Units of Measurement (1.3) & (1.4) Systems of Units.
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Transcript of Units of Measurement (1.3) & (1.4) Systems of Units.
Units of Measurement
(1.3) & (1.4) Systems of Units
Table 1.1 (p. 9)English, Metric, & SI Units English – inch, mile, pound, ounce Metric – base-10, CGS and MKS CGS – Based on centimeter, gram, second
MKS – Based on meter, kilogram, second
SI – International System, modern metric
Problem 6 (p. 29)A pitcher has the ability to throw a baseball at
95 mph. What is the speed in ft/s?
95 mi h
5280 ft mi
* 1 h _ 60 min
* 1 min 60 s
* = ? ftsfts
139.33 ft s
Problem 6 (p. 29) part bHow long does the hitter have to make a
decision about swinging at the ball if the plate and the mound are separated by 60 feet?
v = dt
t = dv
60 ft _139.33 ft/s = ?
= 0.431 s
Problem 6 (p. 29) part c.If the batter wanted a full second to make a
decision, what would the speed in mph have to be?
v = dt
= 60 ft1 s
* 60 s_1 min * 60 min
1 h* 1 mi_
5280 ft= ?
= 40.91 mph
1.5 Significant figures, accuracy, and rounding off
1.2 V and 1.20 V
Imply different levels of accuracy
Accuracy and PrecisionAccuracy = freedom from error (exactness)Precision = The degree of refinement with
which an operation is performed or a measure stated
The precision of a reading can be determined by the number of significant figures (digits) present.
When adding a quantity accurate only to the tenths place to a number accurate to the thousandths place will result in a total having accuracy only to the tenths place.
In the addition or subtraction of approximate numbers, the entry with the lowest level of accuracy determines the format of the solution.
Example 1.1 (p. 12)
a. 532.6 ≈ 536.7 4.02 (as determined by
+ 0.036 532.6) = 536.656
Example 1.1 (p. 12)
b. 0.04 ≈ 0.05 0.003 (as determined + 0.0064 by 0.04) = 0.0494
1.6 Powers of Ten_ 1 _ _ 1 _1000 103
-3 = 10=
__ 1 __ _ 1 _0.00001 10
= 105
-5=
Addition and Subtraction
A * 10 ± B * 10 = (A ± B) * 10
Example: 6300 + 75.000
= (6.3 * 10 ) + (75 * 10 )= (6.3 + 75) * 10= 81.3 * 10
3 3
3
3
n n n
Multiplication
(a* 10 ) (B * 10 ) = (A)(B) * 10
Example: (0.0002) (0.000007)= (2) * 10 * (7) * 10= 14 * 10
n m n + m
-4 -6
-10
DivisionA * 10_ AB * 10 B
* 10 =n
m
n-m
Example:
0.00047 0.002
47 * 10_ 2 * 10
= 23.5 * 10-5
-3
-2
=
Powers(A * 10 ) = A * 10n m m nm
(5 * 10 ) = 5 * 10-5 3 3 -15__1___
0.0005( )3
=
= 125 * 10-15
Example:
1.7: Fixed-Point, Floating Point, Scientific, and Engineering Notation
* Fixed Point – Choose the level of accuracy for the output – example: tenths, hundredths or thousandths place
13 = 0.333
1 16 = 0.063
2300 2 = 1150.000
Floating Point
Number of significant figures varies13 = 0.3333333333…
1 16 = 0.0625
2300 2 = 1150
Scientific NotationScientific notation requires that the decimal
point appear directly after the first digit greater than or equal to 1, but let than 10.
13 = 3.3333333 E-1
1 16 = 6.25 E-2
2300 2 = 1.15 E3
Engineering NotationEngineering notation specifies that all powers of ten
must be multiples of 3, and the mantissa must be greater than or equal to 1 but less than 1000
13 = 333.3333333 E-3
1 16 = 62.5 E-3
2300 2 = 1.15 E3
Engineering Notation and Accuracy
Using engineering notation with two-place accuracy will result in:
13 = 333.33 E-3
1 16 = 62.50 E-3
2300 2 = 1.15 E3
Look at table 1-2 for prefixes
1.8 Conversion Between Levels of Powers of Ten
a. 20 kHz = ______________ MHz
20 * 10_ Hz
3
* 10 = 20 * 10-6 -3
= 0.02 MHz
Conversion: Continued
b. 0.04 ms = ___________ μs
* 10 * 10 = 4 * 10-3 +1
-2
μs
or 40 μs
64 * 10_ s
1.9 Conversion
• 0.5 day = _____ min
0.5 day 24 h 60 min= 720 min
1 day 1 h(
() )
Determine the speed in miles per hour of a competitor who can run a 4-min mile.
1 mi4 min
60 min1 h
60 mi4 hr
15 mih( () ) ==
15 mph
Data is being collected automatically from an experiment at a rate of 14.4 kbps. How long will it take to completely fill a diskette whose capacity is 1.44 MB? Rate = 14.4 kbps, Capacity = 1.44 MB
Rate =
1.44 MB = 1.44 * 2 * 8 bit20
CapacityTime
so Time =Capacity
Rate
t =(1.44 MB) (2 ) (8 _)
(14.4 * 10 )(60 )
20bytesMB
bitsbyte
bitssec
secmin
= 13.98 min3
Number Systems
(N) = [(integer part) . (fractional part)]n
Radix point
Two common number representationsJuxtapositional – placing digit side-by-side
Non-juxtapositional
Juxtapositional
n-1(N) = (a a … a a a a … a )n n-2 1 0. -1 n-2 -m
R = Radix of the number system
Radix point
n = number of digits in the integer portionm = number of digits in the fractional portiona = MSDa = LSD
n-1-m
Base Conversion
19.75 = ( ) 10 2
[(0001 * 1010) + (1001 * 0001) + (0111 * ) + 0101 * + ]
2 ___ 12___12___02___02___1
1 _ 1010
1 _ 1010
1 _ 1010
1994210
100110.75
2 1.50
21.00
______x
______x1
1.11
= 10011.11