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Fundamentals of Physics, 10th Edition Solution Manual by David Halliday, Robert Resnick, Jearl Walker

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The 10th edition of Halliday's Fundamentals of Physics building upon previous issues by offering several new features and additions. The new edition offers most accurate, extensive and varied set of assessment questions of any course management program in addition to all questions including some fo...

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fundamentals of physics
robert resnick
jearl walker
10th edition solution manual by david halliday
the 10th edition of hallidays fundamentals of physics building upon previous issues by offerin

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,Chapter 1

1. THINK In this problem we’re given the radius of Earth, and asked to compute its
circumference, surface area and volume.

EXPRESS Assuming Earth to be a sphere of radius

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

the corresponding circumference, surface area and volume are:
4 3
C  2 RE , A  4 RE2 , V RE .
3
The geometric formulas are given in Appendix E.

ANALYZE (a) Using the formulas given above, we find the circumference to be

C  2 RE  2 (6.37  103 km)  4.00 104 km.

(b) Similarly, the surface area of Earth is

 
2
A  4 RE2  4 6.37  103 km  5.10  108 km2 ,
(c) and its volume is
4 3 4
 
3
V RE  6.37  103 km  1.08  1012 km3.
3 3

LEARN From the formulas given, we see that C RE , A RE2 , and V RE3 . The ratios
of volume to surface area, and surface area to circumference are V / A  RE / 3 and
A / C  2RE .

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 gry2 = 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).

1

,2 CHAPTER 1

(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.

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,

 
1.0 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
 1 inch  6 picas 
0.80 cm =  0.80 cm      1.9 picas.
 2.54 cm  1 inch 
(b) With 12 points = 1 pica, we have

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

5. THINK This problem deals with conversion of furlongs to rods and chains, all of
which are units for distance.

EXPRESS Given that 1 furlong  201.168 m, 1 rod  5.0292 m and 1chain  20.117 m ,
the relevant conversion factors are
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.

ANALYZE Using the above conversion factors, we find

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

, 3

10 chains
(b) and in chains to be d  4.0 furlongs   4.0 furlongs   40 chains.
1 furlong

LEARN Since 4 furlongs is about 800 m, this distance is approximately equal to 160
rods ( 1 rod  5 m ) and 40 chains ( 1 chain  20 m ). So our results make sense.

6. We make use of Table 1-6.

(a) We look at the first (“cahiz”) column: 1 fanega is equivalent to what amount of cahiz?
We note from the already completed part of the table that 1 cahiz equals a dozen fanega.
1
Thus, 1 fanega = 12 cahiz, or 8.33  102 cahiz. Similarly, “1 cahiz = 48 cuartilla” (in the
1
already completed part) implies that 1 cuartilla = 48
cahiz, or 2.08  102 cahiz.
Continuing in this way, the remaining entries in the first column are 6.94  103 and
3.47 103 .

(b) In the second (“fanega”) column, we find 0.250, 8.33  102, and 4.17  102 for the
last three entries.

(c) In the third (“cuartilla”) column, we obtain 0.333 and 0.167 for the last two entries.

1
(d) Finally, in the fourth (“almude”) column, we get 2
= 0.500 for the last entry.

(e) Since the conversion table indicates that 1 almude is equivalent to 2 medios, our
amount of 7.00 almudes must be equal to 14.0 medios.

(f) Using the value (1 almude = 6.94  103 cahiz) found in part (a), we conclude that
7.00 almudes is equivalent to 4.86  102 cahiz.

(g) Since each decimeter is 0.1 meter, then 55.501 cubic decimeters is equal to 0.055501
7.00 7.00
m3 or 55501 cm3. Thus, 7.00 almudes = 12 fanega = 12 (55501 cm3) = 3.24  104 cm3.

7. We use the conversion factors found in Appendix D.

1 acre  ft = (43,560 ft 2 )  ft = 43,560 ft 3

Since 2 in. = (1/6) ft, the volume of water that fell during the storm is

V  (26 km2 )(1/6 ft)  (26 km2 )(3281ft/km) 2 (1/6 ft )  4.66 107 ft 3.

Thus,
4.66  107 ft 3
V  4 3
.  103 acre  ft.
 11
4.3560  10 ft acre  ft

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