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Chapter 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
6.37 10 m^6 10 3 km m 6.37 103 km, RE
the corresponding circumference, surface area and volume are:
2 , 4 2 , 4 3 E E (^) 3 E
C R A R V R
.
The geometric formulas are given in Appendix E.
ANALYZE (a) Using the formulas given above, we find the circumference to be
2 2 (6.37 103 km) 4.00 104 km.
C RE
(b) Similarly, the surface area of Earth is
2 3 2 8 2
A 4 RE 4 6.37 10 km 5.10 10 km ,
(c) and its volume is
6.37 10 km 1.08 10 km. 3^ E 3
V R
LEARN From the formulas given, we see that C RE , A R E^2 , and V R E^3. The ratios
of volume to surface area, and surface area to circumference are V / A RE / 3 and
A C / 2 RE.
- The conversion factors are: 1 gry 1/10 line, 1 line 1/12 inchand 1 point = 1/
inch. The factors imply that
1 gry = (1/10)(1/12)(72 points) = 0.60 point.
Thus, 1 gry^2 = (0.60 point)^2 = 0.36 point^2 , which means that 0.50 gry = 0.18 point^2 .
- The metric prefixes (micro, pico, nano, …) are given for ready reference on the inside front cover of the textbook (see also Table 1–2).
2 CHAPTER 1
(a) Since 1 km = 1 103 m and 1 m = 1 106 m,
1km 10 m^3 10 m^3 106 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 ^2 m 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 10 6 m m 9.1 105 m.
- (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
- 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 mand 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
(a) the distance d in rods to be
40 rods 4.0 furlongs 4.0 furlongs 160 rods, 1 furlong
d
4 CHAPTER 1
- 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
- The volume of ice is given by the product of the semicircular surface area and the thickness. The area of the semicircle is A = r^2 /2, where r is the radius. Therefore, the volume is
2 2
V r z
where z is the ice thickness. Since there are 10^3 m in 1 km and 10^2 cm in 1 m, we have
3 2 2000 km 10 m^ 10 cm 2000 105 cm. 1km 1m
r
In these units, the thickness becomes
2 3000 m 3000 m 10 cm 3000 102 cm 1m
z
which yields
5 2 2 22 3 2000 10 cm 3000 10 cm 1.9 10 cm. 2
V
- 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.
- (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 10^5 seconds. The ratio is therefore 0.864.
- A day is equivalent to 86400 seconds and a meter is equivalent to a million micrometers, so
m m m day s day
m s
b gc h b gb g
- The time on any of these clocks is a straight-line function of that on another, with slopes 1 and y -intercepts 0. From the data in the figure we deduce
2 594 33 662 ,. C (^) 7 B (^) 7 B (^) 40 A 5 t t t t
These are used in obtaining the following results.
(a) We find
495 s B B (^) 40 A A t ^ t t t
when t'A tA = 600 s.
(b) We obtain t (^) C t (^) C t B tB
b g b^495 g^141 s.
(c) Clock B reads tB = (33/40)(400) (662/5) 198 s when clock A reads tA = 400 s.
(d) From tC = 15 = (2/7) tB + (594/7), we get tB 245 s.
14. The metric prefixes (micro ( ), pico, nano, …) are given for ready reference on the
inside front cover of the textbook (also Table 1–2).
(a) 6
100 y 365 day 24 h 60 min 1 century 10 century 52.6 min. 1 century 1 y 1 day 1 h
^
(b) The percent difference is therefore
52.6 min 50 min 4.9%. 52.6 min
- A week is 7 days, each of which has 24 hours, and an hour is equivalent to 3600 seconds. Thus, two weeks (a fortnight) is 1209600 s. By definition of the micro prefix,
this is roughly 1.21 1012 s.
- We denote the pulsar rotation rate f (for frequency).
3
1 rotation 1.55780644887275 10 s
f (^)
the smallest range of correction, B has the next smallest range, and E has the greatest range. From best to worst, the ranking of the clocks is C, D, A, B, E.
CLOCK Sun. Mon. Tues. Wed. Thurs. Fri. -Mon. -Tues. -Wed. -Thurs. -Fri. -Sat. A (^) 16 16 15 17 15 15 B (^) 3 +5 (^) 10 +5 +6 (^) 7 C (^) 58 58 58 58 58 58 D +67 +67 +67 +67 +67 + E +70 +55 +2 +20 +10 +
LEARN Of the five clocks, the readings in clocks A, B and E jump around from one 24- h period to another, making it difficult to correct them.
