Exam (elaborations) TEST BANK FOR Fundamentals of Physics 9th Edition By Resnick, Walker and Halliday (Instructor Solution Manual)

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




Edition By Resnick, Walker and Halliday
m m m m m m




Chapters 1 - 44
m m m m

,Chapter1 m




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




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




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




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




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




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




m m m m m m m m mm mm m m mm m m m mm




3 3

2. The conversion factors are: 1 gry =1 /10 line , 1 line =1 /12 inch and 1 point = 1/72inch.
m m m m m m m m m m m m m m m m m m m m




The factors imply that
m m m m




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




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




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




front cover of the textbook (see also Table 1–2).
m m m m m m m m m




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




1km = 103 m = (103 m)(106 m m)=109 m m m m m m m m
m
m m.

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




resultshould be written as 1.0  109 m.
m m m m m m m m m




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




1cm = 10−2 m = (10−2m)(106  m m) = 104
m m m m m m m m
m
m
m.

We conclude that the fraction of one centimeter equal to 1.0 m is 1.0  10−4.(c)
m m m m m m m m m m m m m m m m




Since 1 yd = (3 ft)(0.3048 m/ft) = 0.9144 m,
m m m m m m m m m m




1

,
, 2 CHAPTER 1 m




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

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




obtain  6 picas
0.80 cm = (0.80 cm)  1inch 
m




  1.9 picas.
m m
m m




m m m m
m m m m




2.54 cm 1inch
m m




m m m m




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




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




2.54 cm 1inch 1 pica
m m m m

m m



m m m m m




   

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




the relevant conversion factors to be
m m m m m




1rod
1.0 furlong =201.168 m = (201.168 m) = 40 rods,
m




m m m m m m m m m m




5.0292 m
and
1chain
=10 chains. 1.0 furlong =201.168 m = (201.168 m)
m




m m m m m m m m m m




20.117 m m




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




conversionfactors, we find
m m m m




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




40 rods
d = 4.0 furlongs =(4.0 furlongs) =160 rods,
m




m m m m m
m

m m




1furlong m




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




10 chains
d = 4.0 furlongs =(4.0 furlongs) = 40 chains.
m m




m m m m m m m




1 furlong m




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




(a) We look at the first (“cahiz”) column: 1 fanega is equivalent to what amount of cahiz? We
m m m m m m m m m m m m m m m m




note from the already completed part of the table that 1 cahiz equals a dozen fanega. Thus, 1
m m m m m m m m m m m m m m m m m m




fanega = 1 cahiz,12or 8.33  10−2cahiz. Similarly, “1 cahiz = 48 cuartilla” (in the
m m m m m m m m m m m m m m m m m




m 2.08  10−2 cahiz. Continuing in
already completed part) implies that 1 cuartilla = 1 cahiz, or48 m m m m m m m m m m m m m m m




m this way, the remaining entries in the first column are 6.94  10−3 and
m m m m m m m m m m m m m




3.4710−3. m