = d(s
m
in q) = = 4.267 × 10–7 m
(955 000 m)–1sin (54.58°)
2
9. 12 510 lines/cm
m = 1
q = 38.77°
l = d(s
m
in q) = = 5.006 × 10–7 m
(1 251 000 m)–1sin (38.77°)
1
Additional Practice 16B
Section Five—Problem Bank V Ch. 16–3
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3. l = 6.93 × 10–7 m
m = 3
q = 0.578°
d = s
m
in
lq
= = 2.06 × 10–4 m
d = 0.206 mm
3(6.93 × 10–7 m)
sin (0.578°)
5. d = 4.43 × 10–6 m
m = 3
q = 21.7 °
10. d = 2.67 × 10–4 m
l = 6.87 × 10–7 m
m = 1
q = sin–1(ml /d)
q = sin–1[(1)(6.87 × 10–7 m) ÷ (2.67 × 10–4 m)] = 0.147°
4. d = 8.04 × 10–6 m
m = 3
q = 13.1 °
l = d(s
m
in q) = = 6.07 × 10–7 m
l = 607 nm
(8.04 × 10–6 m) sin (13.1°)
3
l = d(s
m
in q) = = 5.46 × 10–7 m
l = 546 nm
(4.43 × 10–6 m) sin (21.7°)
3
6. d = 3.92 × 10–6 m
m = 2
q = 13.1 °
l = d(s
m
in q) = = 4.44 × 10–7 m
l = 444 nm
(3.92 × 10–6 m) sin (13.1°)
2
7. d = 2.20 × 10–4 m
l= 5.27 × 10–7 m
m = 1
q = sin–1(ml /d)
q = sin–1[(1)(5.27 × 10–7 m) ÷ (2.20 × 10–4 m)] = 0.137°
8. d = 1.63 × 10–4 m
l = 4.308 × 10–7 m
m = 1
q = sin–1(ml /d)
q = sin–1[(1)(4.308 × 10–7 m) ÷ (1.63 × 10–4 m)] = 0.151°
9. d = 3.29 × 10–4 m
l = 5.83 × 10–7 m
m = 1
q = sin–1(ml /d)
q = sin–1[(1)(5.83 × 10–7 m) ÷ (3.29 × 10–4 m)] = 0.102°
Section Five—Problem Bank V Ch. 17–1
Chapter 17Electric Forces and Fields
V
1. q1 = − 1.30 × 10−5 C
q2 = −1.60 × 10−5 C
Felectric = 12.5 N
kC = 8.99 × 109 N •m2/C2
Felectric = kC
r
q21q2
r = k
FCel
qec
1t
q
r
i
2
c =
r = 0.387 m = 38.7 cm
(8.99 × 109 N •m2/C2)(−1.30 × 10−5 C)(−1.60 × 10−5 C)
12.5 N
Additional Practice 17A
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2. q1 = 9.99 × 10−5 C
q2 = 3.33 × 10−5 C
Felectric = 87.3 N
kC = 8.99 × 109 N •m2/C2
Felectric = kC
r
q21q2
r = k
FCel
qec
1t
q
r
i
2
c =
r = 0.585 m = 58.5 cm
(8.99 × 109 N •m2/C2)(9.99 × 10−5 C)(3.33 × 10−5 C)1
87.3 N
3. q1 = −4.32 × 10−5 C
q2 = 2.24 × 10−5 C
Felectric = −6.5 N
KC = 8.99 × 109 N •m2/C2
Felectric = kC
r
q21q2
r = k
FCel
qec
1t
q
r
i
2
c =
r = 1.15 m
(8.99 × 109 N •m2/C2)(−4.32 × 10−5 C)(2.24 × 10−5 C)
−6.5 N
4. q1 = −5.33 × 10−6 C
q2 = +5.3 × 1−6 C
r = 4.2 × 10−2 m
kC = 8.99 × 109 N •m2/C2
Felectric = kC
r
q21q2
Felectric =
Felectric = −143 N
(8.99 × 109 N •m2/C2)(−5.3 × 10−6 C)(5.3 × 10−6 C)
(4.2 × 10−2 m)2
5. q1 = −1.40 × 10−8 C
q2 = +1.40 × 10−8 C
r = 7.1 × 10−2 m
kC = 8.99 × 109 N •m2/C2
Felectric = kC
r
q21q2
Felectric =
Felectric = 3.50 × 10−4 N
(8.99 × 109 N •m2/C2)(−1.40 × 10−8 C)(1.4 × 10−8 C)
(7.1 × 10−2 m)2
6. q1 = −8.0 × 10−9 C
q2 = +8.0 × 10−9 C
r = 2.0 × 10−2 m
kC = 8.99 × 109 N •m2/C2
Felectric = kC
r
q21q2
Felectric =
Felectric = 1.4 × 10−3 N
(8.99 × 109 N •m2/C2)(−8.0 × 10−9 C)(8.0 × 10−9 C)
(2.0 × 10−2 m)2
Holt Physics Solution ManualV Ch. 17–2
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7. r = 8.3 × 10−10 m
Felectric = 3.34 × 10−10 N
kC = 8.99 × 109 N •m2/C2
Felectric = kC
r
q21q2 =
kC
r2q2
q = Fele
kct
C
ricr2
= q = 1.60 × 10−19 C
(3.34 × 10−10 N)(8.3 × 10−10 N)2
8.99 × 109 N •m2/C2
8. r = 6.4 × 10−8 m
Felectric = 5.62 × 10−14 N
kC = 8.99 × 109 N •m2/C2
Felectric = kC
r
q21q2 =
kC
r2q2
q = Fele
kct
C
ricr2
= q = 1.60 × 10−19 C
(5.62 × 10−14 N)(6.4 × 10−8 m)2
8.99 × 109 N •m2C2
9. r = 9.30 × 10−11 m
Felectric = 2.66 × 10−8 N
kC = 8.99 × 109 N •m2/C2
Felectric = kC
r
q21q2 =
kC
r2q2
q = Fele
kct
C
ricr2
= q = 1.60 × 10−19 C
(2.66 × 10−8 N)(9.30 × 10−11 m)2
8.99 × 109 N •m2/C2
10. r = 6.5 × 10−11 m
Felectric = 9.92 × 10−4 N
kC = 8.99 × 109 N •m2/C2
Felectric = kC
r
q21q2 =
kC
r2q2
q = Fele
kct
C
ricr2
= q = 2.2 × 10−17 C
(9.92 × 10−4 N)(6.5 × 10−11 m)2
8.99 × 109 N •m2/C2
Additional Practice 17B
1. qp = 1.60 × 10−19 C
r4,1 = r2,1 = 1.52 × 10−9 m
kC = 8.99 × 109 N •m2/C2
qp = q1 = q2 = q3 = q4
r3,2 =√
(1.52× 10−9m)2 + (1.52 × 10−9m)2 = 2.15 × 10−9 m
F2,1 = k
rC
2
q
,1
p2
2
= = 9.96 × 10−11 N
F3,1 = k
rC
3
q
,1
p2
2
= = 4.98 × 10−11 N
F4,1 = k
rC
4
q
,1
2
2p = = 9.96 × 10−11 N
j = tan−1 11.
.
5
5
2
2
××
1
1
0
0
−
−
9
9m
m = 45°
F2,1: Fx = 0 N
Fy = 9.96 × 10−11 N
F3,1: Fx = F3,1 cos 45° = (4.98 × 10−11 N)(cos 45°) = 3.52 × 10−11 N
Fy = F3,1 sin 45° = (4.98 × 10−11 N)(sin 45°) = 3.52 × 10−11 N
F4,1: Fx = 9.96 × 10−11 N
Fy = 0 N
(8.99 × 109 N •m2/C2)(1.60 × 10−19 C)2
(1.52 × 10−9 m)2
(8.99 × 109 N •m2/C2)(1.60 × 10−19 C)2
(2.15 × 10−9 m)2
(8.99 × 109 N •m2/C2)(1.60 × 10−19 C)2
(1.52 × 10−9 m)2
Section Five—Problem Bank V Ch. 17–3
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Fx,tot = 0 N + 3.52 × 10−11 N + 9.96 × 10−11 N = 1.3 × 10−10 N
Fy,tot = 9.96 × 10−10 N + 3.52 × 10−11 N + 0 N = 1.3 × 10−10 N
Ftot =√
(Fx,tot)2+ (Fy,tot)2 =√
(1.3 × 10−10 N)2 + (1.3× 10−10 N)2
Ftot =
j = tan−111.
.
3
3
××
1
1
0
0
−
−
1
1
0
0N
N = 45°
1.8 × 10−10 N
2. h = 3.50 m
q1 = q2 = 4.50 C
q3 = 6.30 C
r1,2 = 5.00 m
kC = 8.99 × 109 N • m2/C2
Set the origin at the midpoint.
r1,3 = r2,3 =√
(2.50m)2 + (3.50 m)2 = 4.30 m
F1,3 = kC
r1
q
,3
12q3 = = 1.38 × 1010 N
F2,3 = kC
r2
q
,3
22q3 = = 1.38 × 1010 N
q = tan−1 32.
.
5
5
0
0
m
m = 54.5°
F1,3: Fx = F1,3 cos q = (1.38 × 1010 N) cos (54.5°) = 8.01 × 109 N
Fy = F1,3 sin q = (1.38 × 1010 N) sin (54.5°) = 1.12 × 1010 N
F2,3: Fx = F2,3 cos q = (1.38 × 1010 N) cos (54.5°) = 8.01 × 109 N
The electrical force on q2 points in the −x direction, so Fx must be negative.
Fx = −8.01 × 109 N
Fy = F2,3 sin q = (1.38 × 1010 N) sin (54.5°) = 1.12 × 1010 N
Fx,tot = 8.01 × 109 N − 8.01 × 109 N = 0 N
Fy,tot = 1.12 × 1010 N + 1.12 × 1010 N = 2.24 × 1010 N
There is no x-component of the resultant force, so
Ftot = Fy,tot = 2.24 × 1010 N pointing upward along the y-axis
(8.99 × 109 N •m2/C2)(4.50 C)(6.30 C)
(4.30 m)2
(8.99 × 109 N •m2/C2)(4.50 C)(6.30 C)
(4.30 m)2
3. q1 = −9.00 × 10−9 C
q2 = −8.00 × 10−9 C
q3 = 7.00 × 10−9 C
r1,2 = 2.00 m
r1,3 = 3.00 m
kC = 8.99 × 109 N •m2/C2
F1,2 = kC
r1
q
,2
12q2 = = 1.62 × 10−7 N
F1,3 = kC
r1
q
,3
12q3 = = −6.29 × 10−8 N
F1,2: Fx = 4.05 × 10−8 N
Fy = 0 N
F1,3: Fx = 0 N
Fy = −6.29 × 10−8 N
Ftot =√
(1.62× 10−7N)2 + (−6.29 × 10−8N)2 =
Ftot is negative because the larger, y-component of the force is negative.
j = tan−1 −16.6.229××
1
1
0
0−
−
7
8
N
N = −21.22°
1.74 × 10−7 N
(8.99 × 109 N •m2/C2)(−9.00 × 10−9 C)(−7.00 × 10−9 C)
(3.00 m)2
(8.99 × 109 N •m2/C2)(−9.00 × 10−9 C)(−8.00 × 10−9 C)
(2.00 m)2
Holt Physics Solution ManualV Ch. 17–4
V
4. q1 = −2.34 × 10−8 C
q2 = 4.65 × 10−9 C
q3 = 2.99 × 10−10 C
r1,2 = 0.500 m
r1,3 = 1.00 m
kC = 8.99 × 109 N •m2/C2
F1,2 = kC
r1
q
,2
12q2 =
F1,2 = Fy = −3.91 × 10−6 N
F1,3 = kC
r1
q
,3
12q3 =
F1,3 = Fy = − 6.29 × 10−8 N
Fy,tot = −3.91 × 10−6 N + −6.29 × 10−8 N = 3.97 × 10−6 N
There are no x-components of the electrical force, so the magnitude of the electricalforce is
√(Fy,tot)2.
Ftot = 3.97 × 10−6 N directed along the −y-axis
(8.99 × 109 N •m2/C2)(−2.34 × 10−8 C)(2.99 × 10−10 C)
(1.00 m)2
(8.99 × 109 N •m2/C2)(−2.34 × 10−8 C)(4.65 × 10−9 C)
(0.500 m)2
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5. qe = −1.60 × 10−19 C
r2,3 = r4,3 = 3.02 × 10−5 m
r1,3 =√
2(3.02 × 10−5m)2= 4.27 × 10−5 m
kC = 8.99 × 109 N •m2/C2
qe = q1 = q2 = q3 = q4
F3,2 = Fx = k
rC
3
q
,2
22e
= = 2.52 × 10−19 N
F3,4 = Fy = k
rC
3
q
,4
e2
2
= = 2.52 × 10−19 N
F3,1 = k
rC
3,
q
1
e2
2
= = 1.26 × 10−19 N
j = tan−133.
.
0
0
2
2
××
1
1
0
0
−
−
5
5m
m = 45°
F3,1: Fx = F3,1 cos 45° = (1.26 × 10−19 N) cos 45° = 8.91 × 10−20 N
Fy = F3,1 sin 45° = (1.26 × 10−19 N) sin 45° = 8.91 × 10−20 N
Fx,tot = 8.91 × 10−20 N + 2.52 × 10−19 N + 0 N = 3.41 × 10−19 N
Fy,tot = 8.91 × 10−20 N + 0 N + 2.52 × 10−19 N = 3.41 × 10−19 N
Ftot =√
(Fx,tot)2+ (Fy,tot)2 =√
(3.41× 10−19 N)2 (3.41 × 10−19 N)2
Ftot =
j = tan−1FFx
y,
,
t
t
o
o
t
t = tan−133.
.
4
4
1
1
××
1
1
0
0
−
−
1
1
9
9N
N = 45°
4.82 × 10−19 N
(8.99 × 109 N •m2/C2)(−1.60 × 10−19 C)2
(4.27 × 10−5 m)2
(8.99 × 109 N •m2/C2)(−1.60 × 10−19 C)2
(3.02 × 10−5 m)2
(8.99 × 109 N •m2/C2)(−1.60 × 10−19 C)2
(3.02 × 10−5 m)2
6. q1 = 2.22 × 10−10 C
q2 = 3.33 × 10−9 C
q3 = 4.44 × 10−8 C
r1,2 = 1.00 m
h = 0.250 m
kC = 8.99 × 109 N •m2/C2
r3,1 = r3,2
=√
(0.500 m)2 + (0.250m)2= 0.559 m
F3,1 = kC
r3
q
,1
12q3 = = 2.84 × 10−7 N
F3,2 = kC
r3
q
,2
32q2 = = 4.25 × 10−6 N
j = tan−100.
.
