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

TEST BANK For Fundamentals of Physics 10th
c c c c c c




cEdition By Resnick, Walker and Halliday
c c c c c




Chapters 1 - 44
c c c c

,Chapter 1 c




1. Various geometric formulas are given in Appendix E. c c c c c c c




(a) Expressing the radius of the Earth as c c c c c c




R  6.37  106 m103 km m  6.37  103 km,
c c c c c c c c c c c c




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




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




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


3 3

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



1/72inch. The factors imply that
c c c c c c




1 gry = (1/10)(1/12)(72 points) = 0.60 point.
c c c c c c c




Thus, 1 gry2 = (0.60 point)2 = 0.36 point2, which means that 0.50 gry 2 = 0.18 point 2 .
c c c c c c c c c c c c c c c c c




3. The metric prefixes (micro, pico, nano, …) are given for ready reference on the
c c c c c c c c c c c c c


insidefront cover of the textbook (see also Table 1–2).
c c c c c c c c c c




(a) Since 1 km = 1  103 m and 1 m = 1  106 m,
c c c c c c c c c c c c c c c c




1km  103 m  103 c c c c c  m m  109 m.
c c c c




c m106

The given measurement is 1.0 km (two significant figures), which implies our
c c c c c c c c c c c


cresultshould be written as 1.0  109 m.
c c c c c c c c




(b) We calculate the number of microns in 1 centimeter. Since 1 cm = 102 m,
c c c c c c c c c c c c c c c




1cm = 102 m = c c c c c 102m106  m m  c c c c m.
104
c




We conclude that the fraction of one centimeter equal to 1.0 m is 1.0 
c c c c c c c c c c c c c c




104.(c) Since 1 yd = (3 ft)(0.3048 m/ft) = 0.9144 m,
c c c c c c c c c c c c

,1

, 2 CHAPTER 1 c




1.0 yd = 0.91m106  m m  9.1  105 m.
c c c c c c c c c c




4. (a) Using the conversion factors 1 inch = 2.54 cm exactly and 6 picas = 1 inch,
c c c c c c c c c c c c c c c c


cweobtain
c
 6 picas 
cm = 0.80 cm 1 inch  1.9 picas.
c c c c


0.80 c c c c

  
c c c

 2.54 cm 1 inch  c
c
c c
c
c c


   
(b) With 12 points = 1 pica, we have
c c c c c c c c



0.80 cm = 0.80 cm  1 inch 
 6 picas 
 12 points 
1 pica   23 points.
c c c c c c c c c c c c c
c c c c
2.54 cm 1 inch c c c c c
c c



   


5. Given that 1 furlong  201.168 m , 1 rod  5.0292 m and 1chain  20.117 m, we find
c c c c c c c c c c c c c c c c c c c



the relevant conversion factors to be
c c c c c


1 rod
1.0 furlong  201.168 m  (201.168 m )  40 rods,
c
c c c c c c c c c c

5.0292 m
and
1 chain
1.0 furlong  201.168 m  (201.168 m 10 chains .
c
c c c c c c c c c


) 20.117 m
c
c



Note the cancellation of m (meters), the unwanted unit. Using the given
c c c c c c c c c c c


conversionfactors, we find
c c c c




(a) the distance d in rods to be
c c c c c c

40 rods
d  4.0 furlongs 4.0 furlongs
c
c c c c c c
 160 rods,
c c

1 furlong c




(b) and that distance in chains to be
c c c c c c




10 chains
d  4.0 furlongs 4.0 furlongs  40 chains.
c c
c c c c c c c


1 furlong
c




6. We make use of Table 1-6.
c c c c c




(a) We look at the first (―cahiz‖) column: 1 fanega is equivalent to what amount of
c c c c c c c c c c c c c c


cahiz? We note from the already completed part of the table that 1 cahiz equals a dozen
c c c c c c c c c c c c c c c c c


fanega. Thus, 1 12
c fanega = 1 cahiz, or 8.33  102 cahiz. Similarly, ―1 cahiz = 48
c c c c c c c c c c c c c c c c


cuartilla‖ (in the
c c c




already completed part) implies that 1 cuartilla =
c c c c c c c
1
c

48
c cahiz, or 2.08  c c c c 102 c cahiz.
Continuing in this way, the remaining entries in the first column are 6.94 
c c c c c c c c c c c c c c



103 and
c c



3.47103 . c