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Unformatted text preview: Q1.7 Q1.8 Q1.9 Q1.10 Q1.11 Physics and Measurement ANSWERS TO QUESTIONS Q1.1 Atomic clocks are based on electromagnetic waves which atoms
emit. Also, pulsars are highly regular astronomical clocks. Q1.2 Density varies with temperature and pressure. It would be
necessary to measure both mass and volume very accurately in
order to use the density of water as a standard. Q13 People have different size hands. Defining the unit precisely
would be cumbersome. Q1.4 (a) 0.3 millimeters (b) 50 microseconds (c) 7.2. kilograms Q1.5 and You cannot add or subtract quantities of different
dimension. Q1.6 A dimensionally correct equation need not be true. Example:
1 chimpanzee = 2. chimpanzee is dimensionally correct. If an
equation is not dimensionally correct, it cannot be correct. If I were a runner, I might walk or run 101 miles per day. Since I am a college professor, I walk about 100 miles per day. I drive about 40 miles per day on workdays and up to 2.00 miles per day on
vacation. On February 7, 2.001, I am 55 years and 39 days old. 86 400 s 55 yr[ 365.25 d
+39 d=20128 d[
1yr )=1.74><109s~109s. Many college students are just approaching 1 Gs. Zero digits. An orderof—magnitude calculation is accurate only within a factor of 10.
The mass of the fortysix chapter textbook is on the order of 100 kg . With one datum known to one significant digit, we have 80 million yr + 2.4 yr 2 80 million yr. 2 Physics and Measurement SOLUTIONS TO PROBLEMS Section 1.1 Standards of Length, Mass, and Time No problems in this section Section 1.2 P1.1 Section 1.3 >*P1.2 P1.3 >*P1.4 P1.5 Matter and ModelBuilding From the figure, we may see that the spacing between diagonal planes is half the distance between
diagonally adjacent atoms on a flat plane. This diagonal distance may be obtained from the Pythagorean theorem, Ldiag = \le + L2 . Thus, since the atoms are separated by a distance L = 0.200 nm , the diagonal planes are separated by g L2 +LZ = 0.141 nm . Density and Atomic Mass 3
Modeling the Earth as a sphere, we find its volume as 37173 = $746.37 X 106 m) = 1.08 X 1021 m3. Its 5.98 1024 k
density is then p = E = # = 5.52. X 103 kg / In3 . This value is intermediate between the V 1.08>< 1021 m3
tabulated densities of aluminum and iron. Typical rocks have densities around 2. 000 to 3 000 kg / In3 . The average density of the Earth is significantly higher, so higherdensity material
must be down below the surface. With V 2 (base area)(height) V = (ﬂrz)h and p = %, we have
p_ m _ 1 kg 109 mm3
ﬂrzh 7r(19.5 mm)2 (39.0 mm) 1 In3 p: 2.15><104 kg/ms . Let V represent the volume of the model, the same in p = g for both. Then piron = 9.35 kg/V and m 0 0 m 0 19.3 103 k 3
pgold : g M ,Next, pg 1d 2 g M and mgold =9.35 kg XB—g/ms = 23.0 kg .
piron 9.35 kg 7.86 X 10 kg /m
4
V=V0 —VZ =§ﬂ(1'23—1'13) 3_ 3
P—rgﬁompV—p[471)(rZ3 r13): —4ﬂp(1: Tl) . Chapter 1 3 P1.6 For either sphere the volume is V = $7273 and the mass is m = pV = pgﬂrs. We divide this equation
for the larger sphere by the same equation for the smaller:
ﬂ 2 “177—033 2 i 2 5
ms p47r1'533 1'53
Then r, = r35: 4.50 cm(1.71) = 7.69 cm .
P1.7 Use 1 u = 1.66 X 10’24 g.
1.66 10'24
(a) For He, m0 = 4.00 4%] = 6.64x 10’24 g .
u
1.66 10'24
(b) For Fe, m0 = 55.9 4%] = 9.29 X 10*23 g .
u
1.66 10’24
(c) For Pb, m0 = 207 4%] = 3.44 X 10’22 g .
u
>9P1.8 (a) The mass of any sample is the number of atoms in the sample times the mass m0 of one atom: m = N m0. The first assertion is that the mass of one aluminum atom is
m0 = 270 u = 270 u X 1.66 X 10’27 kg/1 u = 4.48 X 10’26 kg. Then the mass of 6.02 X 1023 atoms is m = Nm0 = 6.02 X 1023 X 4.48 X 10*26 kg = 0.027 0 kg = 27.0 g. Thus the first assertion implies the second. Reasoning in reverse, the second assertion can be
written m = Nmo. 0.027 kg 0.0270k =6.02><1023m ,so m =—
g 0 0 6.02X10Z3 = 4.48 X 10’26 kg, in agreement with the first assertion. (b) The general equaﬁon m = N m0 applied to one mole of any substance gives M g 2 NM u,
where M is the numerical value of the atomic mass. It divides out exactly for all substances, giving 1.000 000 0 X 10’3 kg = N 1.660 540 2 X 10727 kg. With eightdigit data, we can be quite
sure of the result to seven digits. For one mole the number of atoms is N: 4 105+” = 6.022137x1023 .