2
5
5
0
0
0
m
m = 26.6°
F3,1: Fx = F3,1 cos q = (2.84 × 10−7 N) cos (−26.6°) = 2.54 × 10−7 N
Fy = F3,1 sin q = (2.84 × 10−7 N) sin (−26.6°) = 1.27 × 10−7 N
F3,2: Fx = F3,2 cos q = (4.25 × 10−6 N) cos (−26.6°) = 3.80 × 10−6 N
Fy = F3,2 sin q = (4.25 × 10−6 N) sin (−26.6°) = 1.90 × 10−6 N
(8.99 × 109 N •m2/C2)(4.44 × 10−8 C)(3.33 × 10−9 C)
(0.559 m)2
(8.99 × 109 N •m2/C2)(2.22 × 10−10 C)(4.44 × 10−8 C)
(0.559 m)2
Section Five—Problem Bank V Ch. 17–5
V
Fx,tot = 2.54 × 10−7 N + 3.80 × 10−6 N = 4.05 × 10−6 N
Fy,tot = 1.27 × 10−7 N + 1.90 × 10−6 N = 2.03 × 10−6 N
Ftot =√
(Fx,tot2 + (Fy,tot)2 =
√(4.05× 10−6N)2 + (2.03 × 10−6N)
Ftot =
j = tan−1FFx
y,
,
t
t
o
o
t
t = tan−124.
.
0
0
3
5
××
1
1
0
0
−
−
6
6N
N = 26.6°
4.53 × 10−6 N
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7. q1 = q2 = q3 = 2.0 × 10−9 C
r1,2 = 1.0 m
r1,3 =√
(1.0 m)2 + (2.0m)2= 2.24 m
kC = 8.99 × 109 N •m2/C2
F1,2 = kC
r1
q
,2
12q2 = = 3.60 × 10−8 N
F1,3 = kC
r1
q
,3
12q3 = = 7.17 × 10−9 N
F1,2 = Fx = 3.60 × 10−8 N
Fy = 0 N
j = tan−121.
.
0
0
m
m = 63.4°
F1,3: Fx = F1,3 cos q = (7.17 × 10−9 N) cos (63.4°) = 3.21 × 10-−9 N
Fy = F1,3 sin q = (7.17 × 10−9 N) sin (63.4°) = 6.41 × 10−9 N
Fx,tot = 3.60 × 10−8 N + 3.21 × 10−9 N = 3.92 × 10−8 N
Fy,tot = 0 N + 6.41 × 10−9 N = 6.41 × 10−9 N
Ftot =√
(Fx,tot)2+ (Fy,tot)2 =√
(3.92× 10−8N)2 + (6.41 × 10−9N)2Ftot = 3.97 × 10−8 N
j = tan−1FFx
y,
,
t
t
o
o
t
t = tan−163.
.
4
9
1
2
××
1
1
0
0
−
−
9
8N
N = 9.29°
(8.99 × 109 N •m2/C2)(2.0 × 10−9 C)2
(−2.24 m)2
(8.99 × 109 N •m2/C2)(2.0 × 10−9 C)2
(1.0 m)2
8. q1 = −4.0 × 10−3 C
q2 = −8.0 × 10−3 C
q3 = 2.0 × 10−3 C
kC = 8.99 × 109 N •m2/C2
r1,2 = 2.0 m
r1,3 = 2.0 m
F1,2 = kC
r1
q
,
1
2
q2
2 =
F1,2 = Fx,tot =
F1,3 = kC
r1
q
,
1
3
q2
3 =
F1,2 = Fy,tot =
Ftot =√
(Fx,tot)2+ (Fy,tot)2 =√
(7.2 × 104 N)2 + (−1.8× 104 N)2
Ftot =
j = tan−1FFx
y,
,
t
t
o
o
t
t = tan−1–71.2.8×
×1
1
0
04
4
N
N = –14°
7.4 × 104 N
−1.8 × 104 N
(8.99 × 109 N •m2/C2)(−4.0 × 10−3 C)(2.0 × 10−3 C)
(2.0 m)2
7.2 × 104 N
(8.99 × 109 N •m2/C2)(−4.0 × 10−3 C)(−8.0 × 10−3 C)
(2.0 m)2
Holt Physics Solution ManualV Ch. 17–6
V
9. q1 = 9.00 × 10−3 C
q2 = 6.00 × 10−3 C
q3 = 3.00 × 10−3 C
kC = 8.99 × 109 N •m2/C2
x1,2 = 1.00 m
y1,2 = 1.00 m
x1,3 = 1.00 m
y1,3 = 1.00 m
r1,2 = r1,3 =√
x1,22+ y1,22 =
√x1,32+ y1,3
2 =√
(1.00m)2 + (1.00 m)2 = 1.41 m
r2,3 =√
r1,22+ r1,32 =
√(1.41m)2 + (1.41 m)2 = 1.99 m
F1,2 = kC
r1
q
,2
12q2 = = 2.44 × 105 N
F1,3 = kC
r1
q
,3
12q3 = = 1.22 × 105 N
j = tan−1 = 45°
F1,2: Fx = F1,2 cos q = (2.44 × 105 N) cos 45° = 1.73 × 105 N
Fy = F1,2 sin q = (2.44 × 105 N) sin 45° = 1.73 × 105 N
F1,3: Fx = F1,3 cos q = (1.22 × 105 N) cos 45° = 8.63 × 104 N
Fy = F1,3 sin q = (1.22 × 105 N) sin 45° = 8.63 × 104 N
Fx,tot = 1.73 × 105 N + 8.63 × 104 N = 2.59 × 105 N
Fy,tot = 1.73 × 105 N + 8.63 × 104 N = 2.59 × 105 N
Ftot =√
(Fx,tot)2+ (Fy,tot)2 =√
(2.59× 105 N)2 + (2.59 × 105 N)2
Ftot =
j = tan−122.
.
5
5
9
9
××
1
1
0
0
5
5N
N = 45°
3.66 × 105 N
1.00 m1.00 m
(8.99 × 109 N •m2/C2)(9.00 × 10−3 C)(3.00 × 10−3 C)
(1.41 m)2
(8.99 × 109 N •m2/C2)(9.00 × 10−3 C)(6.00 × 10−3 C)
(1.41 m)2
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10. q1 = q2 = q3 = 4.00 × 10−9 C
kC = 8.99 × 109 N •m2/C2
r2,1 = r2,3 = 4.00 m
All forces are along the x-axis, so there are no y-components.
F2,1 = F2,3 = k
r2
C
,
q
12
2
= = 8.99 × 10−9 N
Fx,tot = 2(8.99 × 10−9 N) = 1.80 × 10−8 N
Fy,tot = 0 N
Ftot =√
(Fx,tot)2+ (Fy,tot)2 = 1.80 × 10−8 N along the x-axis
(8.99 × 109 N •m2/C2)(4.00 × 10−9 C)2
(4.00 m)2
1. q1 = 9.0 mC
q2 = −19 mC
q3 = 9.0 mC
kC = 8.99 × 109 N •m2/C2
r2,1 = 3.0 m
The charge, q3, cannot be in electrostatic equilibrium between q1 and q2, because theforces point in the same direction. Because q2 is larger than q1, q3 will be close to q1,and opposite q2.
F3,1 = −F3,2 = (y −
kC
3
q
.03q
m1
)2 = − kC
y
q23q2
q1y2 = −q2(y − 3.0 m)2 = −q2y2 + 6q2y − (9.0 m2)q2
(q1 + q2)y2 − 6q2y + (9.0 m2)q2 = 0
y =
y =
y = 9.6 m = r2,3
F3,2 = kC
r3
q
,2
32q2 =
F3,2 = −1.7 × 10−2 N
(8.99 × 109 N •m2/C2)(−1.9 × 10−5 C)(9.0 × 10−6 C)
(9.6 m)2
6(−19 mC) ±√
(6.0 m)2(−19 mC)2 − 4(9.0 mC − 19mC)(9.0m2)(−19 mC)2(9.0 mC − 19 mC)
6q2 ±√
(6q2)2 − 4(q1+ q2)(9.0 m2)(q2)2(q1 + q2)
Additional Practice 17C
Section Five—Problem Bank V Ch. 17–7
V
2. q1 = 25 mC
q2 = −5.0 mC
q3 = −35 mC
kC = 8.99 × 109 N •m2/C2
r1,2 = 0.25 m
The charge, q3, cannot be in electrostatic equilibrium between q1 and q2, because the forces point in the same direction. Because q1 is larger than q2, q3 will be closest to q2on the side opposite of q1.
F3,2 = −F3,1 = k
(C
−q
x3
)
q22 =
(x +−k
0C
.2
q
53q
m1
)2
(q2 + q1)x2 − (0.50 m)q2x + (0.625 m2)q2 = 0
x =
x =
x = −0.20 m = r3,2
F3,2 = kC
r
q
3,
3
2
q1 = = 39 N(8.99 × 109 N •m2/C2)(−3.5 × 10−5 C)(−5.0 × 10−6 C)
(−0.20 m)2
(0.50 m)(−5.0 mC) ±√
(0.50m)2(−5.0mC)2 − 4(−5.0mC + 25mC)(0.625m2)(−5.0mC)2(−5.0 mC + 25 mC)
(0.50 m)q2 ±√
(0.50m)2q22− 4(q2+ q1)(0.625 m2)q22(q2 + q1)
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3. q1 = 6.0 mC
q2 = −12.0 mC
q3 = 6.0 mC
kC = 8.99 × 109 N •m2/C2
r1,0 = 5.0 × 10−2 m
F2,3 = −F1,2 = −k
rC
1,
q
2
12q2 =
F2,3 = 259 N
−(8.99 × 109 N •m 2/C2)(6.0 × 10−6 C)(−12.0 × 10−6 C)
(5.0 × 10−2 m)2
4. q1 = 7.2 nC
q2 = 6.7 nC
q3 = −3.0 nC
kC = 8.99 × 109 N •m2/C2
r1,2 = 3.2 × 10−1 m = 0.32 m
The charge, q3, must be between the charges to achieve electrostatic equilibrium.
F1,3 + F1,2 = kC
x
q21q3 −
(x −kC
0.
q
32
2
q3
m)2 = 0
(q1 − q2)x2 − (0.64 m)q1x + (0.32 m)2q1x = 0
x =
x = 16 cm
(0.64 m)(7.2 nC) ±√
(0.64m)2 (7.2nC)2 − 4(7.2 nC − 6.7 nC)(0.32 m)2(7.2 nC)2(7.2 nC − 6.7 nC)
5. q1 = 5.5 nC
q2 = 11 nC
q3 = −22 nC
kC = 8.99 × 109 N •m2/C2
r1,2 = 88 cm
The charge, q3, must be between the charges to achieve electrostatic equilibrium.
F1,3 + F1,2 = kC
x
q21q3 −
(x −kC
8
q
82q
cm3
)2 = 0
(q1 − q2)x2 − (176 cm)q1x + (88 cm)2q1x = 0
x =
x = 36 cm
(176 cm)(5.5 nC) ±√
(176 cm)2(5.5 nC)2 − 4(5.5 nC − 11nC)(88 cm)2(5.5 nC)2(5.5 nC − 11 nC)
Holt Physics Solution ManualV Ch. 17–8
V
6. q1 = −2.5 nC
q2 = − 7.5 nC
q3 = 5.0 nC
kC = 8.99 × 109 N •m2/C2
r1,2 = 20.0 cm
The charge, q3, must be between the charges to achieve electrostatic equilibrium.
F1,3 + F1,2 = kC
x
q21q3 −
(x −k
2C
0
q
.2
0
q
c3
m)2 = 0
(q1 − q2)x2 − (40.0 cm)q1x + (20.0 cm)2q1x = 0
x =
x = 7.3 cm
(40.0 cm)(−2.5 nC) ±√
(40.0cm)2(−2.5nC)2 − 4(−2.5nC + 7.5 nC)(20.0 cm)2(−2.5nC)2(−2.5 nC + 7.5 nC)
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7. q1 = −2.3 C
q3 = −4.6 C
r1,2 = r3,1 = 2.0 m
r3,2 = 4.0 m
kC = 8.99 × 109 N •m2/C2
F3,1 + F3,2 = −k
rC
3,
q
1
32q1 −
kC
r
q
3,
3
2
q2 = 0
q2 = −q
r1
3
r
,
3
1
,22
2 = −(−2.
(
3
2.
C
0
)
m
(4
)
.20 m)2
= 9.2 C
8. q1 = 8.0 C
q3 = −4.0 C
r1,2 = 1.0 m
r3,1 = 1.0 m
r3,2 = 2.0 m
F3,1 + F3,2 = −k
rC
3,
q
1
32q1 −
kC
r3
q
,2
32q2 = 0
q2 = −q
r1
3
r
,1
32,2
2
= −(8.0
(1
C
.0
)(
m
2.
)
02
m)2
= −32 C
9. q1 = 49 C
q3 = −7.0 C
r1,2 = 7.0 m
r3,1 = −18.0 m
r3,2 = 25.0 m
F3,1 + F3,2 = − k
rC
3,
q
123q1 −
kC
r3
q
,2
32q2 = 0
q2 = −q
r1
3
r
,1
32,2
2
= −(4
(
9
−C
18
)(
.0
25
m
.0
)2m)2
= −94.5 C
10. q1 = 72 C
q3 = −8.0 C
r1,2 = 15 mm = 1.5 × 10−2 m
r3,1 = −9.0 mm = −9.0 × 10−3 m
r3,2 = 2.4 × 10−2 m
F3,1 + F3,2 = − k
rC
3,
q
123q1 −
kC
r3
q
,2
32q2 = 0
q2 = −q
r1
3
r
,1
32,2
2
= = −512 C−(72 C)(2.4 × 10−2 m)2
(−9.0 × 10−3 m)2
Additional Practice 17D
1. Ex = 9.0 N/C
q = −6.0 CEx =
Fele
qctric
Felectric = Exq = (9.0 N/C)(−6.0 C)
Felectric = −54 N in the −x direction
2. Ey = 1500 N/C
q = 5.0 × 10−9 CEy =
Fele
qctric
Felectric = Eyq = (1500 N/C)(5.0 × 10−9 C)
Felectric = 7.5 × 10−6 N in the +y direction
Section Five—Problem Bank V Ch. 17–9
V
3. m = 3.35 × 10−15 kg
q = 1.60 × 10−19 C
g = 9.81 m/s2
a. Felectric = −Fgravity = −mg = −(3.35 × 10−15 kg)(9.81 m/s2)
Felectric = −3.29 × 10−14 N upward
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4. qq = 3.00 × 10−6 C
q2 = 3.00 × 10−6 C
kC = 8.99 × 109 N •m2/C2
r1 = 0.250 m
r2 =√
(2.00m)2 + (2.00 m)2 = 2.02 m
E1 = Ey = Ey = k
rC
1
q21 =
E1 = Ey,1 = 4.32 × 105 N/C
E2 = k
rC
2
q22 = = 6.61 × 103 N/C
j = tan−1x
y = tan−102.2.0500m
m = 7.12°
Ex,2 = E2 cos 7.12° = (6.61 × 103 N/C)(cos 7.12°) = 6.56 × 103 N/C
Ey,2 = E2 sin 7.12° = (6.61 × 103 N/C)(sin 7.12°) = 8.19 × 103 N/C
Ex,tot = 0 N/C + 6.56 × 103 N/C = 6.56 × 103 N/C
Ey,tot = 4.32 × 105 N/C + 8.19 × 103 N/C = 4.40 × 105 N/C
Etot =√
(Ex,tot2 + (Ey,tot)2
Etot =√
(6.56× 103 N/C)2 + (4.40 × 105 N/C)2
Etot =
tan j = E
E
x
y,
,
t
t
o
o
t
t =
4
6
.
