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

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




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




R = w w (6.37 w  106 m)(10−3 km m) = 6.37  103 km,
w w w w w w w w w




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




(b) The surface area of Earth is A = 4
w w w w w w w w = 4 ( 6.37  103
w w w w w = 5.10 
w w km2.
R2 108
km )
w 2 w

w




4 4
( 6.37  103 km )
3 w ww
(c) The volume of Earth is V = R3 = = 1.08  1012 km 3 .
ww w
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3 3

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



w1/72inch. The factors imply that
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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 gry 2 = 0.18 point 2 .
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3. The metric prefixes (micro, pico, nano, …) are given for ready reference on the
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winsidefront 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 = w w w w w (103  m m) =
w w w m.
109
m)(106
w
w




The given measurement is 1.0 km (two significant figures), which implies
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wour resultshould be written as 1.0  109 m.
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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 = w w w w w (10−2m)(106  m m) = w w w w m.
w104

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

,1

, 2 CHAPTER 1



1.0 yd = (0.91m)(106  m m) = 9.1 
w w w w w w w w w m.
105
w




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

1 inch
cm)
w w w w w

  
w


2.54 cm 1 inch w w w
w
w w w w


  
(b) With 12 points = 1 pica, we
w w w w w w w


whave

0.80
 cm = (0.80 cm) 1 inch  6 picas  12 points 
  23 points.
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  
w
w w

2.54 cm 1 inch 1 pica
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w
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   

5. Given that 1 = 201.168 m , 1 rod = 5.0292 and 1 chain = 20.117 m , we find
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furlong
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the relevant conversion factors to be
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1 rod
1.0 furlong = 201.168 m = (201.168 m = 40 rods,
w
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) w
5.0292 m

and
1 chain
1.0 furlong = 201.168 m = (201.168 =10 chains .
w
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m) w
20.117 m w
w



Note the cancellation of m (meters), the unwanted unit. Using the given
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wconversionfactors, we find w w w




(a) the distance d in rods to be
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40
d = 4.0 furlongs =(4.0 furlongs)
w w w w w w
= 160 rods, w w

wrods
1 furlong
w




(b) and that distance in chains to be
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10 chains
d = 4.0 furlongs =(4.0 furlongs) = 40 chains.
w w
w w w w w w w


1 furlong
w




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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wcahiz? We note from the already completed part of the table that 1 cahiz equals a
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wdozen fanega. Thus,
w
12 1 fanega =
1
cahiz, or 8.33  10−2 cahiz. Similarly, “1 cahiz =
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w48 cuartilla” (in the
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already completed part) implies that 1 cuartilla =
w w w w w w w w
1
wcahiz,
48 w or 2.08 
w w w 10−2 w cahiz.
w Continuing in this way, the remaining entries in the first column are 6.94 
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