- The last day of the 20 centuries is longer than the first day by
20 century^ 0.001 s century^ 0.02 s.
The average day during the 20 centuries is (0 + 0.02)/2 = 0.01 s longer than the first day. Since the increase occurs uniformly, the cumulative effect T is
average increase in length of a day number of days
0.01 s 365.25 day 2000 y day y 7305 s
T
or roughly two hours.
- When the Sun first disappears while lying down, your line of sight to the top of the Sun is tangent to the Earth’s surface at point A shown in the figure. As you stand, elevating your eyes by a height h , the line of sight to the Sun is tangent to the Earth’s surface at point B.
Let d be the distance from point B to your eyes. From the Pythagorean theorem, we have
d^2 r^2 ( r h )^2 r^2^ 2 rh h^2
8 CHAPTER 1
or d^2 2 rh h^2 ,where r is the radius of the Earth. Since r h , the second term can be
dropped, leading to d^2 2 rh. Now the angle between the two radii to the two tangent
points A and B is , which is also the angle through which the Sun moves about Earth
during the time interval t = 11.1 s. The value of can be obtained by using
360 24 h
t
This yields (360 )(11.1 s)
0.. (24 h)(60 min/h)(60 s/min)
Using d r tan , we have d^2 r^2 tan 2 2 rh , or
2
tan
h r
Using the above value for and h = 1.7 m, we have r 5.2 106 m.
- (a) We find the volume in cubic centimeters
3 3 193 gal = 193 gal 231 in^ 2.54 cm 7.31 10 cm^5 1 gal 1in
^
and subtract this from 1 106 cm^3 to obtain 2.69 105 cm^3. The conversion gal in^3 is given in Appendix D (immediately below the table of Volume conversions).
(b) The volume found in part (a) is converted (by dividing by (100 cm/m)^3 ) to 0.731 m^3 , which corresponds to a mass of
c1000 kg m 3 h c0.731 m 2 h= 731 kg
using the density given in the problem statement. At a rate of 0.0018 kg/min, this can be filled in 731kg (^) 4.06 10 min = 0.77 y 5 0.0018 kg min
after dividing by the number of minutes in a year (365 days)(24 h/day) (60 min/h).
- If ME is the mass of Earth, m is the average mass of an atom in Earth, and N is the number of atoms, then ME = Nm or N = ME/m. We convert mass m to kilograms using Appendix D (1 u = 1.661 10 ^27 kg). Thus,
10 CHAPTER 1
m V (1 103 kg/m )(1 m )^3 3 1000 kg.
(b) The total mass of water in the container is
M V (1 103 kg m )(5700 m )^3 3 5.70 106 kg ,
and the time elapsed is t = (10 h)(3600 s/h) = 3.6 104 s. Thus, the mass flow rate R is
6 4
5.70 10 kg 158 kg s. 3.6 10 s
M
R
t
LEARN In terms of volume, the drain rate can be expressed as
3 3 4
5700 m 0.158 m /s 42 gal/s. 3.6 10 s
V
R
t
The greater the flow rate, the less time required to drain a given amount of water.
24. The metric prefixes (micro ( ), pico, nano, …) are given for ready reference on the
inside front cover of the textbook (see also Table 1–2). The surface area A of each grain
of sand of radius r = 50 m = 50 10 ^6 m is given by A = 4 (50 10 ^6 )^2 = 3.14 10 ^8
m^2 (Appendix E contains a variety of geometry formulas). We introduce the notion of
density, m V / , so that the mass can be found from m = V , where = 2600 kg/m^3.
Thus, using V = 4 r^3 /3, the mass of each grain is
3 6 3 9 3
4 kg^4 50 10 m 2600 1.36 10 kg. 3 m 3
r m V
^
^ ^ ^ ^ ^
We observe that (because a cube has six equal faces) the indicated surface area is 6 m^2. The number of spheres (the grains of sand) N that have a total surface area of 6 m^2 is given by 2 8 8 2
6 m 1.91 10. 3.14 10 m
N
Therefore, the total mass M is M Nm 1.91 108 1.36 10 ^9 kg 0.260 kg.