.
4
5
0
6
××
1
1
0
0
5
3N
N
/
/
C
C
j = 89.1°
4.40 × 105 N/C
(8.99 × 109 N •m2/C2)(3.00 × 10−6 C)
(2.02 m)2
(8.99 × 109 N •m2/C2)(3.00 × 10−6 C)
(0.250 m)2
5. q1 = 1.50 × 10−5 C
q2 = 5.00 × 10−6 C
kC = 8.99 × 109 N •m2/C2
r1 = 1.00 m
r2 = 0.500 m
E1 = Ey,1 = k
rC
1
q21 = = 1.35 × 105 N/C
E2 = Ey,2 = k
rC
2
q22 = = 1.80 × 105 N/C
Ey,tot = Etot = 1.35 × 105 N/C + 1.80 × 105 N/C =
The electric field points along the y-axis.
3.15 × 105 N/C
(8.99 × 109 N •m2/C2)(5.00 × 10−6 C)
(0.500 m)2
(8.99 × 109 N •m2/C2)(1.50 × 10−5 C)
(1.00 m)2
Holt Physics Solution ManualV Ch. 17–10
V
6. E = 1663 N/C
Felectric = 8.4 × 10−9 NE =
Fele
qctric
q = = 8.4
1
2
66
×3
1
N
0−
/C
9 N = 5.06 × 10−12 C
FelectricE
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7. E = 4.0 × 103 N/C
Felectric = 6.43 × 10−9 NE =
Fele
qctric
q = = 4
6
.
.
0
43
××10
130−
N
9
/
N
C = 1.61 × 10−12 C
FelectricE
8. q1 = −1.60 × 10−19 C
kC = 8.99 × 109 N •m2/C2
q2 = −1.60 × 10−19 C
q3 = 1.60 × 10−19 C
r1 = 3.00 × 10−10 m
r2 = 2.00 × 10−10 m
Fx,tot = kC
r2q1 +
k
rC
2
q22 +
kC
x2q3 = 0
q1 = q2 = −q3
kCq1r
1
12 +
r
1
22 −
x
12 = 0
kCq1r
1
12 +
r
1
22 =
kC
x2q1
x2 = =
x = 1.66 × 10−10 m
1(3.00 × 1
1
0−10 m)2 + (2.00 × 1
1
0−10 m)
1r
1
12 +
r
1
22
9. q1 = −9.00 C
q2 = 6.00 C
q3 = 3.00 C
r1 = 1.5 mm
r2 = 1.5 mm
kC = 8.99 × 109 N • m2/C2
The electric field of both q2 and q1 point toward q1. To balance the electric field, q3must be placed opposite of q2. Also, because the electric field of q2 points in the −xdirection, E2,x is negative.
Ex,tot = Ex − E2,x + E3,x
Ex,tot = k
rC
1
q21 −
k
rC
2
q22 +
kC
x2q3 = 0
r1 = r2
kC(q
r1
12− q2) +
kC
x2q3 = 0
(q1
r
−
12q2)
= −x
q23
x2 = (q
−
1
q
−3r1
q
2
2) = = 0.45 mm2
x = ±0.67 mm Since q3 must be opposite q2, and the position of q3 is positive, q2 must
be on the negative x-axis, so x = −0.67 mm
−(3.00 C)(1.5 mm)2
(−9.00 C − 6.00 C)
Section Five—Problem Bank V Ch. 17–11
V
10. q1 = 5.50 × 10−8 C
q2 = 1.10 × 10−8 C
q3 = 5.00 × 10−9 C
r1 = −5.00 × 10−7 m
r2 = 5.00 × 10−7 m
kC = 8.99 × 109 N •m2/C2
The electric field of both q2 and q1 point toward q1. To balance the electric field, q3must be on the side opposite q2. Also, because q2 points in the −x direction, E2,x isnegative.
Ex,tot = E1,x − E2,x + E3,x
Ex,tot = k
rC
1
q21 +
k
rC
2
q21 +
kC
x2q3 = 0
r1 = r2, so r12 = r2
2
k
rC
1
q21 −
k
rC
2
q22 =
−k
xC2q3
q1
r
−
12q2 =
−x
q23
x2 = −(q
q
1
3
q
r
2
1
)
2
= = 1.89 × 10−14 m2
x = −1.37 × 10−7 m
−(5.00 × 10−9 C)(−5.00 × 10−7 m)2
(−5.50 × 10−8 C − 1.10 × 10−8 C)
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Section Five—Problem Bank V Ch. 18–1
Chapter 18Electrical Energy and Capacitance
V
1. q = 1.45 × 10−8 C
E = 105 N/C
d = 290 m
PEelectric = −qEd = −(1.45 × 10−8 C)(−105 N/C)(290 m)
PEelectric = 4.42 × 10−4 J
Additional Practice 18A
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2. q1 = 6.4 × 10−8 C
q2 = −6.4 × 10−8 C
r = 0.95 m
PEelectric = kcq
r1q2
PEelectric =
PEelectric = −3.9 × 10−5 J
(8.99 × 109 N•m2/C2)(6.4 × 10−8 C)(−6.4 × 10−8 C)
0.95 m
3. q = 1.60 × 10−19 C
E = 3.0 × 106 N/C
d = 7.3 × 10−7 m
PEelectric = −qEd
PEelectric = −(1.60 × 10−19 C)(3.0 × 106 N/C)(7.3 × 10−7 m)
PEelectric = 3.5 × 10−19 J
4. q1 = −4.2 × 10−8 C
q2 = 6.3 × 10−8 C
PEelectric = −6.92 × 10−4 J
r = P
k
Ec
e
q
le
1
c
q
tr
2
ic
r =
r = 3.4 × 10−2 m = 0.34 cm
(8.99 × 109 N•m2/C2)(−4.2 × 10−8 C)(6.3 × 10−8 C)
−6.92 × 10−4 J
5. q1 = 1.6 × 10−8 C
q2 = 1.4 × 10−8 C
PEelectric = 2.1 × 10−6 J
r = P
k
Ec
e
q
le
1
c
q
tr
2
ic
r =
r = 9.5 × 10−1 m = 96 cm
(8.99 × 109 N•m2/C2)(1.6 × 10−8 C)(1.4 × 10−8 C)
2.1 × 10−6 J
6. q1 = −5.5 × 10−8 C
q2 = 7.7 × 10−8 C
PEelectric = −1.3 × 10−2 J
r = P
k
Ec
e
q
le
1
c
q
tr
2
ic
r =
r = 2.9 × 10−3 m = 2.9 mm
(8.99 × 109N•m2/C2)(−5.5 × 10−8 C)(7.7 × 10−8 C)
−1.3 × 10−2 J
10. r = 1.25 × 10−6 m
PEelectric = 6.3 × 10−6 J
q1 = q2
q2 = rPE
kel
c
ectric
q = rPEk
el
cectric =
q = 2.9 × 10−11 C
a. q1 =
b. q2 = −2.9 × 10−11 C
2.9 × 10−11 C
(1.25 × 10−6 m)(6.3 × 10−6 J)
(8.99 × 109 N•m2/C2)
Additional Practice 18B
1. ∆V = 114.0 V
r = 6.695 × 106 mq =
r∆kc
V = = 8.49 × 10−2 C
(6.695 × 106 m)(114.0 V)
8.99 × 109 N•m2/C2
Holt Physics Solution ManualV Ch. 18–2
V
7. PEelectric = −1.39 × 1011 J
E = 3.4 × 105 N/C
d = 7300 m
q = −PE
Ee
dlectric =
q = 56 C
−(−1.39 × 1011 J)(3.4 × 105 N/C)(7300 m)
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8. r = 1.25 × 10−3 m
PEelectric = 1.25 × 10−4 J
q1 = q2
q2 = rPE
kel
c
ectric
q = rPEk
el
cectric =
q = 4.17 nC
a. q1 =
b. q2 =
c. It is unlikely that the hat would stay on a friend’s head only by charge becausethe charge on the friend would most likely be different than the hat.
−4.2 nC
4.2 nC
(1.25 × 10−3 m)(1.25 × 10–4 J)
(8.99 × 109N•m2/C2
9. r = 9.4 × 10−4 m
PEelectric = 8.89 × 10−10 J
q1 = q2
q2 = rPE
kel
c
ectric
q =rPEk
el
cectric =
q = 9.6 × 10−12 C
a. q1 =
b. q2 = −9.6 × 10−12 C
9.6 × 10−12 C
(9.4 × 10−4 m)(8.89 × 10–10 J)
(8.99 × 109 N•m2/C2)
2. ∆V = 18600 V
r = 1991 mq = = = 4.12 × 10−3 C
(1991 m)(18600 V)8.99 × 109 N•m2/C2
r∆Vkc
Section Five—Problem Bank V Ch. 18–3
V
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4. V = 9.0 V
q = 9.4 × 10−8 Cr =
k
Vcq =
r = 93.4 m
(8.99 × 109 N•m2/C2)(9.4 × 10−8 C)
9.0 V
6. E = 3.0 × 106 N/C
∆d = 6.25 × 10−4 m
10. q1 = 1.60 × 10−19 C
q2 = −1.60 × 10−19 C
q3 = 1.60 × 10−19 C
q4 = −1.60 × 10−19 C
d = 2.82 × 10−10 m
r = r1 + r2 + r3 + r4 = (0.707)d = (0.707)(2.82 × 10−10 m)
r = 1.99 × 10−10 m
V = k
rc [q1 + q2 + q3 + q4] =
k
rc [2q1 + 2q2]
V = 8.9
1
9
.9
×9
1
×09
10
N−•1m0 m
2/C2
[2(1.60 × 10−19 C) + 2(−1.60 × 10−19 C)]
V = 0 V
5. q = 1.28 × 10−18 C
r = 3.95 × 10−2 m∆V =
kc
r
q =
∆V = 2.91 × 10−7 V
(8.99 × 109 N•m2/C2)(1.28 × 10−18 C)
3.95 × 10−2 m
∆V = –E∆d = −(3.0 × 106 N/C)(6.25 × 10−4 m)
∆V = −1.9 × 103 V
7. E = 95 N/C
∆d = 3.0 × 102 m
∆V = −E∆d = −(95 N/C)(3.0 × 102 m)
∆V = −2.8 × 104 V
8. q1 = 8(1.60 × 10−19 C) = 1.28 × 10−18 C
q2 = 1.60 × 1−19 C
d = 9.58 × 10−11 m
q = 105°
f = q2
= = 52.5°
r1 = dcosf = (9.58 × 10−11 m) cos (52.5°) = 5.83 × 10−11 m r2 = dsinf = (9.58 × 10−11 m) sin (52.5°) = 7.60 × 10−11 m
V = kc
r
q
1
1 + 2k
rc
2
q2 = kcq
r1
1 + 2
r
q
2
2V = (8.99 × 109 N•m2/C2) + V = 197 V + 37.9 V = 235 V
2(1.60 × 10−19 C)
7.60 × 10−11 m
1.28 × 10−18 C5.83 × 10−11 m
105°
2
9. q1 = 3.04 × 10−18 C
q2 = 5.60 × 10−18 C
r1 = 1.89 × 10−10 m
r2 = −9.30 × 10−11 m
V = kc
r
q
1
1 + k
rc
2
q2 = kcq
r1
1 + q
r2
2V = (8.99 × 109 N•m2/C2) 13.
.
8
0
9
4
××
1
1
0
0−
−
1
1
0
8
m
C +
−5
9
.
.
6
3
0
0
××
1
1
0
0
−
−
1
1
8
1C
m
V = 145 V − 541 V = −396 V
3. V = 1.0 V
q = –1.60 × 10–19 Cr =
k
Vcq =
r = 1.4 × 10−9 m
(8.99 × 109 N•m2/C2)(−1.60 × 10−19 C)
1.0 V
Holt Physics Solution ManualV Ch. 18–4
V
1. A = 4.8 × 10−3 m2
C = 1.8 × 10−5 F
k = 1
d = ke
C0A =
d = 2.4 × 10−9 m
(1)(8.85 × 10−12 C2/N•m2)(4.8 × 10−3 m2)
1.8 × 10−5 F
Additional Practice 18C
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2. A = 6.4 × 10−3 m2
C = 4.55 × 10−9 Fd =
eC0A =
d = 1.2 × 10−5 m
(8.85 × 10−12 C2/N•m2)(6.4 × 10−3 m2)
4.55 × 10−9 F
3. R = 6.4 × 106 m Csphere = k
R
c =
8.99
6
×.4
1
×09
1
N
06
•m
m2/C2 = 7.1 × 10−4 F
8. C = 1.0 × 10−6 F
Q = 3.0 × 10−2 C
9. C = 2.0 × 10−6 F
Q = 4.0 × 10−4 C
∆V = Q
C =
3
1
.
.
0
0
××
1
1
0
0
−
−
2
6C
F = 3.0 × 104 V = 30 kV
∆V = Q
C =
4
2
.
.
0
0
××
1
1
0
0
−
−
4
6C
F = 2.0 × 102 V
10. C = 5.0 × 10−5 F
Q = 6.0 × 10−4 C
∆V = Q
C =
6
5
.
.
0
0
××
1
1
0
0
−
−
4
5C
F = 12 V
4. R = 0.10 mCsphere =
k
R
c =
8.99 ×0
1
.
0
190
N
m
•m2/C2 = 1.1 × 10−11 F
5. C = 1.4 × 10−5 F
∆V = 1.5 × 104 V
Q = C∆V = (1.4 × 10−5 F)(1.5 104 V) =
PEelectric = 2
Q
C
2
= 2(1
(
.
0
4
.2
×1
1
C
0−)2
5 F) = 1.6 × 103 J
0.21 C
6. C = 1.0 × 10−9 F
∆V = 600 V
Q = C∆V = (1.0 × 10−9 F)(600 V) =
PEelectric = Q
2C
2
= (
2
6
(
.
