Physics 231 - Directorybrown/231/Review_Exam3.pdf · MSU Physics 231 Fall 2015 3 Some Housekeeping...

MSU Physics 231 Fall 2015 1

Physics 231Exam 3: Review

Alex Brown2015


PHYSICS 231Review for midterm 3

Topic 9: Gravitation – Kepler’s laws

Topic 10: Solids & Fluids

Topic 11: Waves & Sound

Topic 12: Temperature, Thermal Expansion,

and Ideal Gases


Some Housekeeping• The 3nd exam will be Wednesday December 2.

• The exam will take place right here in BPS 1410.

• We will have assigned seating, so show up early.

• If you are sick, you MUST have a note from a doctor on the doctor’s letterhead or prescription pad.

• You CANNOT use cell phones during the exam.

• You can (should!) bring your equation sheets - three two-sided -on8.5x11 paper

3


The gravitational force

2221

rMmG

rmmGF

G=6.673·10-11 N m2/kg2

N m2/kg2 = m3/(kg s2)

Newton:

The gravitational force works between every two massiveparticles in the universe.


Gravitational potential energy

PEgravity = -GMm/r

Near surface of a planet: PEgravity = mgph

where gp = GMp/Rp2

Mp = mass of planet and Rp is the radius of the planet

for earth gp = 9.8


Orbiting condition (period)Consider an object in circular motion around large mass M

332

2

22

2

2

4

2

KrrGM

T

rT

mrm

rmv

rGMm

maF c

/1099earth)( /10297sun)(

4

3215

3221

2

3

2

msKmsK

GMK

rT

M

m

r

v

rGMv 2


slide previous in the rR


Some definitionsFirst cosmic speed: speed of a satellite of mass m on a low-lying circular orbit with rsatellite ≈ Rp (p=planet)

v1 =(gpRp) where gp = GMp/Rp2 Earth: v1=7.91x103 m/s

Second cosmic speed: speed needed to break free from a planet:

v2 = (2gpRp) Earth: v2=1.12x104 m/s

Synchronous orbit of a satellite: rotation period of satellite of mass m is the same as rotation period of the planet

For earth: period T = 24 hours = 86400 s

r= 42,000 km R=6,400 km /1099earth)( 3215

3

2

msK

KrT


Quantities that are constant for a given orbit

ellipse area

2 momentumangular

abATAmL

a

b

32 period aKT

aGMmMETE

2energy mechanical total

2/12/32/1)()(22

ab

aKabm

TmAL


Quantities that depend on distance r

a

b

2)2/1(energy kinetic

velocity

energy potential

mvKE

mrLv

rMmGPE

r


An Example

A

BstarTwo planets are orbiting a star. The orbit of A has a radius of 108 km. The distance of closest approach of B to the star is 5x107 km and its maximum distance from the star is 109 km. If A has a rotational period of 1 year, what is the rotational period of B?

aA = 108 kmaB = ½(Rmin+Rmax) = ½(5x107+109) = 5.25x108 km

Rmin = distance of closest approach.= perihelion

Rmax= maximum distance= aphelion

a3/T2 = constant aA3/TA

2= aB3/TB

2 so TB2=(aB

3/aA3)TA

2

So TB=(5.253 x (1 yr)2) = 12 years


The Young’s modulus (tensile)

positive) is ( //

] Unitless![ L/L :strain tensile](Pa) Pascal[N/mF/A :stress tensile

strain tensilestress tensile

0

0

0

2

LLA

FLLLAFY

Y

Beyond the elastic limit an object is permanently deformed (it does not return to its original shape if the stress is removed).


Pascal’s principleA change in pressure applied to a fluid that is enclosed is transmitted to the whole fluid and all the walls of the container that hold the fluid.

Hydraulic Lift

If we apply a smallforce F1, we can exerta very large Force F2.

P = F1/A1 = F2/A2

If A2>>A1 then F2>>F1.


P = P0+ g hh = distance between liquid surface

and the point where you measure P

P0

Ph

B = Vo g = Mf g The buoyant force equals the weight of the amount of fluid that can be put in the volume Vo taken by the object.If object is not moving: B = W o =

Pressure at depth h

Buoyant force for submerged object

Buoyant force for floating objecthBThe buoyant force equals the weight of the amount of fluid

that can be put in the part of the volume (Vd) of the object that is under water. Vd = Mo/ (any object)for a block area A, (Vd = Ah), then h = Mo/(A)

W=mg

W=mg


Measurement of Pressure

The open-tube manometer.The pressure at A and B is the same:P = PA = PBP = P0 + ghso h = (P-P0)/(g)

If the pressure P = 1.1 atm, what is h? (the liquid is water)h = (1.10-1.00)x105/(1.0x103x9.81) = 1.0 m


T1 T2

T1 = Mog = oVg T2 = Mog – Mfg = oVg - Vg

(T2 /T1 ) = R = 1 - ( /o ) = o (1-R)

(o / ) = 1/(1-R) = specific density when the fluid is water = w

o = density of object = density of fluidV = volume of object


Bernoulli’s equation

P: pressure½v2:kinetic Energy per unit volume gy: potential energy per unit volume

P1,A1,y1

P2,A2,y2

v1

v2

P1 + ½v12 + gy1= P2 + ½v2

2 + gy2 P + ½v2 + gy = constant

P: pressure y: height: height g: gravitational v: velocity acceleration


1) P + ½v2 + gy = constant2) A1v1 =A2v2

equal

A. Incompressible fluid, so density is constant

B. AA>AB so vA<vB use 2)vA<vB so PA>PB use 1)

greater C. Must be the same, else the liquid would get ‘stuck’ between A and B, or a ‘hole’ would open between A and B

equalless than

See B.


