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Extended, International Adaptation, 12th
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Edition Halliday (All Chapters included)
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Chapter 1: MEASUREMENT
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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 ty ty
Ans: E ty t y
2. A nanosecond is:
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A. 109 s ty
B. 10−9 s ty
C. 10−10 s t y
D. 10−10 s t y
E. 10−12
Ans: B t y
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 ty ty ty ty ty
C. wavelength of light emitted by Kr86 ty ty ty ty ty
D. a precision meter stick in Paris
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E. the speed of light ty ty ty
Ans: E ty t y
4. In 1866, the U. S. Congress defined the U. S. yard as exactly 3600/3937
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international 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 ty
B. 0.1m
C. 1 m ty
D. 100 m ty
E. 1000 m ty
Ans: C t y
Ans:
B t y
2 Chapter 1: MEASUREMENT ty
,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 quantity
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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 t y
8. A ty gram is: ty
A. 10−6 kg t y
B. 10−3 kg t y
C. 1 kg t y
D. 103 kg t y
E. 106 kg t y
Ans: B t y
9. Which of the following weighs about a pound?
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A. 0.05 kg ty
B. 0.5 kg ty
C. 5 kg ty
D. 50 kg ty
E. 500 kg ty
Ans: D t y
10. (5.0 × 104) × (3.0 × 106) =
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A. 1.5 × 109 ty ty
B. 1.5 × 1010 ty ty
C. 1.5 × 1011 ty ty
D. 1.5 × 1012 ty ty
E. 1.5 × 1013 ty ty
Ans: C t y
11. (5.0 × 104) × (3.0 × 10−6) =
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A. 1.5 × 10−3 ty ty
B. 1.5 × 10−1 ty ty
C. 1.5 × 101 ty ty
D. 1.5 × 103 ty ty
E. 1.5 × 105 ty ty
Chapter 1: MEASUREMENT
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, 12. 5.0 × 105 + 3.0 × 106
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t y
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t y
=
A. 8.0 × 105 ty ty
B. 8.0 × 106 ty ty
C. 5.3 × 105 ty ty
D. 3.5 × 105 ty ty
E. 3.5 × 106 ty ty
Ans: E t y
13. (7.0 × t y t y 106)/(2.0 t y × t y 10−6) t y =
A. 3.5 × ty ty 10−12
B. 3.5 × ty ty 10−6
C. 3.5
D. 3.5 × ty ty 106
E. 3.5 × ty ty 1012
Ans: t y E
14. The number of significant figures in 0.00150 is:
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A. 2
B. 3
C. 4
D. 5
E. 6
Ans: B t y
15. The number of significant figures in 15.0 is:
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A. 1
B. 2
C. 3
D. 4
E. 5
Ans: C t y
16. 3.2 × 2.7 =
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A. 9
B. 8
C. 8.6
D. 8.64
E. 8.640
Ans: C t y
Ans:
B t y
2 Chapter 1: MEASUREMENT ty