1
0
.0
××10
1
−
0
7
−9C
F
)
)
2
= 1.8 × 10−4 J
6.0 × 10−7 C
7. C = 5.0 × 10−13 F
∆V = 1.5 V
Q = C∆V = (5 × 10−13 F)(1.5 V) = 7.5 × 10−13 C
Holt Physics Solution ManualV Ch. 19–1
19ChapterCurrent and Resistance
V
1. ∆Q = 76 C
∆t = 19 sI =
∆∆Q
t =
7
1
6
9
C
s = 4.0 A
Additional Practice 19A
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2. ∆Q = 1.14 × 10−4 C
∆t = 0.36 sI =
∆∆Q
t =
1.14
0
×.3
1
6
0
s
−4 C = 0.32 mA
3. ∆Q = 2.9 × 10−2 C
∆t = 11 sI =
∆∆Q
t =
2.9 ×11
10
s
−2 C = 2.6 mA
4. ∆Q = 98 C
I = 1.4 A∆t =
∆I
Q =
1
9
.
8
4
C
A = 70 s
1. R = 1.0 × 105 ΩI1 = 1.0 × 10−3 A
I2 = 1.5 × 10−2 A
∆V1 = I1R = (1.0 × 10−3 A) (1.0 × 105 Ω) =
∆V2 = I2R = (1.5 × 10−2 A) (1.0 × 105 Ω) = 1.5 × 103 V
1.0 × 102 V
5. ∆Q = 30.9 C
I = 9.65 A∆t =
∆I
Q =
9
3
.
0
6
.
5
9 C
A = 3.20 s
6. ∆Q = 56 C
I = 7.8 A∆t =
∆I
Q =
7
5
.
6
8
C
A = 7.2 s
7. ∆t = 15 s
I = 9.3 A∆Q = I∆t = (9.3 A) (15 s) = 1.4 × 102 C
8. ∆t = 2.0 min = 120 s
I = 3.0 A∆Q = I∆t = (3.0 A) (120 s) = 3.6 × 102 C
9. ∆t = 2.0 s
I = 0.70 A∆Q = I∆t = (0.70 A) (2.0 s) = 1.4 C
Additional Practice 19B
10. ∆t = 4.3 s
I = 5.6 A∆Q = I∆t = (5.6 A) (4.3 s) = 24 C
Section Five—Solution Manual V Ch. 19–2
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3. I = 4.66 A
R = 25.0 Ω∆V = IR = (4.66 A) (25.0 Ω) = 116 V
4. ∆V = 120 V
R = 12.2 ΩI =
∆R
V =
1
1
2
2
.
0
2
V
Ω = 9.84 A
5. ∆V = 650 V
R = 1.0 × 102 ΩI =
∆R
V =
1.0
6
×50
10
V2 Ω
= 6.5 A
6. R = 40.0 Ω∆V1 = 120 V
V2 = 240 V
I1 = ∆
R
V1 = 4
1
0
2
.
0
0
V
Ω =
I2 = ∆
R
V2 = 4
2
0
4
.
0
0
V
Ω = 6.00 A
3.00 A
Additional Practice 19C
7. I = 0.75 A
∆V = 120 VR =
∆I
V =
0
1
.
2
7
0
5
V
A = 1.6 × 102 Ω
8. I = 0.89 A
∆V = 5.00 × 102 VR =
∆I
V =
5.0
0
0
.8
×9
1
A
02 V = 5.6 × 102 Ω
9. I = 0.545 A
∆V = 120 VR =
∆I
V =
0
1
.5
2
4
0
5
V
A = 220 Ω
10. I = 0.65 A
∆V = 117 VR =
∆I
V =
0
1
.
1
6
7
5
V
A = 1.8 × 102 Ω
2. I = 0.75 A
R = 6.4 Ω∆V = IR = (0.75 A) (6.4 Ω) = 4.8 V
1. ∆V = 2.5 × 104 V
I = 20.0 A
P = I∆V = (20.0 A) (2.5 × 104 V) = 5.0 × 105 W
2. ∆V = 720 V
R = 0.30 ΩP =
(∆V
R
)2
= (7
0
2
.3
0
0
V
Ω)2
= 1.7 × 106 W
3. ∆V = 120 V
R1 = 144 ΩR2 = 240 Ω
P1 = (∆
R
V
1
)2
= (1
1
2
4
0
4
V
Ω)2
=
P2 = (∆
R
V
2
)2
= (1
2
2
4
0
0
V
Ω)2
=
The brighter bulb is the 60.0-W bulb, which has the 240−Ω resistance.
60.0 W
100 W
4. ∆V = 120 V
P = 1750 WR =
(∆V
P
)2
= (
1
1
7
2
5
0
0
V
W
)2
= 8.22 Ω
Holt Physics Solution ManualV Ch. 19–3
V
5. ∆V = 120 V
P = 650 WR =
(∆V
P
)2
= (1
6
2
5
0
0
V
W
)2
= 22.2 Ω
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6. ∆V = 120 V
P = 370 W R =
(∆V
P
)2
= (1
3
2
7
0
0
V
W
)2
= 38.9 Ω
7. P = 350 W
R = 75 ΩI2 =
R
P I =
R
P =
3
7
55
0ΩW
= 2.16 A
8. P = 230 W
R = 91 ΩI2 =
R
P I =
R
P =
2
9
31
0ΩW
= 1.59 A
9. I = 8.0 × 106 A
P = 6.0 × 1013 W ∆V =
P
I =
6
8
.0
.0
××1
1
0
0
1
6
3
A
W = 7.5 × 106 V
10. I = 16.3 A
P = 1.06 × 104 W ∆V =
P
I =
1.06
16
×.3
10
A
4 W = 6.50 × 102 V
1. ∆t = 1.0 h
Energy = 2.7 × 108 J P = En
∆er
t
gy = = 75 kW
2.7 × 108 J(1.0 h) (3.6 × 106 J/kW•h)
2. ∆t = 3.0 h
Energy = 4.86 × 108 J P = En
∆er
t
gy = = 45 kW
4.86 × 108 J(3.0 h) (3.6 × 106 J/kW•h)
3. P = 1200 W = 1.200 kW
Energy = 1.512 × 108 J ∆t = En
P
ergy = = 35.00 h
1.512 × 108 J(1.200 kW) (3.6 × 106 J/kW•h)
Additional Practice 19D
4. P = 600 W = 0.600 kW
Energy = 8.64 × 109 J ∆t = En
P
ergy = = 4.00 × 103 h
8.64 × 109 J(0.600 kW) (3.6 × 106 J/kW•h)
5. Cost of electricity = $0.0650/kW•h
Energy = 200.0 kW•h
Cost = (Energy) ($0.065/kW•h)
Cost = (200.0 kW•h) ($0.0650/kW•h) = $13.00
6. Cost of electricity = $0.078/kW•h
Final Meter Reading = 24422 kW•h
Previous Meter Reading = 24204 kW•h
Energy = Final Meter Reading – Previous Meter Reading
Energy = 24422 kW•h – 24204 kW•h = 218 kW•h
Cost = (Energy) ($0.078/kW•h)
Cost = (218 kW•h) ($0.078/kW•h) = $17.00
Section Five—Solution Manual V Ch. 19–4
V
7. ∆t = 8.0 h
P = 0.125 kWEnergy = P∆t = (0.125 kW) (8.0 h) = 1.0 kW•h
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8. ∆t = 24.0 h
P = 0.75 kWEnergy = P∆t = (0.75 kW) (24.0 h) (3.6 × 106 J/kW•h) = 6.5 × 107 J
9. ∆t = 10.0 min = 0.167 h
P = 0.55 kWEnergy = P∆t = (0.55 kW) (0.167 h) (3.6 × 106 J/kW•h) = 3.3 × 105 J
10. ∆t = 3.0 min
= 5.0 × 10–2 h
P = 0.85 kW
Energy = P∆t = (0.85 kW) (5.0 × 10–2h) (3.6 × 106 J/kW•h)
Energy = 1.5 × 105 J
Section Five—Problem Bank V Ch. 20–1
Chapter 20Circuits and Circuit Elements
V
1. ∆V = 12 V
R1 = 16 Ω
I = 0.42 A
R2 = ∆
I
V − R1 =
0
1
.4
2
2
V
A − 16 Ω = 29 Ω − 16 Ω = 13 Ω
Additional Practice 20A
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2. ∆V = 3.0 V
R1 = 24Ω
I = 0.062 A
R2 = ∆
I
V − R1 =
0
3
.0
.0
62
V
A − 24 Ω = 48 Ω − 24 Ω = 24 Ω
3. ∆V = 9.0 V
R1 = 9.1 Ω
I = 0.33 A
R2 = ∆
I
V − R1 =
0
9
.
.
3
0
3
V
A − 9.1 Ω = 27 Ω − 9 Ω = 18 Ω
4. 73 bulbs
Reach bulb = 3.0 Ω
Req = ΣReach bulb All bulbs have equal resistance.
Req = (73) (3.0 Ω) = 219 Ω
5. 25 speakers
Reach speaker = 12.0 Ω
Req = ΣReach speaker All speakers have equal resistance.
Req = (25)(12.0 Ω) = 3.0 × 102 Ω
6. 57 lights
Reach light = 2.0 Ω
Req = ΣReach light All lights have equal resistance.
Req = (57)(2.0 Ω) = 114 Ω
7. 4 speakers
Reach speaker = 4.1 Ω
∆V = 12 V
Req = ΩReach speaker All speakers have equal resistance.
Req = (4)(4.1 Ω) = 16.4 Ω
I = R
∆
e
V
q =
1
1
6
2
.4
V
Ω = 7.3 × 10−1 A
8. 10 bulbs
Reach bulb = 10 Ω
∆V = 115 V
Req = ΣReach bulb All bulbs have equal resistance.
Req = (10)(10 Ω) = 100 Ω
I = R
∆
e
V
q =
1
1
0
0
0
0
ΩV
= 1 A
9. R1 = 96 Ω
R2 = 48 Ω
R3 = 29 Ω
∆V = 115 V
Req = ΣR = R1 + R2 + R3 = 96 Ω + 48 Ω + 29 Ω = 173 Ω
I = R
∆
e
V
q =
1
1
7
1
3
5
ΩV
= 665 mA
1. ∆V = 3.0 V
R1 = 3.3 Ω
I = 1.41 A
∆V = IReq
I = R
∆
e
V
q =
∆R
V
1 +
∆R
V
2
∆R
V
2 = I −
∆R
V
1
R2 = = = [1.41 A
3.0
−V
0.91 A] = 6.0 Ω
3.0 V
1.41 A − 3
3
.
.
3
0
ΩV
∆V
I − ∆R
V
1
Holt Physics Solution ManualV Ch. 20–2
V
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10. R1 = 56 Ω
R2 = 82 Ω
R3 = 24 Ω
∆V = 9.0 V
Req = ΣR = R1 + R2 + R3 = 56Ω + 82Ω + 24Ω = 162Ω
I = R
∆
e
V
q =
1
9
6
.0
2
V
Ω = 55.6 mA
2. ∆V = 12 V
R1 = 56 Ω
I = 3.21 A
∆V = IReq
I = R
∆
e
V
q =
∆R
V
1 +
∆R
V
2
∆R
V
2 = I −
∆R
V
1
R2 = = = = 4.0 Ω12 V
[3.21 A − 0.21 A]
12 V
3.21 A − 5
1
6
2
ΩV
∆V
I − ∆R
V
1
3. ∆V = 1.5 V
R1 = 18 Ω
I = 0.103 A
∆V = IReq
I = R
∆
e
V
q =
∆R
V
1 +
∆R
V
2
∆R
V
2 = I −
∆R
V
1
R2 = = = = 75 Ω1.5 V
[0.103 A − 0.083 A]
1.5 V
0.103 A − 1
1
.
8
5
ΩV
∆V
I − ∆R
V
1
Additional Practice 20B
4. R1 = 39 Ω
R2 = 82 Ω
R3 = 12 Ω
R4 = 22 Ω
∆V = 3.0 V
R
1
eq =
R
1
1 +
R
1
2 +
R
1
3 +
R
1
4 =
39
1
Ω +
82
1
Ω +
12
1
Ω +
22
1
Ω
R
1
eq =
0
1
.0
Ω26 +
0
1
.0
Ω12 +
0
1
.0
Ω83 +
0
1
.0
Ω45 =
01.1Ω7
Req = 6.0 Ω
Section Five—Problem Bank V Ch. 20–3
V
Givens Solutions
Cop
yrig
ht ©
by H
olt,
Rin
ehar
t and
Win
ston
.All
right
s re
serv
ed.
5. R1 = 10.0 Ω
R2 = 12 Ω
R3 = 15 Ω
R4 = 18 Ω
∆V = 12 V
R
1
eq =
R
1
1 +
R
1
2 +
R
1
3 +
R
1
4 =
10.
1
0 Ω +
12
1
Ω +
15
1
Ω +
18
1
Ω
R
1
eq =
0
1
.1
Ω0
+ 0
1
.0
Ω83 +
0
1
.0
Ω67 +
0
1
.0
Ω56
Req = 3.3 Ω
6. R1 = 33 Ω
R2 = 39 Ω
R3 = 47 Ω
R4 = 68 Ω
V = 1.5 V
R
1
eq =
R
1
1 +
R
1
2 +
R
1
3 +
R
1
4 =
33
1
Ω +
39
1
Ω +
47
1
Ω +
68
1
Ω
R
1
eq =
0
1
.0
Ω30 +
0
1
.0
Ω26 +
0
1
.0
Ω21 +
0
1
.0
Ω15
Req = 11 Ω
7. ∆V = 120 V
R1 = 75 Ω
R2 = 91 Ω
I1 = R
V
1 I2 =
R
V
2
I1 = 1
7
2
5
0
ΩV
=
I2 = 1
9
2
1
0
ΩV
= 1.3 A
1.6 A
8. ∆V = 120 V
R1 = 82 Ω
R2 = 24 Ω
I1 = R
V
1 I2 =
R
V
2
I1 = 1
8
2
2
0
ΩV
=
I2 = 1
2
2
4
0
ΩV
= 5.0 A
1.5 A
9. ∆V = 120 V
R1 = 11 Ω
R2 = 36 Ω
I1 = R
V
1 I2 =
R
V
2
I1 = 1
1
2
1
0
ΩV
=
I2 = 1
3
2
6
0
ΩV
= 3.3 A
11 A
10. ∆V = 1.5 V
R1 = 3.3 Ω
R2 = 4.3 Ω
I1 = R
V
1 I2 =
R
V
2
I1 = 3
1
.
.
3
5
ΩV
=
I2 = 4
1
.
.
3
5
ΩV
= 0.35 A
0.45 A
Holt Physics Solution ManualV Ch. 20–4
V
Givens Solutions
Cop
yrig
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olt,
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ehar
t and
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ston
.All
right
s re
serv
ed.
2. I = 0.375 A
V = 9.00 V
R1 = 18.0 Ω
R2 = 3.0 Ω
R3 = 4.0 Ω
R4 = 8.00 Ω
R5 = 20.0 Ω
R6 = 22.0 Ω
R7 = 6.0 Ω
Series:
Group (a): Req,a = R2 + R3 = 3.0Ωeq + 4.0Ωeq = 7.0Ω
Group (b): Req,b = R5 + R6 = 20.0 Ω + 22.0 Ω = 42.0 Ω
Parallel:
Group (c): R
1
eq =
18.
