Test Bank for Fundamentals of Physics 10th Edition by Resnick, Walker and Halliday

TEST BANK For Fundamentals of Physics 10th
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Edition By Resnick, Walker and Halliday
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Chapters 1 - 44
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,Chapter 1 sw




1. Various geometric formulas are given in Appendix E.
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(a) Expressing the radius of the Earth as sw sw sw sw sw sw




R  s w s w 6.37 s w  106 m103 km m  6.37  103
sw sw sw sw sw s w s w sw s w km,

its circumference is s  2 R  2 (6.37  103 km)  4.00104 km.
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s w
s sw sw s w sw sw w
s s w




(b) The surface area of Earth is A  4
sw sw sw sw sw sw sw sw  4  6.37  103
sw sw sw sw sw  5.10  s w sw km2.
R2 108
km 
w
s 2 sw


sw




4 4
 6.37  103 km 
3 sw sw sw
(c) The volume of Earth is V  R3   1.08  1012 km 3 .
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3 3

2. The conversion factors are: 1 gry  1/10 line , 1 line  1/12 inch and 1 point
sw sw s w sw sw sw w
s sw w
s sw sw sw w
s s w sw sw sw




= 1/72inch. The factors imply that
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s sw sw sw sw




1 gry = (1/10)(1/12)(72 points) = 0.60 point.
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Thus, 1 gry2 = (0.60 point)2 = 0.36 point2, which means that 0.50
sw sw sw sw sw sw sw sw sw sw sw sw s w gry 2 = sw s w 0.18 s w point 2 .
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3. The metric prefixes (micro, pico, nano, …) are given for ready reference on the
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insidefront cover of the textbook (see also Table 1–2).
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(a) Since 1 km = 1  103 m and 1 m = 1  106 m,
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1km  103 m  s w sw sw s w s w 103  m m 
sw s w sw
m.
109
m106
sw

sw




The given measurement is 1.0 km (two significant figures), which
s w s w s w s w s w s w s w s w s w



implies our resultshould be written as 1.0  109 m.
s w s w s w w
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(b) We calculate the number of microns in 1 centimeter. Since 1 cm = 102 m,
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1cm = 102 m = s w sw sw sw s w 102m106 sw m sw m.
s w m  104 sw sw




We conclude that the fraction of one centimeter equal to 1.0 m is 1.0 
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, 104.(c) Since 1 yd = (3 ft)(0.3048 m/ft) = 0.9144 m,
sw w
s sw sw sw sw sw sw sw sw sw sw




1

, 2 CHAPTER 1 sw




1.0 yd = sw s w s w 0.91m106  m sw sw s w m  9.1  m. sw s w sw




105
sw




4. (a) Using the conversion factors 1 inch = 2.54 cm exactly and 6 picas = 1
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inch, weobtain
sw s w w
s
 6 picas 
 = 0.80  1.9 picas.
sw sw sw s w

0.80 cm 1 inch
cm    
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1 inch 
sw s w

2.54 cm s w sw
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   
(b) With 12 points = 1 pica, we
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have
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


0.80 cm = 0.80 cm 1 inch  6 picas  12 points 
    23 points.
sw sw s w sw sw sw s w sw sw s w sw s w



sw
sw

2.54 cm 1 inch 1 pica
sw  sw
s w

sw s w
sw s w

sw
s w s w




   


5. Given that 1  201.168 m , 1 rod  5.0292 and 1 chain  20.117 m , we find
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s sw w
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s sw sw



furlong
sw m sw




the relevant conversion factors to be
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1 rod
1.0 furlong  201.168 m  (201.168  40 rods,
sw

sw sw w
s sw sw sw s w sw



m) 5.0292 m
sw w
s




and
1 chain
1.0 furlong  201.168 m  (201.168 10 chains .
sw

s w s w sw s w sw sw sw sw



m) 20.117 m
s w sw
sw




Note the cancellation of m (meters), the unwanted unit. Using the
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given conversionfactors, we find
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s sw sw




(a) the distance d in rods to be
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40
d  4.0 furlongs 4.0 furlongs
s w sw sw sw sw s w
 160 rods,
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rods
sw




1 furlong
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(b) and that distance in chains to be
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10 chains
d  4.0 furlongs 4.0 furlongs  40 chains.
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s w sw sw sw sw sw sw



1 furlong
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6. We make use of Table 1-6.
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(a) We look at the first (―cahiz‖) column: 1 fanega is equivalent to what amount of
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cahiz? We note from the already completed part of the table that 1 cahiz equals a
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s sw sw sw sw sw sw sw sw sw sw sw sw sw sw




dozen fanega. Thus,
sw
12
1 fanega = 1 cahiz, or 8.33  102 cahiz. Similarly, ―1
sw sw sw sw sw s w s w sw sw sw sw sw s w sw




cahiz = 48 cuartilla‖ (in the
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already completed part) implies that 1 cuartilla =
sw sw sw sw sw sw sw
1
sw
cahiz, or 2.08 
sw
48
sw sw sw s w 102 s w cahiz.
sw Continuing s w in s w this s w way, s w the s w remaining s w entries s w in s w the s w first s w column s w are