1.660 540 2 (c) The atomic mass of hydrogen is 1.008 0 u and that of oxygen is 15.999 u. The mass of one
molecule of H20 is 2(1.008 0)+ 15.999 u = 18.0 u. Then the molar mass is 18.0 g . (d) For COZ we have 12.011 g +2(15.999 g) = 44.0 g as the mass of one mole. 4 Physics and Measurement P1.9 P1.10 P1.11 1 kg
103 g Mass of gold abraded: ,Am i = 3.80 g — 3.35 g = 0.45 g = (0.45 g)[ J: 4.5 X 10 1.66 X 10’27 kg 1u Each atom has mass m0  197 u  197 u[ ] 3.27 X 10’25 kg. Now, Am I = ‘ AN ‘mo, and the number of atoms missing is Am 4.5><10’4 k
I _ g _ —25 = 1.38 X 1021 atoms.
m0 3.27 X 10’ kg MN= The rate of loss is ’4 kg. IAN‘_1.38><1021 atoms 1yr [1d)[ 1h )[1 min
At 50 yr 365.25 d 24 h 60 min 60 s J AN  11
— = 8.72 X 10 atoms/s .
At
3
(a) 1’11sz3 =(7.86 g/Cm3)(5.00 X 10*6 cm) = 9.83 X 10*16 g 29.83 X 10*19 kg
9.83 10*19 k
(b) N— m — X g — 1.06x107 atoms
mo 55.9 u(1.66 x 10’27 kg /1 u)
(a) The crosssectional area is +15.0 cm+\_
A = 2(0.150 m)(0.010 m)+(0.340 m)(0.010 m) 1.00+ + I
_ 73 2  cm
—6.40><10 m . 36.0
cm
The volume of the beam is 1.00
cm
V = AL = (6.40 X 10’3 m2)(1.50 m) = 9.60 X 10’3 m3. _
Thus, its mass is
FIG. P1.11
m = pV = (7.56 X 103 kg/m3)(9.60 X 10’3 m3) = 72.6 kg .
727
(b) The mass of one typical atom is m0 = (55.9 = 9.28 X 10’26 kg . Now
u
72.6 k
m = Nmo and the number of atoms is N = i = —j§ = 7.82 x 1026 atoms .
m0 9.28 X 10 kg Chapter 1 5 1.66 X 10’27 kg P1.12 (a) The mass of one molecule is m0 = 18.0 u[ 1
u J: 2.99 X 10’26 kg. The number of molecules in the pail is 1.20 k
m g  4.02>< 1025 molecules . N — = — —
m0 2.99 X 10’26 kg pail : (b) Suppose that enough time has elapsed for thorough mixing of the hydrosphere. m a 1.20 k
Nboth = Npaﬂ P 11 = (4.02 x 1025 molecules) —2% ,
Mtotal X or
N both 2 3.65 x 104 molecules .
Section 1.4 Dimensional Analysis
P1.13 The term x has dimensions of L, a has dimensions of LT’Z, and t has dimensions of T. Therefore, the equation x = kamt” has dimensions of
L = (LT’Z)m(T)” or L1T° = L’”T”’2m. The powers of L and T must be the same on each side of the equation. Therefore, L1=Lm and m=1. Likewise, equating terms in T, we see that n — 2m must equal 0. Thus, 11 = 2 . The value of k, a dimensionless constant, cannot be obtained by dimensional analysis . >*P1.14 (a) Circumference has dimensions of L.
(b) Volume has dimensions of L3.
(c) Area has dimensions of L2. 1/2
Expression has dimension L(LZ) 2 L2, so this must be area Expression (ii) has dimension L, so it is (a).
Expression (iii) has dimension L(LZ) 2 L3, so it is Thus, (a) 2 ii; (b) : iii, (c) = i . 6 Physics and Measurement P1.15 (a) This is incorrect since the units of [ax] are mZ/sZ , while the units of [v] are m/s. (b) This is correct since the units of [y] are In, and cos(kx) is dimensionless if [k] is in mil.
21: 2F . . .