1
0 Ω +
7.0
1
Ω =
0.
1
05
Ω56
+ 0
1
.1
Ω43
Req = 5.03 Ω
Group (d): R
1
eq =
42.
1
0 Ω +
6.0
1
Ω =
0.
1
02
Ω38
+ 0
1
.1
Ω67
Req = 5.24 Ω
Series:
Group (e): Req,e = Req,c + R4 + Req,d + R
= 5.03 Ω + 8.00 Ω + 5.24 Ω + R = 18.27 Ω + R
R = ∆
I
V − 18.27 Ω =
0
9
.
.
3
0
7
0
5
V
A − 18.27 Ω = 5.7 Ω
1. I = 0.680 A
V = 15.0 V
R1 = 15 Ω
R2 = 11 Ω
R3 = 6.0 Ω
R4 = 7.0 Ω
R5 = 12.0 Ω
Series:
Group (a): Req,a = R1 + R2 = 15 Ω + 11 Ω = 26 Ω
Group (b): Req,b = R3 + R4 = 6.0 Ω + 7.0 Ω = 13.0 Ω
Parallel:
Groups (a) + (b): R
1
eq =
26
1
Ω +
13.
1
0 Ω =
0
1
.0
Ω38 +
0
1
.0
Ω77 =
0
1
.1
Ω2
Req = 8.3 Ω
Series:
Req = 12.0 Ω + 8.3 Ω + R = 20.3 Ω + R
∆V = IReq = I(20.3 Ω + R)
R = ∆
I
V − 20.3 Ω =
0
1
.
5
6
.
8
0
0
V
A − 20.3 Ω = 1.8 Ω
Additional Practice 20C
Section Five—Problem Bank V Ch. 20–5
V
3. I = 0.185 A
V = 12.0 V
R1 = 2.0 Ω
R2 = 3.0 Ω
R3 = 3.0 Ω
R4 = 4.0 Ω
R5 = 8.0 Ω
R6 = 2.0 Ω
R7 = 3.0 Ω
R8 = 4.0 Ω
R9 = 4.0 Ω
R10 = 5.0 Ω
R11 = 6.0 Ω
Series:
Group (a): Req,a = R1 + R2 = 2.0 Ω + 3.0 Ω = 5.0 Ω
Group (b): Req,b = R3 + R4 = 3.0 Ω + 4.0 Ω = 7.0 Ω
Group (c): Req,c = R6 + R7 + R8 = 2.0 Ω + 3.0 Ω + 4.0 Ω = 9.0 Ω
Group (d): Req,d = R9 + R10 + R11 = 4.0 Ω + 5.0 Ω + 6.0 Ω = 15.0 Ω
Parallel:
Group (e): Re
1
q,e =
Re
1
q,a +
Re
1
q,b =
5.0
1
Ω +
7.0
1
Ω =
0
1
.2
Ω0
+ 0
1
.1
Ω4
Req,e = 2.9 Ω
Group (f): R
1
eq,f =
Re
1
q,c +
Re
1
q,d =
9.0
1
Ω +
15.
1
0 Ω =
0
1
.1
Ω11 +
0
1
.0
Ω67
Req,f = 5.6 Ω
Series:
Req = 4R5 + 2Req,e + Req,f + R = 4(8.0 Ω) + 2(2.9 Ω) + 5.6 Ω + R
Req = 32.0 Ω + 5.8 Ω + 5.6 Ω + R = 43.4 Ω + R
R = ∆
I
V − 43.4 Ω =
0
1
.
2
1
.
8
0
5
V
A − 43.4 Ω = 21.5 Ω
Givens Solutions
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yrig
ht ©
by H
olt,
Rin
ehar
t and
Win
ston
.All
right
s re
serv
ed.
4. each resistor = 10.0 Ω Parallel:
Group (a): Re
1
q,a =
10.
1
0 Ω +
10.
1
0 Ω =
0
1
.1
Ω00 +
0
1
.1
Ω00
Req,a = 5.00 Ω
Group (b): Re
1
q,b =
10.
1
0 Ω +
10.
1
0 Ω +
10.
1
0 Ω +
10.
1
0 Ω
Re
1
q,b =
0
1
.1
Ω00 +
0
1
.1
Ω00 +
0
1
.1
Ω00 +
0
1
.1
Ω00
Req,b = 2.50 Ω
Group (c): Req,c = 10.0 Ω + 10.0 Ω + 5.00 Ω + 2.50 Ω + 5.00 Ω + 2.50 Ω + 10.0 Ω + 10.0 Ω
Req,c = 55.0 Ω
5. each resistor = 2.0 Ω Series:
Group (a): Req,a = 2.0 Ω + 2.0 Ω = 4.0 Ω
Group (b): Req,b = 2.0 Ω + 2.0 Ω + 2.0 Ω = 6.0 Ω
Group (c): Req,c = 2.0 Ω + 2.0 Ω + 2.0 Ω = 2.0 Ω = 8.0 Ω
Parallel:
Group (d): Re
1
q,d =
2.0
1
Ω +
4.0
1
Ω +
6.0
1
Ω +
8.0
1
Ω =
0
1
.5
Ω0
+ 0
1
.2
Ω5
+ 0
1
.1
Ω7
+ 0
1
.1
Ω3
Req,d = 0.95 Ω
Series:
Req = 2.0 Ω + 2.0 Ω + 0.95 Ω = 5.0 Ω
Holt Physics Solution ManualV Ch. 20–6
V
6. each resistor = 10.0 Ω Parallel:
Group (a): Re
1
q,a =
10.
1
0 Ω +
10.
1
0 Ω +
10.
1
0 Ω =
0
1
.1
Ω00 +
0
1
.1
Ω00 +
0
1
.1
Ω00
Req,a = 3.33 Ω
Series:
Req = 4(10.0 Ω) + 3.33 Ω + 2(10.0 Ω) + 3.33 Ω + 2(10.0 Ω) + 3.33 Ω + 3(10.0 Ω)
Req = 1.20 × 102 Ω
Givens Solutions
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yrig
ht ©
by H
olt,
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ehar
t and
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ston
.All
right
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serv
ed.
7. ∆Vtot = 12.0 V
R1 = 3.0 Ω
R2 = 5.0 Ω
R3 = 2.0 Ω
R4 = 4.0 Ω
R5 = 5.0 Ω
R6 = 6.0 Ω
R7 = 1.5 Ω
Parallel:
Group (a): Re
1
q,a =
R
1
2 +
R
1
3 =
5.0
1
Ω +
2.0
1
Ω =
0
1
.2
Ω0
+ 0
1
.5
Ω0
Req,a = 1.4 Ω
Group (b): Re
1
q,b =
R
1
5 +
R
1
6 =
5.0
1
Ω +
6.0
1
Ω =
0
1
.2
Ω0
+ 0
1
.1
Ω7
Req,b = 2.7 Ω
Series:
Group (c): Req,c = R1 + Req,a + R4 + Req,b + R7
Req,c = 3.0 Ω + 1.4 Ω + 4.0 Ω + 2.7 Ω + 1.5 Ω = 13 Ω
I = ∆R
V
eq
t
,
o
c
t = 1
1
2
3
.0
ΩV
= 0.92 A
8. ∆Vtot = 15.0 V
R1 = 5.0 Ω
R2 = 5.0 Ω
R3 = 5.0 Ω
R4 = 3.0 Ω
Parallel:
Group (a): Re
1
q,a =
R
1
1 +
R
1
2 +
R
1
3 =
5.0
1
Ω +
5.0
1
Ω +
5.0
1
Ω =
0
1
.2
Ω0
3Req,a = 1.7 Ω
Series:
Req = 2R4 + 2Req,a = (2)(3.0 Ω) + (2)(1.7 Ω) = 9.4 Ω
I = ∆R
V
e
t
q
ot = 1
9
5
.4
.0
ΩV
= 1.6 A
Section Five—Problem Bank V Ch. 20–7
V
Cop
yrig
ht ©
by H
olt,
Rin
ehar
t and
Win
ston
.All
right
s re
serv
ed.
9. ∆Vtot = 24.0 V
R1 = 4.0 Ω
R2 = 4.0 Ω
R3 = 4.0 Ω
R4 = 3.0 Ω
R5 = 1.0 Ω
R6 = 2.0 Ω
R7 = 4.0 Ω
R8 = 2.0 Ω
R9 = 4.0 Ω
R10 = 2.0 Ω
R11 = 2.0 Ω
Series:
Group (a): Req,a = R6 + R7 + R8 = 2.0 Ω + 4.0 Ω + 2.0 Ω = 8.0 Ω
Group (b): Req,b = R9 + R10 = 4.0 Ω + 2.0 Ω = 6.0 Ω
Parrallel:
Group (c): Re
1
q,c =
R
1
1 +
R
1
2 =
4.0
1
Ω +
4.0
1
Ω =
0
1
.2
Ω5
+ 0
1
.2
Ω5
Req,c = 2.0 Ω
Group (d): Re
1
q,d =
Re
1
q,a +
Re
1
q,b =
8.0
1
Ω +
6.0
1
Ω =
0
1
.1
Ω3
+ 0
1
.1
Ω7
Req,d = 3.4 Ω
Series:
Group (e): Req,e = Req,c + R3 + R4 + R5 + Req,d + R11
Req,e = 2.0 Ω + 4.0 Ω + 3.0 Ω + 1.0 Ω + 3.4 Ω + 2.0 Ω = 15 Ω
I = ∆R
V
eq
t
,
o
e
t = 2
1
4
5
.0
ΩV
= 1.6 A
10. ∆V = 24.0 V
R1 = 4.0 Ω
R2 = 8.0 Ω
R3 = 2.0 Ω
R4 = 5.0 Ω
R5 = 3.0 Ω
R6 = 2.0 Ω
R7 = 3.0 Ω
Parallel:
Group (a): Re
1
q,a =
R
1
1 +
R
1
2 =
4.0
1
Ω +
8.0
1
Ω =
0
1
.2
Ω5
+ 0
1
.1
Ω3
Req,a = 2.6 Ω
Group (b): Re
1
q,b =
R
1
6 +
R
1
7 =
2.0
1
Ω +
3.0
1
Ω =
0
1
.5
Ω0
+ 0
1
.3
Ω3
Req,b = 1.2 Ω
Series:
Req = 2Req,a + R3 + R4 + R5 + Req,b = (2)(2.7 Ω) + 2.0 Ω + 5.0 Ω + 1.2 Ω + 3.0 Ω
Req = 17 Ω
I = ∆R
V
e
t
q
ot = 2
1
4
7
.0
ΩV
= 1.4 A
Givens Solutions
1. ∆Vtot = 12.0 V
R1 = 3.0 Ω
R2 = 10.0 Ω
R3 = 10.0 Ω
R4 = 10.0 Ω
R5 = 4.0 Ω
Parallel:
Re
1
q,a =
R
1
2 +
R
1
3 +
R
1
4 =
10.0
1
Ω +
10.
1
0 Ω +
10.
1
0 Ω =
0
1
.1
Ω00 +
0
1
.1
Ω00 +
0
1
.1
Ω00
Req,a = 3.3 Ω
Series:
Req,b = R1 + Req,a + R5 = 3.0 Ω + 3.3 Ω + 4.0 Ω = 10 Ω
I = ∆R
V
eq
t
,
o
b
t = 1
1
2
0
.0
ΩV
=
∆V = IR5
∆V = (1.2 A)(4.0 Ω) = 4.8 V
1.2 A
Additional Practice 20D
3. ∆Vtot = 9.0 V
R1 = 5.0 Ω
R2 = 4.0 Ω
R3 = 7.0 Ω
R4 = 6.0 Ω
R5 = 3.0 Ω
R6 = 2.0 Ω
Parallel:
Re
1
q,a =
R
1
3 +
R
1
4 =
7.0
1
Ω +
6.0
1
Ω =
0
1
.1
Ω4
+ 0
1
.1
Ω7
Req,a = 3.2 Ω
Series:
Req = R1 + R2 + Req,a + R5 + R6 = 5.0 Ω + 4.0 Ω + 3.2 Ω + 3.0 Ω + 2.0 Ω = 17 Ω
I = ∆R
V
e
t
q
ot = 9
1
.
7
0
ΩV
= 0.52 A
∆V = I Req,a = (0.52 A)(3.2 Ω) =
I = ∆R
V
4 =
6
1
.
.
0
7
ΩV
= 0.28 A
1.7 V
Givens Solutions
4. ∆Vtot = 9.0 V
R1 = 7.0 Ω
R2 = 5.0 Ω
R3 = 4.0 Ω
R4 = 3.0 Ω
R5 = 2.0 Ω
R6 = 10.0 Ω
R7 = 6.0 Ω
Series:
Req,a = R2 + R3 + R4 = 5.0 Ω + 4.0 Ω + 3.0 Ω = 12.0 Ω
Req,b = R5 + R6 + R7 = 2.0 Ω + 10.0 Ω + 6.0 Ω = 18.0 Ω
Parallel:
Re
1
q,b =
Re
1
q,a +
Re
1
q,b =
12.
1
0 Ω +
18.
1
0 Ω =
0
1
.0
Ω83 +
0
1
.0
Ω56
Req,b = 7.2 Ω
Series:
Req,d = R1 + Req,c = 7.0 Ω + 7.2 Ω = 14 Ω
I = ∆R
V
e
t
q
ot = 9
1
.
4
0
ΩV
= 0.63 A
∆V = IReq,c = (0.63 A)(7.2 Ω) = 4.5 V
I = ∆R
V
6 =
1
4
8
.
.
5
0
V
Ω =
∆V = IR6 = (0.25 A)(10.0 Ω) = 2.5 V
0.25 A
Holt Physics Solution ManualV Ch. 20–8
V
Cop
yrig
ht ©
by H
olt,
Rin
ehar
t and
Win
ston
.All
right
s re
serv
ed.
2. ∆Vtot = 1.5 V
R1 = 6.0 Ω
R2 = 2.0 Ω
R3 = 4.0 Ω
R4 = 8.0 Ω
R5 = 9.0 Ω
Parallel:
Re
1
q,a =
R
1
4 +
R
1
5 =
8.0
1
Ω +
9.0
1
Ω =
0
1
.1
Ω25 +
0.1
Ω11
Req,a = 4.2 Ω
Series:
Req = R1 + R2 + R3 + Req,a = 6.0 Ω + 2.0 Ω + 4.0 Ω + 4.2 Ω = 16 Ω
I = ∆R
V
e
t
q
ot = 1
1
.
6
5
ΩV
= 0.094 A
∆V = IReq,a = (0.094 A)(4.2 Ω) =
I = ∆R
V
5 =
0
9
.
.
