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? 139.33 ft s ? ft s 95 mi h 5280 ft mi 1 h _ 60 min 1 min 60 s = * * *

Problem 6 (p. 29) part b How 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 = d t t = d v 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? d t 60 ft 1 s 60 s_ 1 min 60 min 1 h 1 mi_ 5280 ft ? v = = * * * = = 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 Precision Accuracy = 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) 532.6 ≈ 536.7 4.02 (as determined by 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 10 = = 10 __ 1 __ _ 1 _ 0.00001 10 -3 = 10 3 __ 1 __ _ 1 _ 0.00001 10 5 = = 10 -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

Multiplication n m n + m (a* 10 ) (B * 10 ) = (A)(B) * 10 Example: (0.0002) (0.000007) = (2) * 10 * (7) * 10 = 14 * 10 -4 -6 -10

Division A * 10_ A B * 10 B = * 10 Example: 0.00047 0.002 47 * 10_ n-m = * 10 m Example: -5 0.00047 0.002 47 * 10_ 2 * 10 -2 = 23.5 * 10 = -3

( ) Powers = 125 * 10 (A * 10 ) = A * 10 Example: __1___ 0.0005 = n m nm Example: ( ) __1___ 0.0005 3 3 -5 3 -15 = (5 * 10 ) = 5 * 10 -15 = 125 * 10

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 1 3 1 16 = 0.333 = 0.063 2300 2 = 1150.000

Floating Point Number of significant figures varies 1 3 = 0.3333333333… 1 16 = 0.0625 2300 2 = 1150

Scientific Notation Scientific notation requires that the decimal point appear directly after the first digit greater than or equal to 1, but let than 10. 1 3 = 3.3333333 E-1 1 16 = 6.25 E-2 2300 2 = 1.15 E3

Engineering Notation Engineering 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 1 3 = 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: 1 3 = 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 20 kHz = ______________ MHz 20 * 10_ Hz 3 -6 -3 * 10 = 20 * 10 = 0.02 MHz

Conversion: Continued 0.04 ms = ___________ μs -2 4 * 10_ s +1 -3 6 * 10 * 10 = 4 * 10 μs or 40 μs

( ) ( ) 1.9 Conversion 0.5 day = _____ min 0.5 day 24 h 60 min

Determine the speed in miles per hour of a competitor who can run a 4-min mile. ( ) ( ) 1 mi 4 min 60 min 1 h 60 mi 4 hr 15 mi h = = 15 mph

Rate = 14.4 kbps, Capacity = 1.44 MB 1.44 MB = 1.44 * 2 * 8 bit 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 20 1.44 MB = 1.44 * 2 * 8 bit Capacity Rate Capacity Time so Time = Rate = bytes 20 bits (1.44 MB) (2 ) (8 _) (14.4 * 10 )(60 ) MB byte = 13.98 min t = bits 3 sec sec min

Number Systems (N) = [(integer part) . (fractional part)] Radix point (N) = [(integer part) . (fractional part)] n Two common number representations Juxtapositional – placing digit side-by-side Non-juxtapositional

Juxtapositional (N) = (a a … a a a a … a ) Radix point (N) = (a a … a a a a … a ) n n-1 n-2 1 0. -1 n-2 -m R = Radix of the number system n = number of digits in the integer portion m = number of digits in the fractional portion a = MSD n-1 a = LSD -m

Base Conversion 19.75 = ( ) 10 2 1 _ 1010 1 _ 1010 1 _ 1010 [(0001 * 1010) + (1001 * 0001) + (0111 * ) + 0101 * + ] 2 ___ 1 2___1 2___0 19 9 4 2 1 10011 0.75 2 1.50 1.00 ______x = 10011.11 1 ______x 1 .11