- The volume of the section is (2500 m)(800 m)(2.0 m) = 4.0 106 m^3. Letting “ d ” stand for the thickness of the mud after it has (uniformly) distributed in the valley, then its volume there would be (400 m)(400 m) d. Requiring these two volumes to be equal, we can solve for d. Thus, d = 25 m. The volume of a small part of the mud over a patch of area of 4.0 m^2 is (4.0) d = 100 m^3. Since each cubic meter corresponds to a mass of
1900 kg (stated in the problem), then the mass of that small part of the mud is
1.9 10^5 kg.
26. (a) The volume of the cloud is (3000 m) (1000 m)^2 = 9.4 109 m^3. Since each cubic
meter of the cloud contains from 50 106 to 500 106 water drops, then we conclude that the entire cloud contains from 4.7 1018 to 4.7 1019 drops. Since the volume of
each drop is 43 (10 10 ^6 m)^3 = 4.2 10 ^15 m^3 , then the total volume of water in a cloud
is from 2 103 to 2 104 m^3.
(b) Using the fact that 1 L 1 10 cm 3 3 1 10 ^3 m^3 , the amount of water estimated in
part (a) would fill from 2 106 to 2 107 bottles.
(c) At 1000 kg for every cubic meter, the mass of water is from 2 106 to 2 107 kg. The coincidence in numbers between the results of parts (b) and (c) of this problem is due to the fact that each liter has a mass of one kilogram when water is at its normal density (under standard conditions).
27. We introduce the notion of density, m V / , and convert to SI units: 1000 g = 1 kg,
and 100 cm = 1 m.
(a) The density of a sample of iron is
3 7.87 g cm^3 1 kg^ 100 cm 7870 kg/m.^3 1000 g 1 m
If we ignore the empty spaces between the close-packed spheres, then the density of an individual iron atom will be the same as the density of any iron sample. That is, if M is the mass and V is the volume of an atom, then
26 29 3 3 3
9.27 10 kg 1.18 10 m. 7.87 10 kg m
M
V
(b) We set V = 4 R^3 /3, where R is the radius of an atom (Appendix E contains several geometry formulas). Solving for R , we find
1 3^29 3 1 3 3 3 1.18^10 m 1.41 10 10 m. 4 4
V
R
^
^
^ ^
The center-to-center distance between atoms is twice the radius, or 2.82 10 ^10 m.
4 3 4 3 3
0.0200 g 4.00 10 g/mm 4.00 10 kg/cm. 50.0 mm
m V
^
If we neglect the volume of the empty spaces between the candies, then the total mass of
the candies in the container when filled to height h is M Ah , where
A (14.0 cm)(17.0 cm) 238 cm^2 is the base area of the container that remains
unchanged. Thus, the rate of mass change is given by
( ) (^) (4.00 10 4 kg/cm )(238 cm )(0.250 cm/s) 3 2
0.0238 kg/s 1.43 kg/min.
dM d Ah dh A dt dt dt
- The total volume V of the real house is that of a triangular prism (of height h = 3.0 m and base area A = 20 12 = 240 m^2 ) in addition to a rectangular box (height h ´ = 6.0 m and same base). Therefore,
(^1) 1800 m. 3 2 2
h V hA h A ^ h A
(a) Each dimension is reduced by a factor of 1/12, and we find
V doll m^3 m^3 F H G
I K
(^1800) J
3 c h.^.
(b) In this case, each dimension (relative to the real house) is reduced by a factor of 1/144. Therefore,
V miniature m^3 F 6.0 10 m H G I K
(^1800) J
3 c h 4 3.
- THINK In this problem we are asked to differentiate between three types of tons: displacement ton, freight ton and register ton, all of which are units of volume.
EXPRESS The three different tons are defined in terms of barrel bulk , with
1 barrel bulk 0.1415 m^3 4.0155 U.S. bushels (using 1 m^3 28.378 U.S. bushels ). Thus, in terms of U.S. bushels, we have
4.0155 U.S. bushels 1 displacement ton (7 barrels bulk) 28.108 U.S. bushels 1 barrel bulk
4.0155 U.S. bushels 1 freight ton (8 barrels bulk) 32.124 U.S. bushels 1 barrel bulk 4.0155 U.S. bushels 1 register ton (20 barrels bulk) 80.31 U.S. bushels 1 barrel bulk
14 CHAPTER 1
ANALYZE (a) The difference between 73 “freight” tons and 73 “displacement” tons is
73(freight tons displacement tons) 73(32.124 U.S. bushels 28.108 U.S. bushels) 293.168 U.S. bushels 293 U.S. bushels
V
(b) Similarly, the difference between 73 “register” tons and 73 “displacement” tons is
3
73(register tons displacement tons) 73(80.31 U.S. bushels 28.108 U.S. bushels)
3810.746 U.S. bushels 3.81 10 U.S. bushels
V
LEARN With 1 register ton 1 freight ton 1displacement ton,we expect the difference
found in (b) to be greater than that in (a). This is indeed the case.