3
0
9
ΩV
= 0.043 A
0.39 V
Section Five—Problem Bank V Ch. 20–9
V
Cop
yrig
ht ©
by H
olt,
Rin
ehar
t and
Win
ston
.All
right
s re
serv
ed.
5. ∆Vtot = 3.0 V
R1 = 2.0 Ω
R2 = 4.0 Ω
R3 = 6.0 Ω
R4 = 1.0 Ω
R5 = 3.0 Ω
R6 = 5.0 Ω
Series:
Req,a = R1 + R2 + R3 = 2.0 Ω + 4.0 Ω + 6.0 Ω = 12 Ω
Req,b = R4 + R5 = 1.0 Ω + 3.0 Ω = 4.0 Ω
Parallel:
Re
1
q,c =
Re
1
q,a +
Re
1
q,b +
R
1
eq =
12
1
Ω +
4.0
1
Ω +
5.0
1
Ω =
0
1
.0
Ω83 +
0
1
.2
Ω5
+ 0
1
.2
Ω0
Req,c = 1.9 Ω
I = ∆R
V
eq
t
,
o
c
t = 1
3
.
.
9
0
ΩV
= 1.6 A
I = R
∆
e
V
q,a =
3
1
.
2
0
ΩV
=
∆V = IR2 = (0.25 A)(4.0 Ω) = 1.0 V
0.25 A
Givens Solutions
6. ∆Vtot = 3.0 V
R1 = 2.0 Ω
R2 = 3.0 Ω
R3 = 2.0 Ω
R4 = 4.0 Ω
R5 = 2.0 Ω
R6 = 2.0 Ω
Series:
Req,a = R1 + R2 = 2.0 Ω + 3.0 Ω = 5.0 Ω
Req,b = R3 + R4 = 2.0 Ω + 4.0 Ω = 6.0 Ω
Parallel:
Re
1
q,c =
Re
1
q,a +
Re
1
q,b =
5.0
1
Ω +
6.0
1
Ω =
0
1
.2
Ω0
+ 0
1
.1
Ω7
Req,c = 2.7 Ω
Re
1
q,d =
R
1
5 +
R
1
6 =
2.0
1
Ω +
2.0
1
Ω
Req,d = 1.0 Ω
Series:
Req = Req,c + Req,d = 2.7 Ω + 1.0 Ω = 3.7 Ω
I = ∆R
V
e
t
q
ot = 3
3
.
.
7
0
ΩV
= 0.81 A
∆V = IReq,c = (0.81 A)(2.7 Ω) = 2.2 V
I = R
∆
e
V
q,a =
5
2
.
.
0
2
ΩV
=
∆V = IR2 = (0.44 A)(3.0 Ω) = 1.3 V
0.44 A
8. ∆Vtot = 12.0 V
R1 = 4.0 Ω
R2 = 5.0 Ω
R3 = 2.0 Ω
R4 = 3.0 Ω
R5 = 7.0 Ω
R6 = 3.0 Ω
Parallel:
Re
1
q,a =
R
1
3 +
R
1
4 +
R
1
5 =
2.0
1
Ω +
3.0
1
Ω +
7.0
1
Ω =
0
1
.5
Ω0
+ 0
1
.3
Ω3
+ 0
1
.1
Ω4
Req,a = 1.0 Ω
Re
1
q,b =
R
1
1 +
R
1
2 =
4.0
1
Ω +
5.0
1
Ω =
0
1
.2
Ω5
+ 0
1
.2
Ω0
Req,b = 2.2 Ω
Series:
Req,c = Req,a + Req,b + R4 = 1.0 Ω + 2.2 Ω + 3.0 Ω = 6.2 Ω
I = ∆R
V
eq
t
,
o
c
t = 1
6
2
.2
.0
ΩV
= 1.9 A
∆V = IReq,a = (1.9 A)(1.0 Ω) = 1.9 V
Givens Solutions
Holt Physics Solution ManualV Ch. 20–10
V
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7. ∆Vtot = 12.0 V
R1 = 5.0 Ω
R2 = 5.0 Ω
R3 = 5.0 Ω
R4 = 2.0 Ω
R5 = 5.0 Ω
R6 = 5.0 Ω
R7 = 5.0 Ω
Parallel:
Re
1
q,a =
R
1
1 +
R
1
2 =
5.0
1
Ω +
5.0
1
Ω =
0
1
.2
Ω0
+ 0
1
.2
Ω0
Req,a = 2.5 Ω
Series:
Req,b = R4 + R5 = 2.0 Ω + 5.0 Ω = 7.0 Ω
Req,c = R6 + R7 = 5.0 Ω + 5.0 Ω = 10 Ω
Parallel:
Re
1
q,d =
Re
1
q,b +
Re
1
q,c =
7.0
1
Ω +
10
1
Ω =
0
1
.1
Ω4
+ 0
1
.1
Ω0
Req,d = 4.2 Ω
Series:
Req = Req,a + R3 + Req,d = 2.5 Ω + 5.0 Ω + 4.2 Ω = 12 Ω
I = ∆R
V
e
t
q
ot = 1
1
2
2
.0
ΩV
= 1.0 A
∆V = IReq,d = (1.0 A)(4.2 Ω) =
I = R
∆
e
V
q,b =
7
4
.
.
0
2
ΩV
=
∆V = IR4 = (0.6 A)(2.0 Ω) = 1.2 V
0.6 A
4.2 V
I = ∆R
V
5 =
7
1
.
.
0
9
ΩV
= 0.27 A
Section Five—Problem Bank V Ch. 20–11
V
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9. ∆Vtot = 1.5 V
R1 = 4.0 Ω
R2 = 5.0 Ω
R3 = 6.0 Ω
R4 = 4.0 Ω
R5 = 6.0 Ω
R6 = 12.0 Ω
R7 = 6.0 Ω
R8 = 3.0 Ω
R9 = 3.0 Ω
R10 = 3.0 Ω
Series:
Req,a = R3 + R4 = 6.0 Ω + 4.0 Ω = 10 Ω
Req,b = R5 + R6 + R7 = 6.0 Ω + 12.0 Ω + 6.0 Ω = 24 Ω
Req,c = R8 + R9 = 3.0 Ω + 3.0 Ω = 6.0 Ω
Parallel:
Re
1
q,d =
Re
1
q,a +
Re
1
q,b +
Re
1
q,c =
0
1
.1
Ω0
+ 0
1
.0
Ω42 +
0
1
.1
Ω7
Req,d = 3.2 Ω
Series:
Req = R1 + R2 + Req,d + R10 = 4.0 Ω + 5.0 Ω + 3.2 Ω + 3.0 Ω = 15 Ω
I = ∆R
V
e
t
q
ot = 1
1
.
5
5
ΩV
= 0.10 A
∆V = IReq,d = (0.10 A)(3.2 Ω) = 0.32 V
I = R
∆
e
V
q,b =
0
2
.3
4
2
ΩV
=
∆V = IR6 = (0.013 A)(12 Ω) = 0.16 V
0.013 A
10. ∆Vtot = 12.0 V
R1 = 5.0 Ω
R2 = 6.0 Ω
R3 = 15.0 Ω
R4 = 30.0 Ω
Parallel:
Re
1
q,a =
R
1
3 +
R
1
4 =
15.
1
0 Ω +
30.
1
0 Ω =
0
1
.0
Ω67 +
0
1
.0
Ω33
Req,a = 10 Ω
Series:
Req = R1 + R2 + Req,a = 5.0 Ω + 6.0 Ω + 10.0 Ω = 21 Ω
I = ∆R
V
e
t
q
ot = 1
2
2
1
.0
ΩV
= 0.57 A
∆V = IReq,a = (0.57 A)(10 Ω) = 5.7 V
I = ∆R
V
3 =
1
5
5
.
.
7
0
V
Ω = 0.38 A
Holt Physics Solution ManualV Ch. 21–1
21ChapterMagnetism
V
1. q = 1.60 × 10−19 C
v = 1.2 × 106 m/s
Fmagnetic = 1.2 × 10−17 N
B = Fma
q
g
v
netic = = 60.3 × 10−5 T1.2 × 10−17 N
(1.60 × 10−19 C)(1.2 × 106 m/s)
Additional Practice 21A
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2. q = 1.60 × 10−19 C
v = 3.9 × 106 m/s
Fmagnetic = 1.9 × 10−22 N
B = Fma
q
g
v
netic = = 3.0 × 10−10 T1.9 × 10−22 N
(1.60 × 10−19 C)(3.9 × 106 m/s)
3. q = 1.60 × 10−19 C
v = 7.8 × 106 m/s
Fmagnetic = 3.7 × 10−13 N
B = Fma
q
g
v
netic = = 03.0 T3.7 × 10−13 N
(1.60 × 10−19 C)(7.8 × 106 m/s)
4. q = 1.60 × 10−19 C
r = 1.0 km = 1.0 × 103 m
m = 1.67 × 10−27 kg
B = 3.3 T
m
r
v2
= qv B
v = q
m
r B = = 3.2 × 1011 m/s
(1.60 × 10−19 C)(1.0 × 103 m)(3.3 T)
1.67 × 10−27 kg
5. q = 1.60 × 10−19 C
B = 5.0 × 10−5 T
Fmagnetic = 6.1 × 10−17 N
v = Fm
q
ag
B
netic = = 7.6 × 106 m/s6.1 × 10−17 N
(1.60 × 10−19 C)(5.0 × 10−5 T)
6. B = 1 × 10−8 T
q = 1.60 × 10−19 C
Fmagnetic = 3.2 × 10−22 N
v = Fm
q
ag
B
netic = = 2 × 105 m/s3.2 × 10−22 N
(1.60 × 10−19 C)(1 × 10−8 T)
7. q = 1.60 × 10−19 C
v = 6 × 106 m/s to the right
q = 45°
B = 3 × 10−4 T upward
Fmagnetic = qvBsin q = (1.60 × 10−19 C)(6 × 106 m/s)(3 × 10−4 T)sin 45°
Fmagnetic = 2 × 10−16 N
8. q = 1.60 × 10−19 C
B = 0.8 T
v = 3.0 × 107 m/s
Fmagnetic = qvB = (1.60 × 10−19 C)(3.0 × 107 m/s)(0.8 T) = 4 × 10−12 N
9. q = 1.60 × 10−19 C
v = 2.2 × 106 m/s
B = 1.1 × 10−2 T
Fmagnetic = qvB = (1.60 × 10−19 C)(2.2 × 106 m/s)(1.1 × 10−2 T) = 3.9 × 10−15 N
Section Five—Solution Manual V Ch. 21–2
V
Fmagnetic = qvB = (1.60 × 10−19 C)(9.3 × 105 m/s)(4.1 × 10−10 T) = 6.1 × 10−23 N
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10. q = 1.60 × 10−19 C
v = 9.3 × 105 m/s
B = 4.1 × 10−10 T
1. B = 4.6 × 10−4 T
Fmagnetic = 2.9 × 10−3 N
T = 10.0 A
l = Fma
Bg
Inetic = = 0.63 m
2.9 × 10−3 N(4.6 × 10−4 T)(10.0 A)
2. I = 1 A
B = 2.8 × 10−5 T
Fmagnetic = 5.6 × 10−5 N
l = Fma
Bg
Inetic =
(2.8
5
×.6
1
×0−
150−
T
5
)(
N
1 A) = 2 m
3. l = 12 m
I = 12 A
Fmagnetic = 7.3 × 10−2 N
B = Fma
I
g
l
netic = (
7
1
.
2
3
A
×)
1
(
0
1
−
2
2
m
N
) = 5.1 × 10−4 T
4. Fmagnetic = 7.8 × 105 N
l = 12 m = 1.2 × 104 m
I = 1.8 × 104 A
B = Fma
I
g
l
netic = = 3.6 × 10−3 T7.8 × 105 N
(1.8 × 104 A)(1.2 × 104 m)
5. I = 14.32 A
l = 15.0 cm = 0.150 m
Fmagnetic = 6.62 × 10−4 N
B = Fma
I
g
l
netic = (14
6
.3
.6
2
2
A
×)(
1
0
0
.
−
1
4
50
N
m) = 3.08 × 10−4 T
Additional Practice 21B
6. l = 10 m
m = 75 kg
B = 4.8 × 10−4 T
g = 9.81 m/s2
mg = BIl
I = mBl
g = = 1.5 × 105 A(75 kg)(9.81 m/s2)(4.8 × 10−4 T)(10 m)
7. l = 1.0 mFmagnetic = 9.1 × 10−5 N
B = 1.3 × 10−4 T
I = Fma
B
g
l
netic = = 0.7 A9.1 × 10−5 N
(1.3 × 10−4 T)(1.0 m)
8. I = 1.5 × 103 A
l = 15 km = 1.4 × 104 m
q = 45°
B = 2.3 × 10−5 T
Fmagnetic = Bcos qI l = (2.3 × 10−5 T)cos 45°(1.5 × 103 A)(1.5 × 104 m)
Fmagnetic = 3.7 × 102 N
9. I = 14 A
l = 2 m
B = 3.6 × 10−4 T
Fmagnetic = BI l = (3.6 × 10−4 T)(14 A)(2 m) = 1 × 10−2 N
10. I = 0.5 A
l = 5 cm = 5 × 10−2 m
B = 1.3 × 10−4 T
Fmagnetic = BI l = (1.3 × 10−4 T)(0.5 A)(5 × 10−2 m) = 3 × 10−6 N
Holt Physics Solution ManualV Ch. 22–1
22ChapterInduction and Alternating Current
V
1. N = 540 turns
A = 0.016 m2
qi = 0°
qf = 90.0°
∆t = 0.05 s
emf = 3.0 V
B = −N
em
A∆f
c
∆o
t
sq =
B =
B = 1.7 × 10−2 T
(3.0 V)(0.05 s)−(540)(0.016 m2)[cos 90.0° − cos 0°]
emf ∆t−NA[cosqf − cosqi]
Additional Practice 22A
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2. N = 320 turns
A = 0.068 m2
qi = 0°
qf = 90°
∆t = 0.25 s
emf = 4.0 V
B = −N
em
A∆f
c
∆o
t
sq =
B =
B = 4.6 × 10−2 T
(4.0 V)(0.25 s)−(320)(0.068 m2)(cos 90.0° − cos 0°)
emf ∆t−NA(cosqf − cosqi)
3. N = 628 turns
A = 0.93 m2
qi = 0°
∆t = 0.30 s
qf = 30.0°
emf = 62 V
B = =
B =
B = 0.24 T
(62 V)(0.30 s)−(628)(0.93 m2)(cos 30.0° − cos 0°)
emf ∆t−NA(cosqf − cosqi)
emf ∆t−NA∆cosq
4. N = 550 turns
A = 5.0 × 10−5 m2
∆B = 2.5 × 10−4 T
∆t = 2.1 × 10−5 s
q = 0°
emf = −NA∆
∆B
t
cos q
emf =
emf = 0.33 V
−(550)(5.0 × 10−5 m2)(2.5 × 10−4 T)(cos 0°)
2.1 × 10−5 s
5. N = 220 turns
A = 6.0 × 10−6 m2
∆B = 9.7 × 10−4 T
∆t = 1.7 × 10−6 s
q = 0°
emf = −NA∆
∆B
t
cos q
emf =
emf = 0.75 V
−(220)(6.0 × 10−6 m2)(9.7 × 10−4 T)(cos 0°)
1.7 × 10−6 s
Section Five—Solution Manual V Ch. 22–2
V
emf = −NA∆
∆B
t
cos q
emf =
emf = 0.11 V
−(148)(1.25 × 10−8 m2)(5.2 × 10−4 T)(cos 0°)
8.5 × 10−9 s
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6. N = 148 turns
A = 1.25 × 10−8 m2
q = 0°
∆B = 5.2 × 10−4 T
∆t = 8.5 × 10−9 s
7. emf = 220 V
R = 120 ΩI =
em
R
f =
1
2
2
2
0
0
ΩV
= 1.8 A
8. N = 180 turns
A = 5.0 × 10−5 m2
∆B = 5.2 × 10−4 T
q = 0°
∆t = 1.9 × 10−5 s
R = 1.0 × 102Ω
emf = −NA∆
∆B
t
cos q
emf =
emf =
I = em
R
f =
1.0
0.