motion ofdirection in the

22)(cos),(

)(sin),(

fT

v

kT

tkxAtxvtkxAtxy

MSU Physics 231 Fall 2015 22PHY 231

22

Sound / Decibel Level =10log (I/I0) I0=10-12 W/m2

y = log10x = log(x) inverse of x = 10y

( not this: y=ln(x) x=ey ) log(ab) = log(a) + log(b)log(a/b) = log(a) - log(b)log(an) = n log(a)

0

20

40

60

80

100

120

10-12

10-11

10-10

10-9

10-8

10-7

10-6

10-5

10-4

10-3

10-2

10-1

1

Intensity I (W/m2)

soun

d le

vel (

dB)


Decibels (units are called dB)

For an increase of n dB:the intensity of the sound is multiplied by a factor of ?.

n = 2-1= 10 log(I2/I0) – 10 log(I1/I0)

n = 10 log(I2/I1)

(n/10) = log(I2/I1) n (I2/I1)10 10

10 (n/10) = (I2/I1) 20 10030 1000

=10 log(I/I0) I0=10-12 W/m2


Intensity and Distance

The amount of energy passingthrough a spherical surfaceat distance r from the sourceis constant, but the surfacebecomes larger.

I = Power/Surface = P/A=P/(4r2)

r=1 I = P/(4r2) = P/(4) 1r=2 I = P/(4r2) = P/(16) 1/4r=3 I = P/(4r2) = P/(36) 1/9

I1/I2 = r22/r1

2


ExampleA person living on campus (c) 300 m from the rail trackis tired of the noise of the passing trains. They decide tomove to Abbott (a) (3.5 km from the rail track). If the Sound level of the trains was originally 70dB (vacuum cleaner), what is the sound level at Abbott?

Campus (c): 70 dB = 10log (Ic/I0) Ic= 107 I0=10-5 W/m2

Ia/Ic = rc2/ra

2 = 0.0073

Ia = Ic (rc2/ra

2) = 7.3x10-8 W/m2

Sound level: 49dB (normal conversation)


Doppler Effect: In General

source you

s

o

vvvv

ff

vo =vobserver: positive if moving towards to sourcevs = vsource: positive if moving towards the observer

Observed frequency is:1) higher if wavefronts are being bunched together (source & observer getting closer)2) lower if wavefronts are gaining extra separation (source & observer moving apart)


sound of speedplane theof speed M

anglemach theis andnumber Mach theis M where1sin 1

M


Interference: Beats!

Amplitude of the “beat” changes with time, so the intensity of the sound changes as a function of time. fbeat = |fA-fB|


.0,1,2,3... n 12n

4

L

.0,1,2,3... n 1n

2

L

.0,1,2,3... n 1n

2

L

number harmonic theis 1)(n

number harmonic theis 1)(2n

number harmonic theis 1)(n


Standing Wave

Just like with sound, the velocity of the standing wave depends on the density of the material.

Linear mass density of a string:μ = mass/length

Also depends on the string’s tension: T

Higher string density: velocity goes downHigher tension: velocity goes up

21 E P

poweraverage

22 AvT


Thermal expansionL= Lo T

L0

L

T=T0 T=T0+T

A = AoT = 2

V = Vo T = 3

length

surface

volume

Some examples: = 24x10-6 1/K Aluminum = 1.2x10-4 1/K Alcohol

: coefficient of linear expansiondifferent for each material


Molecular mass

kg 102.00 1002.6

012.0(carbon)

kg 0.0120 g 12.0 (carbon) examplefor

mass (molar)molecular

numbers sAvagodro' 1002.6

molecule)(or atomoneof mass

26-23

molar

23

m

M

NmM

N

m

molar

A

A


molar

molecules)(or atoms all of mass total

molecules)(or atoms contains mole one so

or moles ofnumber

molecules)(or atoms ofnumber total

MnM

NmM

N

NnNNNn

N

A

AA

Number of atoms and moles


Boyle & Charles & Gay-LussacIDEAL GAS LAW

n = number of molesR = universal gas constant 8.31 J/mol·K

If the number of moles is fixed

2

22

1

11 constant TVP

TVP

TPV

nRTPV Does not depend on what type or atom or molecule


P1 = 1 atm P2 = 2 atmV1 = 2 m3 V2 = 10 m3

T1 = 100 K N1 = NA N2 = NA

T2 = ? K

A) 10B) 100C) 500D) 1000


Relationship to ideal gas law

32 KNPV

The objects bounce off of each other and the walls of the container (elastic). One can derive the following result

TkKTkNPV BB 23get to with combine


2

21 vmK with

23 combine TkK B

root-mean-square (rms) velocity for one object

molar

Brms M

RTm

Tkvv 332


d is the number of “degrees of freedom” for the motion

d = 3 for an atom (motion in x, y, z directions) like helium gas

d = 5 for a diatomic molecule (motion in x, y, z and two ways to rotate)like nitrogen gas N2

) (since 2223

nRNkPVdnRTdTkdNKNdE BBth

(Homework question for “one degree of freedom” use d = 1)

Total thermal energy


Questions from the Fall 2014 3rd midterm exam

Two of these will be on our exam

Physics 231 - Directorybrown/231/Review_Exam3.pdf · MSU Physics 231 Fall 2015 3 Some Housekeeping...

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