- The customer expects a volume V 1 = 20 7056 in^3 and receives V 2 = 20 5826 in.^3 ,
the difference being V V 1 (^) V 2 24600 in.^3 , or
3 3 3
2.54cm 1L 24600 in. 403L 1 inch 1000 cm
V
^
where Appendix D has been used.
- The first two conversions are easy enough that a formal conversion is not especially called for, but in the interest of practice makes perfect we go ahead and proceed formally:
(a)
2 peck 11 tuffets = 11 tuffets 22 pecks 1 tuffet
(b)
0.50 Imperial bushel 11 tuffets = 11 tuffets 5.5 Imperial bushels 1 tuffet
(c)
36.3687 L
11 tuffets = 5.5 Imperial bushel 200 L 1 Imperial bushel
- Table 7 can be completed as follows:
(a) It should be clear that the first column (under “wey”) is the reciprocal of the first
row – so that 9 10 = 0.900,^
3 40 =^ 7.50^ ^10
(^2) , and so forth. Thus, 1 pottle = 1.56 10 (^3) wey
and 1 gill = 8.32 10 ^6 wey are the last two entries in the first column.
(b) In the second column (under “chaldron”), clearly we have 1 chaldron = 1 chaldron (that is, the entries along the “diagonal” in the table must be 1’s). To find out how many
16 CHAPTER 1
- THINK This problem compares the U.K. gallon with U.S. gallon, two non-SI units for volume. The interpretation of the type of gallons, whether U.K. or U.S., affects the amount of gasoline one calculates for traveling a given distance.
EXPRESS If the fuel consumption rate is R (in miles/gallon), then the amount of gasoline (in gallons) needed for a trip of distance d (in miles) would be
(miles) (gallon) (miles/gallon)
d V R
Since the car was manufactured in U.K., the fuel consumption rate is calibrated based on U.K. gallon, and the correct interpretation should be “40 miles per U.K. gallon.” In U.K., one would think of gallon as U.K. gallon; however, in the U.S., the word “gallon” would naturally be interpreted as U.S. gallon. Note also that since1 U.K. gallon 4.5460900 L
and 1 U.S. gallon 3.7854118 L, the relationship between the two is
1 U.S. gallon 1 U.K. gallon (4.5460900 L) 1.20095 U.S. gallons 3.7854118 L
ANALYZE (a) The amount of gasoline actually required is
750 miles 18.75 U.K. gallons 18.8 U.K. gallons 40 miles/U.K. gallon
V
This means that the driver mistakenly believes that the car should need 18.8 U.S. gallons.
(b) Using the conversion factor found above, this is equivalent to
1.20095 U.S. gallons 18.75 U.K. gallons 22.5 U.S. gallons 1 U.K. gallon
V
LEARN One U.K. gallon is greater than one U.S gallon by roughly a factor of 1.2 in volume. Therefore, 40 mi/U.K. gallon is less fuel-efficient than 40 mi/U.S. gallon.
- Equation 1-9 gives (to very high precision!) the conversion from atomic mass units to kilograms. Since this problem deals with the ratio of total mass (1.0 kg) divided by the mass of one atom (1.0 u, but converted to kilograms), then the computation reduces to simply taking the reciprocal of the number given in Eq. 1-9 and rounding off appropriately. Thus, the answer is 6.0 1026.
- THINK This problem involves converting cord , a non-SI unit for volume, to SI unit.
EXPRESS Using the (exact) conversion 1 in. = 2.54 cm = 0.0254 m for length, we have
0.0254 m 1 ft 12 in (12 in.) 0.3048 m 1in
Thus, 1 ft^3 (0.3048 m)^3 0.0283 m^3 for volume (these results also can be found in
Appendix D).
ANALYZE The volume of a cord of wood is (^) V (8 ft) (4 ft) (4 ft) 128 ft^3. Using
the conversion factor found above, we obtain 3 3 3 3 3
0.0283 m 1 cord 128 ft (128 ft ) 3.625 m 1 ft
V
which implies that^3
1 m cord 0.276 cord 0.3 cord
^
LEARN The unwanted units ft^3 all cancel out, as they should. In conversions, units obey the same algebraic rules as variables and numbers.