×25
10
V2Ω
= 2.5 × 10−3 A = 25 mA
0.25 V
−(180)(5.0 × 10−5 m2)(5.2 × 10−4 T)(cos 0°)
1.9 × 10−5 s
9. N = 246 turns
A = 0.40 m2
q = 0°
Bi = 0.237 T
Bf = 0.320 T
∆t = 0.9 s
emf = 9.1 V
∆t = −NA∆
em
B
f
cos q =
−NA[Bf
e
−m
B
fi] cos q
∆t =
∆t = 0.90 s
−(246)(0.40 m2)[0.320 T − 0.237 T](cos 0°)
9.1 V
1. N = 220 turns
A = 0.080 m2
B = 4.8 × 10−3 T
maximum emf = 12 V
w = maxi
N
m
A
um
B
emf =
w = 1.4 × 102 rad/s
12 V(220)(0.080 m2)(4.8 × 10−3 T)
2. N = 140 turns
D = 0.33 m
B = 9.3 × 10−2 T
maximum emf = 150 V
w = maxi
N
m
A
um
B
emf
A = πD
22
= π0.3
2
3 m2
= 8.6 × 10−2 m2
Additional Practice 22B
10. N = 785 turns
A = 7.3 × 10−2 m2
∆B = 6.9 × 10−3 T
emf = 2.8 V
q = 0°
∆t = −NA∆
em
B
f
cos q =
∆t =
∆tearthquake = (0.14 s/oscillation)(120 oscillations)
∆t =
The earthquake lasted for 16.8 s.
16.8 s
0.14 s/oscillation
−(785)(7.3 × 10−2 m2)(6.9 × 10−3 T)(cos 0°)
2.8 V
Holt Physics Solutions ManualV Ch. 22–3
w = = 130 rad/s150 V
(140)(8.6 × 10−2 m2)(9.3 × 10−2 T)
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VV
3. N = 195 turns
A = 0.052 m2
B = 3.2 × 10−3 T
Imax = 1.2 A
R = 16 Ω
maximum emf = Imax R = (1.2 A)(16 Ω) =
w = = 590 rad/s19 V
(195)(0.052 m2)(3.2 × 10−3 T)
19 V
4. N = 385 turns
A = 0.38 m2
B = 9.4 × 10−3 T
f = 45 Hz
w = 2 π fmaximum emf = NABw = NAB2πf
maximum emf = (385)(0.38 m2)(9.4 × 10−3 T)(2π)(45 Hz)
maximum emf = 390 V
5. Imax = 14 A
R = 5 Ωmaximum emf = Imax R = (14 A)(5 Ω) = 70 V
6. N = 119 turns
A = 4.9 × 10−2m2
B = 9.4 × 10−3 T
w = 345 rad/s
maximum emf = NABwmaximum emf = (119)(4.9 × 10−2 m2)(9.4 × 10−3 T)(345 rad/s)
maximum emf = 19 V
7. maximum emf = 40 VR = 8 Ω Imax =
maxim
R
um emf =
4
8
0
ΩV
= 5 A
8. N = 425 turns
A = 2.16 × 10−2m2
B = 3.9 × 10−2 T
f = 33 Hz
R = 25 Ω
w = 2 π fmaximum emf = NABw = NAB2πf
maximum emf = (425)(2.16 × 10−2 m2)(3.9 × 10−2 T)(2π)(33 Hz)
maximum emf =
Imax = maxim
R
um emf =
2
7
5
4
ΩV
= 3.0 × 101 A
74 V
9. A = 1.20 × 10−2m2
B = 6.0 × 10−2 T
w = 393 rad/s
maximum emf = 213 V
N = maxim
AB
u
wm emf =
N = 750 turns
213 V(1.2 × 10−2 m2)(6.0 × 10−2 T)(393 rad/s)
10. A = 0.60 m2
B = 0.012 T
f = 44 Hz
maximum emf = 320 V
N = maxim
AB
u
wm emf =
max
A
im
B
u
2
m
πf
emf
N =
N = 160 turns
320 V(0.60 m2)(0.012 T)(2π)(44 Hz)
Section One—Pupil’s Edition Solutions V Ch. 22–4
1. ∆Vrms = 320 V
R = 100 Ω∆Vmax =
∆0.
V
7r
0m
7s =
3
0
2
.7
0
0
V
7 =
Irms = ∆V
Rrms =
1
3
0
2
0
0
ΩV
=
Imax = 0
I
.r
7m
0s
7 =
0
3
.7
A
07 = 4 A
3 A
450 V
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V
Additional Practice 22C
Givens Solutions
2. Irms = 1.3 AImax =
0
I
.r
7m
0s
7 =
0
1
.
.
7
3
0
A
7 = 1.8 A
3. Irms = 2.5 A
∆Vrms = 2.2 × 104 VImax =
0
I
.r
7m
0s
7 =
0
2
.
.
7
5
0
A
7 =
R = ∆
I
V
rm
rm
s
s = 2.2
2
×.5
1
A
04 V = 8.8 × 104Ω = 88 kΩ
3.5 A
4. ∆Vrms = 220 V
Irms = 1.7 A∆Vmax =
∆0.
V
7r
0m
7s =
2
0
2
.7
0
0
V
7 =
Imax = 0
I
.r
7m
0s
7 =
0
1
.
.
7
7
0
A
7 = 2.4 A
311 V
5. Imax = 1.2 A
∆Vmax = 211 VIrms = 0.707 Imax = 0.707(1.2 A) =
∆Vrms = 0.707 Vmax = 0.707(211 V) =
R = ∆
I
V
rm
rm
s
s = 0
1
.
4
8
9
5
V
A = 175 Ω
149 V
0.85 A
6. Vmax = 170 V ∆Vrms = 0.707 Vmax = 0.707(170 V) = 120 V
7. ∆Vrms = 115 V
R = 50.0 ΩIrms =
∆V
Rrms =
5
1
0
1
.
5
0
V
Ω =
Imax = 0
I
.r
7m
0s
7 =
2
0
.
.
3
7
0
07
A = 3.25 A
2.30 A
8. Irms = 2.1 A
R = 16 k Ω = 1.6 × 104 ΩImax =
0
I
.r
7m
0s
7 =
0
2
.
.
7
1
0
A
7 =
P = (Irms)2R = (2.1 A)2(1.6 × 104 W) = 7 × 104 Ω P = 70 kW
3.0 A
9. ∆Vrms = 1.56 × 104 V
Irms = 1.3 A
R = 1.2 × 104 Ω
∆Vmax = ∆0
V
.7
r
0
m
7
s =
1.56
0.
×70
1
7
04 V =
Imax = 0
I
.r
7m
0s
7 =
0
1
.
.
7
3
0
A
7 =
P = (Irms)2R = (1.3 A)2(1.2 × 104 Ω) = 2.0 × 104 W
1.8 A
2.2 × 104 V = 22kV
Holt Physics Solutions ManualV Ch. 22–5
10. Irms = 2.2 × 1010 A
R = 6.1 × 10−10 ΩImax =
0
I
.r
7m
0s
7 =
2.2
0
×.7
1
0
0
7
10 A =
P = (Irms)2 R = (2.2 × 1010 A)2(6.1 × 10−10 Ω) = 2.9 × 1011 W
3.1 × 1010 A
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V
1. ∆V2 = 6.9 × 103 V
N1 = 1400 turns
N2 = 140 turns
∆V1 = ∆V2 N
N1
2 = (6.9 × 103 V)
1
1
4
4
0
0
0 = 6.9 × 104 V
Additional Practice 22D
2. ∆V2 = 3.4 × 103 V
N1 = 90 turns
N2 = 2250 turns
∆V1 = ∆V2 N
N1
2 = (3.4 × 103 V)
2
9
2
0
50 = 1.4 × 102 V
3. ∆V1 = 4.6 × 104 V
N1 = 1250 turns
N2 = 250 turns
∆V2 = ∆V1 N
N2
1 = (4.6 × 104 V)
1
2
2
5
5
0
0 = 9.2 × 103 V
4. ∆V1 = 5600 V
N1 = 140 turns
N2 = 840 turns
∆V2 = ∆V1 N
N2
1 = (5600 V)
8
1
4
4
0
0 = 3.36 × 104 V
5. ∆V1 = 9200 V
N1 = 120 turns
N2 = 1200 turns
∆V2 = ∆V1 N
N2
1 = (9200 V)
1
1
2
2
0
0
0 =
6. ∆V1 = 3.6 × 104 V
9.20 × 104 V
∆V2 = 7.2 × 103 V
N1 = 55 turns
N2 = N1 ∆∆
V
V2
1 = (55)73.
.
2
6
××
1
1
0
0
3
4V
V = 11 turns
7. ∆V1 = 240 V
∆V2 = 5.0 VN
N1
2 =
∆∆
V
V1
2 =
2
5
4
.0
0
V
V = 48:1
8. ∆V1 = 1800 V
∆V2 = 3600 V
N1 = 58 turns
N2 = N1 ∆∆
V
V2
1 = (58)316
8
0
0
0
0
V
V = 1.2 × 102 turns
9. ∆V1 = 4900 V
∆V2 = 4.9 × 104 V
N2 = 480 turns
N1 = N2 ∆∆
V
V1
2 = (480)4.9
49
×0
1
0
0
V4 V
= 48 turns
10. P = 1.38 × 106 W
∆V2 = 3.4 × 103 V
N1 = 340 turns
N2 = 17 turns
∆V1 = ∆V2 N
N1
2 = (3.4 × 103 V)
3
1
4
7
0 =
P = ∆V1I1
I1 = ∆
P
V1 =
1
6
.3
.8
8
××
1
1
0
04
6
V
W = 2.0 × 101 A
6.8 × 104 V
Holt Physics Solution ManualV Ch. 23–1
23ChapterAtomic Physics
V
1. λ = 527 nm = 5.27 × 10−7 mE = hf =
h
λc = = 3.77 × 10−19 J
(6.63 × 10−34 J•s)(3.00 × 108 m/s)
5.27 × 10−7 m
Additional Practice 23A
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2. λ = 430.8 nm
= 4.308 × 10−7 m E = hf = h
λc = = 4.62 × 10−22 J
(6.63 × 10−34 J•s)(3.00 × 108 m/s)
4.308 × 10−7 m
3. E = 20.7 eVf =
E
h = = 5.00 × 1015 Hz
(20.7 eV)(1.60 × 10−19 J/eV)
6.63 × 10−34 J•s
4. E = 1.24 × 10−3 eVf =
E
h = = 2.99 × 1011 Hz
(1.24 × 10−3 eV)(1.60 × 10−19 J/eV)
6.63 × 10−34 J•s
5. E = 1.78 eVf =
E
h = = 4.30 × 1014 Hz
(1.78 eV)(1.60 × 10−19 J/eV)
6.63 × 10−34 J•s
6. E = 12.4 MeV
= 1.24 × 107 eV λ = h
E
c = = 1.00 × 10−13 m
(6.63 × 10−34 J•s)(3.00 × 108 m/s)(1.24 × 107 eV)(1.60 × 10−19 J/eV)
7. E = 939.57 MeV
= 9.3957 × 108 eV λ = h
E
c = =
1.32 × 10−15 m = 1.32 × 10−6 nm
If a photon were to have this wavelength, it would not lie within the visible part ofthe spectrum.
