,
,
,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:
C 2 R , A 4
4 R2 , V R3 .
E E E
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
A 4 R2 4 6.37 103 km
2
5.10 108 km2 ,
E
(c) and its volume is 4 3
V 6.37 103 km 1.08 1012 km3.
4
R3
E
3 3 2 3
LEARN From the formulas given, we see that C A RE , and V RE . The
RE ratios
,
of volume to surface area, and surface area to V/A RE / and
circumference are 3
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 m.
m 109
m 106
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.
m 104
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.
m 9.1 105
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 1 1.9 picas.
cm inch
2.54 cm 1 inch
(b) With 12 points = 1 pica, we
have
0.80 cm =1 inch
0.80 6 picas 12 points
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 1 201.168 m, 1 rod 5.0292 m and 1 chain
furlong 20.117 m,
the relevant conversion factors are
1.0 furlong 201.168 m (201.168 1 rod 40 rods,
m) 5.0292
and m
1.0 furlong 201.168 m (201.168 1 chain 10 chains .
m) 20.117 m
Note the cancellation of m (meters), the unwanted unit.
ANALYZE Using the above conversion factors, we find
(a) ) the distance d in rods to be d 4.0 furlongs 4.0 160 rods,
40 rods
furlongs
1 furlong
, 3
10 chains
d 4.0 furlongs 4.0 furlongs 40 chains.
(b) and in chains to 1 furlong
be
LEARN Since 4 furlongs is about 800 m, this distance is approximately equal
to 160
rods (1 5 m) and 40 chains (1 20 m ). So our results make sense.
rod chain
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 = cahiz, or 8.33 10 2 cahiz. Similarly, “1 cahiz = 48 cuartilla” (in
1 12
the
already completed part) implies that 1 cuartilla cahiz, or 2.08 10 2
1 48 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.
(d) Finally, in the fourth (“almude”) column, we = 0.500 for the last entry.
1 2
get
(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 m3 or 55501 cm3. Thus, 7.00 almudes = 7.00 fanega = 7.00 (55501 cm3)
= 3.24 104 cm3. 12 12
7. We use the conversion factors found in Appendix D.
1 acre ft = (43,560 ft2 ) ft = 43,560 ft3
Since 2 in. = (1/6) ft, the volume of water that fell during the storm is
V (26 km2 )(1/6 ft) (26 km 2 )(3281ft/km) 2 (1/6 4.66 1 ft3.
7
ft ) 0
Thus, V
,4 CHAPTER 1
3
1.1 10 acre ft. 4.66 107 ft 3
4.3560 104 ft 3 acreft
, 5
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
3
r 2000 km10 m 102 cm 2000 10 cm.
1km 1m 5
In these units, the thickness becomes 2 10 cm
z 3000 m 3000 m 3000 102 cm
1m
2
which yields 2000 105 cm 3000 102 cm 1.9 1022 cm3.
V 2
10. Since a change of longitude equal to 360 corresponds to a 24 hour change,
then one expects to change longitude by 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
, 6 CHAPTER 1
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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:
C 2 R , A 4
4 R2 , V R3 .
E E E
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
A 4 R2 4 6.37 103 km
2
5.10 108 km2 ,
E
(c) and its volume is 4 3
V 6.37 103 km 1.08 1012 km3.
4
R3
E
3 3 2 3
LEARN From the formulas given, we see that C A RE , and V RE . The
RE ratios
,
of volume to surface area, and surface area to V/A RE / and
circumference are 3
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 m.
m 109
m 106
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.
m 104
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.
m 9.1 105
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 1 1.9 picas.
cm inch
2.54 cm 1 inch
(b) With 12 points = 1 pica, we
have
0.80 cm =1 inch
0.80 6 picas 12 points
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 1 201.168 m, 1 rod 5.0292 m and 1 chain
furlong 20.117 m,
the relevant conversion factors are
1.0 furlong 201.168 m (201.168 1 rod 40 rods,
m) 5.0292
and m
1.0 furlong 201.168 m (201.168 1 chain 10 chains .
m) 20.117 m
Note the cancellation of m (meters), the unwanted unit.
ANALYZE Using the above conversion factors, we find
(a) ) the distance d in rods to be d 4.0 furlongs 4.0 160 rods,
40 rods
furlongs
1 furlong
, 3
10 chains
d 4.0 furlongs 4.0 furlongs 40 chains.
(b) and in chains to 1 furlong
be
LEARN Since 4 furlongs is about 800 m, this distance is approximately equal
to 160
rods (1 5 m) and 40 chains (1 20 m ). So our results make sense.
rod chain
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 = cahiz, or 8.33 10 2 cahiz. Similarly, “1 cahiz = 48 cuartilla” (in
1 12
the
already completed part) implies that 1 cuartilla cahiz, or 2.08 10 2
1 48 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.
(d) Finally, in the fourth (“almude”) column, we = 0.500 for the last entry.
1 2
get
(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 m3 or 55501 cm3. Thus, 7.00 almudes = 7.00 fanega = 7.00 (55501 cm3)
= 3.24 104 cm3. 12 12
7. We use the conversion factors found in Appendix D.
1 acre ft = (43,560 ft2 ) ft = 43,560 ft3
Since 2 in. = (1/6) ft, the volume of water that fell during the storm is
V (26 km2 )(1/6 ft) (26 km 2 )(3281ft/km) 2 (1/6 4.66 1 ft3.
7
ft ) 0
Thus, V
,4 CHAPTER 1
3
1.1 10 acre ft. 4.66 107 ft 3
4.3560 104 ft 3 acreft
, 5
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
3
r 2000 km10 m 102 cm 2000 10 cm.
1km 1m 5
In these units, the thickness becomes 2 10 cm
z 3000 m 3000 m 3000 102 cm
1m
2
which yields 2000 105 cm 3000 102 cm 1.9 1022 cm3.
V 2
10. Since a change of longitude equal to 360 corresponds to a 24 hour change,
then one expects to change longitude by 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
, 6 CHAPTER 1
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TO DOWNLOAD THE FULL PDF
CLICK ON THE L.I.N.K
ON THE NEXT PAGE
