Test Bank – Fundamentals of Physics, 9th Edition (Resnick, Halliday & Walker) Verified Test Bank | Complete Newest Version | Physics • Mechanics • Electromagnetism • Exam Prep

,Fundamentals of Physics 10th Edition By Resnick, Walker and Halliday Chapters 1 - 44

Chapter 1

1. Various geometric formulas are given in Appendix E.

(a) Expressing the radius of the Earth as

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

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

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




4 4
 6.37  103 km 
3
(c) The volume of Earth is V  R3   1.08  1012 km3 .
3 3

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

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

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

3. The metric prefixes (micro, pico, nano, …) are given for ready reference on the inside
front cover of the textbook (see also Table 1–2).

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

1km  103 m  103 m106  m m  109 m.

The given measurement is 1.0 km (two significant figures), which implies our result
should be written as 1.0  109 m.

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

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

,
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We conclude that the fraction of one centimeter equal to 1.0 m is 1.0  104.

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


1


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

4. (a) Using the conversion factors 1 inch = 2.54 cm exactly and 6 picas = 1 inch, we
obtain   6 picas 
0.80 cm = 0.80 cm  1 inch    1.9 picas.
2.54 cm 1 inch
  
(b) With 12 points = 1 pica, we have

0.80 cm = 0.80 cm  1 inch   12 points 
 6 picas 
2.54 cm 1 inch 1 pica  23 points.
   

5. Given that 1 furlong  201.168 m , 1 rod  5.0292 m and 1 chain  20.117 m , we find
the relevant conversion factors to be
1 rod
1.0 furlong  201.168 m  (201.168 m )  40 rods,
5.0292 m
and
1 chain
1.0 furlong  201.168 m  (201.168 m ) 10 chains .
20.117 m
Note the cancellation of m (meters), the unwanted unit. Using the given conversion
factors, we find

(a) the distance d in rods to be
40 rods
d  4.0 furlongs 4.0 furlongs  160 rods,
1 furlong