Fundamentals of Physics Extended, Halliday - Solutions, summaries, and outlines. 2022 updated

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 .
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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).



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(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

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


cahiz, or 2.08  102 cahiz.
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already completed part) implies that 1 cuartilla = 48
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.

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(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   11
.  103 acre  ft.
4.3560  10 ft acre  ft
4 3

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8. From Fig. 1-4, we see that 212 S is equivalent to 258 W and 212 – 32 = 180 S is
equivalent to 216 – 60 = 156 Z. The information allows us to convert S to W or Z.

(a) In units of W, we have
 258 W 
50.0 S   50.0 S    60.8 W
 212 S 

(b) In units of Z, we have
 156 Z 
50.0 S   50.0 S    43.3 Z
 180 S 

9. The volume of ice is given by the product of the semicircular surface area and the
thickness. The area of the semicircle is A = r2/2, where r is the radius. Therefore, the
volume is

V  r2 z
2

where z is the ice thickness. Since there are 103 m in 1 km and 102 cm in 1 m, we have

 103 m   102 cm 
r   2000 km       2000  10 cm.
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 1km   1m 

In these units, the thickness becomes
 102 cm 
z  3000 m   3000 m     3000  10 cm
2

 1m 


  3000  10 
2
which yields V  2000  105 cm 2
cm  1.9  1022 cm3.
2

10. Since a change of longitude equal to 360 corresponds to a 24 hour change, then one
expects to change longitude by 360 / 24 15 before resetting one's watch by 1.0 h.

11. (a) Presuming that a French decimal day is equivalent to a regular day, then the ratio
of weeks is simply 10/7 or (to 3 significant figures) 1.43.

(b) In a regular day, there are 86400 seconds, but in the French system described in the
problem, there would be 105 seconds. The ratio is therefore 0.864.

12. A day is equivalent to 86400 seconds and a meter is equivalent to a million
micrometers, so