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