tt MEASUREMENT
, Chapter 1: tt MEASUREMENT
1. The SI standard of time is based on:
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A. the daily rotation of the earth
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B. the frequency of light emitted by Kr86
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C. the yearly revolution of the earth about the sun
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D. a precision pendulum clock
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E. none of these tt tt
Ans: E tt t t
2. A nanosecond is:
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A. 109 s tt
B. 10−9 s tt
C. 10−10 s tt
D. 10−10 s tt
E. 10−12
Ans: B tt
3. The SI standard of length is based on:
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A. the distance from the north pole to the equator along a meridian passing through Paris
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B. wavelength of light emitted by Hg198 tt tt tt tt tt
C. wavelength of light emitted by Kr86 tt tt tt tt tt
D. a precision meter stick in Paris
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E. the speed of tt tt
light Ans: Ett tt t t
4. In 1866, the U. S. Congress defined the U. S. yard as exactly 3600/3937 international
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meter. This was done primarily because:
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A. length can be measured more accurately in meters than in yards
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B. the meter is more stable than the yard
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C. this definition relates the common U. S. length units to a more widely used system
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D. there are more wavelengths in a yard than in a meter
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E. the members of this Congress were exceptionally
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intelligent Ans: C
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5. Which of the following is closest to a yard in length?
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A. 0.01 m tt
B. 0.1m tt
C. 1 m tt
D. 100 m tt
E. 1000 m tt
Ans: C tt
Ans:
B
2 Chapter 1: MEASUREMENT tt
, 6. There is no SI base unit for area because:
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A. an area has no thickness; hence no physical standard can be built
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B. we live in a three (not a two) dimensional world
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C. it is impossible to express square feet in terms of meters
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D. area can be expressed in terms of square meters
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E. area is not an important physical
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quantity Ans: D
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7. The SI base unit for mass is:
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A. gram
B. pound
C. kilogram
D. ounce
E. kilopound
Ans: C
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8. A gram is:
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A. 10−6 kg tt
B. 10−3 kg t t
C. 1 kg tt
D. 103 kg tt
E. 106 kg tt
Ans: B tt
9. Which of the following weighs about a pound?
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A. 0.05 kg tt
B. 0.5 kg tt
C. 5 kg tt
D. 50 kg tt
E. 500 kg tt
Ans: D tt
10. (5.0 × 104) × (3.0 × 106) =
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A. 1.5 × 109 tt tt
B. 1.5 × 1010 tt tt
C. 1.5 × 1011 tt tt
D. 1.5 × 1012 tt tt
E. 1.5 × 1013 tt tt
Ans: C tt
11. (5.0 × 104) × (3.0 × 10−6) =
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A. 1.5 × 10−3 tt tt
B. 1.5 × 10−1 tt tt
C. 1.5 × 101 tt tt
D. 1.5 × 103 tt tt
E. 1.5 × 105 tt tt
Chapter 1: tt MEASUREMENT
, 12. 5.0 tt × 105 + 3.0 × 106 =
tt tt tt tt tt tt
A. 8.0 × 105 tt tt
B. 8.0 × 106 tt tt
C. 5.3 × 105 tt tt
D. 3.5 × 105 tt tt
E. 3.5 × 106 tt tt
Ans: E tt
13. (7.0 × 106)/(2.0 × 10−6) =
tt tt tt tt tt tt
A. 3.5 × 10−12 tt tt
B. 3.5 × 10−6 tt tt
C. 3.5
D. 3.5 × 106 tt tt
E. 3.5 × 1012 tt tt
Ans: E tt
14. The number of significant figures in 0.00150 is:
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A. t t2 t t
B. 3
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C. t t4 t t
D. 5
t t t t
E. 6
t t t t
Ans: B tt
15. The number of significant figures in 15.0 is:
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A. tt1 tt
B. 2
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C. 3
t t t t
D. 4
t t t t
E. 5
t t t t
Ans: C tt
16. 3.2 tt × 2.7 = tt tt tt
A. 9
B. 8
C. 8.6
D. 8.64
E. 8.640
Ans: C tt
Ans:
B
2 Chapter 1: MEASUREMENT tt