- (a) In atomic mass units, the mass of one molecule is (16 + 1 + 1)u = 18 u. Using Eq. 1-9, we find
27 18u = 18u 1.6605402^10 kg 3.0 10 26 kg. 1u
(^)
(b) We divide the total mass by the mass of each molecule and obtain the (approximate) number of water molecules: 21 46 26
N
- A million milligrams comprise a kilogram, so 2.3 kg/week is 2.3 106 mg/week. Figuring 7 days a week, 24 hours per day, 3600 second per hour, we find 604800 seconds are equivalent to one week. Thus, (2.3 106 mg/week)/(604800 s/week) = 3.8 mg/s.
- The volume of the water that fell is
^
2 2 2
6 2 6 3
1000 m 0.0254 m 26 km 2.0 in. 26 km 2.0 in. 1 km 1 in. 26 10 m 0.0508 m 1.3 10 m.
V
We write the mass-per-unit-volume (density) of the water as: 1 103 kg m.^3
m V
The mass of the water that fell is therefore given by m = V :
m 1 103 kg m^3 1.3 106 m^3 1.3 109 kg.
75 g 10 g
m
Therefore, in atomic mass units, the average mass of one atom in the common Eastern mole is
23 23
10 g 1.66 10 g 10 u. A 6.02^10
m N
- (a) Squaring the relation 1 ken = 1.97 m, and setting up the ratio, we obtain
1 1
ken^2 m
m m
2 2
2 2
(b) Similarly, we find
1 1
ken^3 m
m m
3 3 3
(c) The volume of a cylinder is the circular area of its base multiplied by its height. Thus,
2 2 3
r h 3.00 5.50 156 ken.
(d) If we multiply this by the result of part (b), we determine the volume in cubic meters: (155.5)(7.65) = 1.19 103 m^3.
- According to Appendix D, a nautical mile is 1.852 km, so 24.5 nautical miles would be 45.374 km. Also, according to Appendix D, a mile is 1.609 km, so 24.5 miles is 39.4205 km. The difference is 5.95 km.
- (a) For the minimum (43 cm) case, 9 cubits converts as follows:
0.43m 9cubits 9cubits 3.9m. 1cubit
And for the maximum (53 cm) case we have
0.53m 9cubits 9cubits 4.8m. 1cubit
(b) Similarly, with 0.43 m 430 mm and 0.53 m 530 mm, we find 3.9 103 mm and 4.8 103 mm, respectively.
(c) We can convert length and diameter first and then compute the volume, or first compute the volume and then convert. We proceed using the latter approach (where d is diameter and is length).
20 CHAPTER 1
3 2 3 3 3 cylinder, min
0.43m 28 cubit 28 cubit 2.2 m. 4 1 cubit
V d
^
Similarly, with 0.43 m replaced by 0.53 m, we obtain V cylinder, max = 4.2 m^3.
- Abbreviating wapentake as “wp” and assuming a hide to be 110 acres, we set up the ratio 25 wp/11 barn along with appropriate conversion factors:
2
28 2
36
100 hide 110 acre 4047 m 1 wp 1 hide 1 acre 1 10 m 1 barn
25 wp 1 10. 11 barn
^
- THINK The objective of this problem is to convert the Earth-Sun distance (1 AU) to parsecs and light-years.
EXPRESS To relate parsec (pc) to AU, we note that when is measured in radians, it is
equal to the arc length s divided by the radius R. For a very large radius circle and small
value of , the arc may be approximated as the straight line-segment of length 1 AU.
Thus,
1 arcmin 1 2 radian 1 arcsec 1 arcsec 4.85 10 rad 60 arcsec 60 arcmin 360
^
Therefore, one parsec is
5 6
1 AU
1 pc 2.06 10 AU 4.85 10
s
Next, we relate AU to light-year (ly). Since a year is about 3.16 107 s,
1ly 186,000mi s 3.16 10 s^7 5.9 1012 mi.
ANALYZE (a) Since 1 pc 2.06 105 AU, inverting the relation gives
6 5
1 pc 1 AU 1 AU 4.9 10 pc. 2.06 10 AU
^ ^
(b) Given that 1 AU 92.9 106 mi and 1 ly 5.9 1012 mi, the two expressions together lead to
6 6 5 12
1 ly 1 AU 92.9 10 mi (92.9 10 mi) 1.57 10 ly 5.9 10 mi
^ ^