1.32 × 10−15 m(6.63 × 10−34 J•s)(3.00 × 108 m/s)(9.3957 × 108 eV)(1.60 × 10 −19 J/eV)
8. E = 3.1 × 10−6 eV λ = h
E
c = = 0.401 m
(6.63 × 10−34 J•s)(3.00 × 108 m/s)(3.1 × 10−6 eV)(1.60 × 10−19 J/eV)
Additional Practice 23B
1. λ = 240 nm = 2.4 × 10−7 m
hft = 2.3 eV
KEmax = h
λc − hft
KEmax = − 2.3 eV
KEmax = 5.2 eV − 2.3 eV = 2.9 eV
(6.63 × 10−34 J•s)(3.00 × 108 m/s)2.4 × 10−7 m)(1.60 × 10−19 J/eV)
Section Five—Solution Manual V Ch. 23–2
V
KEmax = h
λc − hft
KEmax = − 2.16 eV
KEmax = 2.40 eV − 2.16 eV = 0.24 eV
(6.63 × 10−34 J•s)(3.00 × 108 m/s)(5.19 × 10−7 m)(1.60 × 10−19 J/eV)
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2. λ = 519 nm = 5.19 × 10−7 m
hft = 2.16 eV
3. f = 6.5 × 1014 Hz
KEmax = 0.20 eVft =
hf − K
h
Emax
ft =
ft = 6.0 × 1014 Hz
[(6.63 × 10−34 J•s)(6.5 × 1014 Hz) − (0.20 eV)(1.60 × 10−19 J/eV)]
6.63 × 10−34 J•s
4. f = 9.89 × 1014 Hz
KEmax = 0.90 eVft =
hf − K
h
Emax
ft =
ft = 7.72 × 1014 Hz
(6.63 × 10−34 J•s)(9.89 × 1014 Hz) − (0.90 eV)(1.60 × 10−19 J/eV)
6.63 × 10−34 J•s
5. ft = 1.36 × 1015 Hzhft = = 5.64 eV
(6.63 × 10−34 J•s)(1.36 × 1015 Hz)
1.60 × 10−19 J/eV
6. ft = 1.1 × 1015 Hzhft = = 4.6 eV
(6.63 × 10−34 J•s)(1.1 × 1015 Hz)
1.60 × 10−19 J/eV
7. hft = 4.1 eVλ =
h
E
c = = 3.0 × 10−7 m = 300 nm
(6.63 × 10−34 J•s)(3.00 × 108 m/s)
(4.1 eV)(1.60 × 10−19 eV)
8. hft = 5.0 eVλ =
h
E
c = = 2.5 × 10−7 m = 250 nm
(6.63 × 10−34 J•s)(3.00 × 108 m/s)
(5.0 eV)(1.60 × 10−19 eV)
9. KEmax = 0.62 V
me = 9.109 × 10−31 kgKEmax = hf − hft = 1
2mev
2
v = 2Km
Em
eax =
v = 4.7 × 105 m/s
2(0.62 eV)(1.60 × 10−19 J/eV)
9.109 × 10−31 kg
10. KEmax = 1.2 eV
me = 9.109 × 10−31 kgKEmax = hf − hft = 1
2 mev
2
v = 2Km
Em
eax =
v = 6.5 × 105 m/s
2(1.2 eV)(1.60 × 10−19 eV)
9.109 × 10−31 kg
Holt Physics Solutions ManualV Ch. 23–3
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1. v = 28 m/s
λ = 8.97 × 10−37 mm =
λh
v = = 26 kg
6.63 × 10−34 J•s(8.97 × 10−37 m)(28 m/s)
2. v = 7.1 × 102 m/s
λ = 5.8 × 10−42 mm =
λh
v = = 1.6 × 105 kg
6.63 × 10−34 J•s(5.8 × 10−42 m)(7.1 × 102 m/s)
3. v = 5.6 × 10−6 m/s
λ = 2.96 × 10−8 mm =
λh
v = = 4.0 × 10−21 kg
6.63 × 10−34 J•s(2.96 × 10−8 m)(5.6 × 10−6 m/s)
4. v = 12 m/s
λ = 2.6 × 10−29 mm =
λh
v = = 2.1 × 10−6 kg
6.63 × 10−34 J•s(2.6 × 10−29 m)(12 m/s)
5. me = 9.109 × 10−31 kg
v = 2.19 × 106 m/sλ =
m
h
v = = 3.3 × 10−10 m
6.63 × 10−34 J•s(9.109 × 10−31 kg)(2.19 × 106 m/s)
6. m = 7.6 × 107 kg
v = 35 m/sλ =
m
h
v = = 2.5 × 10−43 m
6.63 × 10−34 J•s(7.6 × 107 kg)(35 m/s)
7. m = 5.94 × 1024 kg
v = 3.0 × 104 m/sλ =
m
h
v = = 3.7 × 10−63 m
6.63 × 10−34 J•s(5.94 × 1024 kg)(3.0 × 104 m/s)
8. m = 4.0 × 1041 kg
v = 1.7 × 104 m/sλ =
m
h
v = = 9.7 × 10−80 m
6.63 × 10−34 J•s(4.0 × 1041 kg)(1.7 × 104 m/s)
Additional Practice 23C
Givens Solutions
9. me = 9.109 × 10−31 kg
λ = 9.87 × 10−14 mv =
m
h
λ = = 7.37 × 109 m/s
6.63 × 10−34 J•s(9.109 × 10−31 kg)(9.87 × 10−14 m)
10. mn = 1.675 × 10−27 kg
λ = 5.6 × 10−14 mv =
m
h
λ = = 7.1 × 106 m/s
6.63 × 10−34 J•s(1.675 × 10−27 kg)(5.6 × 10−14 m)
Holt Physics Solution ManualV Ch. 25–1
25ChapterSubatomic Physics
V
1. Z = 19
A = 39
atomic mass of K-39 = 38.963 708 u
atomic mass of H = 1.007 825 u
mn = 1.008 665 u
N = A − Z = 39 − 19 = 20
∆m = Z (atomic mass of H) + Nmn − atomic mass of K-39
∆m = 19(1.007 825) + 20 (1.008 665 u)–38.963 708 u
∆m = 19.156 75 u + 20.1733 u – 38.963 708 u
∆m = 0.3653 u
Ebind = (0.3653 u)(931.50 MeV/u) = 340.3 MeV
Problem Bank Solutions Chapter 25A
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2. Z = 50
A = 120
atomic mass of Sn-120 = 119.902 197
atomic mass of H = 1.007 825 u
mn = 1.008 665 u
N = A − Z = 120 − 50 = 70
∆m = Z(atomic mass of H) + Nmn − atomic mass of Sn-120
∆m = 50(1.007 825 u) + 70(1.008 665 u) − 119.902 197 u
∆m = 50.391 25 u + 70.606 55 u − 119.902 197 u
∆m = 1.095 60 u
Ebind = (1.095 60 u)(931.50 MeV/u) = 1020.55 MeV
3. For 10747 Ag:
Z = 47
A = 107
atomic mass of Ag-107 = 106.905 091 u
atomic mass of H = 1.007 825 u
mn = 1.008 665 u
For 6329 Cu:
Z = 29
A = 63
atomic mass of Cu-63 = 62.929 599 u
N = A − Z = 107 − 47 = 60
∆m = Z(atomic mass of H) + Nmn − atomic mass of Ag-107
∆m = 47(1.007 825 u) + 60(1.008 665 u) − 106.905 091 u
∆m = 47.367 775 u + 60.5199 u − 106.905 091 u
∆m = 0.9826 u
Ebind = (0.9826 u)(931.50 MeV/u) = 915.29 MeV
N = A − Z = 63 − 29 = 34
∆m = Z(atomic mass of H) + Nmn − atomic mass of Cu-63
∆m = 29(1.007 825 u) + 34(1.008 665 u) − 62.929 599 u
∆m = 29.226 925 u + 34.2946 u − 62.929 599 u
∆m = 0.5919 u
Ebind = (0.5919 u)(931.50 MeV/u) = 551.4 MeV
The difference in binding energy is 915.29 MeV − 551.4 MeV = 363.9 MeV
Section Five—Solution Manual V Ch. 25–2
V
N = A − Z = 12 − 6 = 6
∆m = Z (atomic mass of H) + Nmn − atomic mass of C-12
∆m = 6(1.007 825 u) + 6(1.008 665u) − 12.000 000 u
∆m = 6.046 95 u + 6.051 99 u − 12.000 000 u
∆m = 0.098 94 u
Ebind = (0.098 94 u)(931.50 MeV/u) = 92.163 MeV
N = A − Z = 16 − 8 = 8
∆m = Z (atomic mass of H) + Nmn − atomic mass of O-16
∆m = 8(1.007 825 u) + 8(1.008 665u) − 15.994 915 u
∆m = 8.0626 u + 8.06932 u − 15.994 915 u
∆m = 0.1370 u
Ebind = (0.1370 u)(931.50 MeV/u) = 127.62 MeV
The difference in binding energy is
127.62 MeV − 92.163 MeV = 35.46 MeV
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d.4. For 12
6 C:
Z = 6
A = 12
atomic mass of C-12 = 12.000 000 u
atomic mass of H= 1.007 825 u
mn = 1.008 665 u
For 168 O:
Z = 8
A = 16
atomic mass of O-16 = 15.994 915
5. Z = 17
A = 35
atomic mass of Cl-35 = 34.968 853 u
atomic mass of H = 1.007 825 u
mn = 1.008 665 u
N = A − Z = 35 − 17 = 18
∆m = Z (atomic mass of H) + Nmn − atomic mass of CI-35
∆m = 17(1.007 825 u) + 18(1.008 665 u) − 34.968 853 u
∆m = 17.133 025 u + 18.155 97 u − 34.968 853 u
∆m = 0.320 14 u
Ebind = (0.320 14 u)(931.50 MeV/u) = 298.21 MeV
7. A = 58
Z = 28
atomic mass of Ni-58 = 57.935 345 u
atomic mass of H = 1.007 825 u
mn = 1.008 665 u
N = A − Z = 58 − 28 = 30
∆m = Z(atomic mass of H) + Nmn − atomic mass of Ni-58
∆m = 28(1.007 825 u) + 30(1.008 665 u) − 57.935 345 u
∆m = 28.2191 u + 30.259 95 u − 57.935 345 u
∆m = 0.5437 u
6. Z = 1
A = 2
atomic mass of H-2 = 2.014 102 u
atomic mass of H = 1.007 825 u
mn = 1.008 665 u
N = A − Z = 2 − 1 = 1
∆m = Z(atomic mass of H) + Nmn − atomic mass of H-2
∆m = 1(1.007 825 u) + 1(1.008 665 u) − 2.014 102 u
∆m = 2.388 × 10−3 u
Ebind = (2.388 × 10−3 u)(931.50 MeV/u) = 2.2244 MeV
Holt Physics Solutions ManualV Ch. 25–3
9. A = 90
Z = 40
atomic mass of Zr-90 = 89.904 702 u
atomic mass of H = 1.007 825 u
mn = 1.008 665 u
N = A − Z = 90 − 40 = 50
∆m = Z(atomic mass of H) + Nmn − atomic mass of Zr-90
∆m = 40(1.007 825 u) + 50(1.008 665 u) − 89.904 702 u
∆m = 40.313 u + 50.433 25 u − 89.940 702 u
∆m = 0.842 u
Ebind = (0.842 u)(931.50 MeV/u) = 784 MeV
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d.
VV
10. A = 32
Z = 16
atomic mass of S-32 = 31.972 071 u
atomic mass of H = 1.007 825 u
mn = 1.008 665 u
N = A − Z = 32 − 16 = 16
∆m = Z(atomic mass of H) + Nmn − atomic mass of S-32
∆m = 16(1.007 825 u) = 16(1.008 665 u) − 31.972 071 u
∆m = 16.1252 u + 16.1386 u − 31.972 071 u
∆m = 0.2918 u
Problem Bank Solutions Chapter 25B
Givens Solutions
1. 21084 Po → ? + 4
2He A = 210 − 4 = 206
Z = 84 − 2 = 82
? = 20682 Pb
2. 167 N → ? + 0
−1e + v A = 16 − 0 = 16
Z = 7 − (−1) = 8
? = 168 O
3. 14762 Sm → 143
60 Nd + ? + v A = 147 − 143 = 4
Z = 62 − 60 = 2
? = 42 He
4. 1910 Ne → ? + 0
1e + v A = 19 − 0 = 19
Z = 10 − 1 = 9
? = 199 F
8. A = 64
Z = 30
atomic mass of Zn-64 = 63.929 144 u
atomic mass of H= 1.007 825 u
mn = 1.008 665 u
N = A − Z = 64 − 30 = 34
∆m = Z(atomic mass of H) + Nmn − atomic mass of Zn-64
∆m = 30(1.007 825 u) + 34(1.008 665 u) − 63.929 144 u
∆m = 30.234 75 u + 34.2946 u − 63.929 144 u
∆m = 0.6002 u
Section Five—Solution Manual V Ch. 25–4
7. 16074 W → 156
72 Hf + ? A = 160 − 156 = 4
Z = 74 − 72 = 2
? = 42 He
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d.
V
Additional Practice 22C
Givens Solutions
8. ? → 10752 Te + 4
2He A = 107 + 4 = 111
Z = 52 + 2 = 54
? = 11154 Xe
9. 15772 Hf → 153
70 Yb + ? A = 157 − 153 = 4
Z = 72 − 70 = 2
? = 42 He
10. 14158 Ce → ? + 0
−1e + v A = 141 − 0 = 141
Z = 58 − (−1) = 59
? = 14159 Pr
Problem Bank Solutions Chapter 25C
Givens Solutions
1. mi = 5.25 × 10−3 g
mf = 3.28 × 10−4 g
∆t = 12 h
m
m
i
f = 3
5
.
.
2
2
8
5
××
1
1
0
0
−
−
4
3g
g =
1
1
6
If 1
1
6 of the sample remains after 12 h, then 1
8 of the sample must have remained after
6.0 h, 14
of the sample must have remained after 3.0 h, and 12
of the sample must have
remained after 1.5 h. So T1/2 = 1.5 h
2. mi = 3.29 × 10−3 g
mf = 8.22 × 10−4 g
∆t = 30.0 s
m
m
i
f = 8
3
.
.
2
2
2
9
××
1
1
0
0
−
−
4
3g
g =
1
4
If 14
of the sample remains after 30.0 s, then 12
of the sample must have remained after
15.0 s, so T1/2 = .15.0 s
3. mi = 4.14 × 10−4 g
mf = 2.07 × 10−4 g
∆t = 1.25 days
m
m
i
f = 2
4
.
.
0
1
7
4
××
1
1
0
0
−
−
4
4g
g =
1
2
If 12
of the sample remains after 1.25 days, then T1/2 = 1.25 days
5. ? → 13154 Xe + 0
−1e + v A = 131 + 0 = 131
Z = 54 + (−1) = 53
? = 13153 I
6. ? → 9039 Y + 0
−1e + v A = 90 + 0 = 90
Z = 39 + (−1) = 38
? = 9038 Sr
Holt Physics Solutions ManualV Ch. 25–5
6. T1/2 = 2.7 y
N = 3.2 × 109λ =
0
T
.6
1
9
/2
3 = = 8.1 × 10−9 s−10.693
(2.7 y)(3.156 × 107 s/y)
Givens Solutions
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d.
V
7. activity = 765.3 mCi
T1/2 = 22 hλ =
0
T
.6
1
9
/2
3 =
(22.0 h)
0
(
.
3
6
6
9
0
3
0 s/h) = 8.75 × 10−6 s−1
N = = = 3.2 × 1015 nuclei(0.7653 Ci)(3.7 × 1010 s−1/Ci)
8.75 × 10−6 s−1activity
λ
8. T1/2 = 21.6 h
N = 6.5 × 106
λ = 0
T
.6
1
9
/2
3 =
(21.6 h
0
)
.
(
6
3
9
6
3
00 s/h) =
activity = N λ = = 1.5 × 10−9 Ci(8.9 × 10−6 s−1)(6.5 × 106)
3.7 × 1010 s−1/Ci
8.90 × 10−6 s−1
9. T1/2 = 12.33 y
N = 4.8 × 108λ =
0
T
.6
1
9
/2
3 = = 1.78 × 10–9 s–10.693
(12.33 y)(3.56 × 107 s/y)
10. activity = 0.3600 Ci
T1/2 = 17.2 s
λ = 0
T
.6
1
9
/2
3 =
1
0
7
.6
.2
93
s =
N = acti
λvity = = 3.3 × 104 nuclei
(0.3600 Ci)(3.7 × 1010 s−1/Ci)
(4.03 × 10−2 s−1)
4.03 × 10−2 s−1
4. T1/2 = 10.64 h For the sample to reach 12
its original strength, it takes 10.64 h. For the sample to
reach 14
its original strength, it takes 2(10.64 h) = 21.28 h. For the sample to reach 8
1 its
original strength, it takes 3(10.64 h) = 31.92 h
5. T1/2 = 462 days For the sample to reach 12
its original strength, it takes 462 days. For the sample to
reach 14
its original strength, it takes 2(462 days) = 924 days