>9P1.16 (a) a cx: — or a = k represents the proportlonallty of acceleratlon to resultant force and
m m
the inverse proportionality of acceleration to mass. If k has no dimensions, we have
[P] L M L
— k , — 1 , F — .
[a] [1M T. M [1 T2 ML l<gm T2 s2 (b) In units, , so 1 newton: 1 kg'm/s2 . P1.17 Inserting the proper units for everything except G, [kg m] _ CW]2 32 _ [H112 ' Multiply both sides by [m]2 and divide by [kg]Z ; the units of G are Section 1.5 Conversion of Units >9P1.18 Each of the four walls has area (8.00 ft)(12.0 ft) = 96.0 ftz. Together, they have area 2
4(96.0ft2)[ 1”“ j = 35.7 m2 .
3.28 ft P1.19 Apply the following conversion factors: 1in=2..54cm,1d=86400s,100cm=1m,and109 nm=1m = 9.19 nm/s . 1 . (2.54 cm/in)(10’Z m/cm)(109 nm/m)
[E m/dayl 86 400 s/day This means the proteins are assembled at a rate of many layers of atoms each second! = 1.39>< 10’4 m3 1in 0.025 4 3
*P1.20 8.50 in3 28.50 in3[—m) Chapter 1 7 P1.21 Conceptualize: We must calculate the area and convert units. Since a meter is about 3 feet, we should
expect the area to be about A z (30 m)(50 m) = 1 500 m2 . Categorize: We model the lot as a perfect rectangle to use Area = Length x Width. Use the
conversion: 1 m: 3.281 ft. 3.2.81 ft Analyze: A=LW=(100ft)[ 1”“ )(15011)[ 1”“ j: 1390m2 = 1.39x103 m2 . Finalize: Our calculated result agrees reasonably well with our initial estimate and has the proper
units of m2 . Unit conversion is a common technique that is applied to many problems. P1.22 (a) V = (40.0 m)(20.0 m)(12.0 m) = 9.60 X 103 m3
V:9.60 X 103 m3(3.28 11/1 m)3 = 3.39 X 105113 (b) The mass of the air is
m = pairV= (1.20 kg/m3)(9.60 X 103 m3): 1.15 X 104 kg.
The student must look up weight in the index to find
Pg = mg: (1.15 X 104 kg)(9.80 m/sz) = 1.13 X 105 N. Converting to pounds, Pg = (1.13 X 105 N)(11b/4.45 N) = 2.54 X 104 1b . P1.23 (a) Seven minutes is 42.0 seconds, so the rate is
r 2 300 gal = 7.14>< 10’Z gal/s .
42.0 s
(b) Converting gallons first to liters, then to m3,
73 3
r = (7.14 X 10’2 gal/s) 3786 L 10—1“
1 gal 1 L 1': 2.70x10’4 m3/s . (c) At that rate, to fill a 1m3 tank would take 3
t— 1H}, 3 1h —1.03h.
2.70><10 m /s 3600 8 Physics and Measurement >*P1.24 P1.25 P1.26 >*P1.27 P1.28 (a) Length of Mammoth Cave 2 348 = 560 km 2 5.60 x 105 m = 5.60 x 107 cm .
(b) Height of Ribbon Falls 2 1 612 ft[%) = 491 m = 0.491 km 2 4.91 x 104 cm . (c) Height of Denali = 20 320 ft[%) = 6.19 km = 6.19 x 103 m = 6.19 x 105 cm . (d) Depth of King's Canyon 2 8 200 ft[%) = 2.50 km 2 2.50 x 103 In 2 2.50 x 105 cm . From Table 1.5, the density of lead is 1.13 x 104 kg / In3 , so we should expect our calculated value to be close to this number. This density value tells us that lead is about 11 times denser than water,
which agrees with our experience that lead sinks. Density is defined as mass per volume, in p = We must convert to SI units in the calculation.
23.94 1 k 3 p=_g3 g [100 cm) = 1.14><104 kg/ms
2.10 cm 1 000 g 1 m At one step in the calculation, we note that one million cubic centimeters make one cubic meter. Our
result is indeed close to the expected value. Since the last reported significant digit is not certain, the
difference in the two values is probably due to measurement uncertainty and should not be a
concern. One important commonsense check on density values is that objects which sink in water must have a density greater than 1 g / cm3 , and objects that float must be less dense than water. It is often useful to remember that the 1 600m race at track and field events is approximately 1 mile
in length. To be precise, there are 1 609 meters in a mile. Thus, 1 acre is equal in area to '2 1609 2
(1 acre)[1i][ m) = 4.05><103 mZ . 640 acres mi 2 000 lb '
The weight flow rate is 1 200 ton [ 1 h. )[1 mm)  667 lb/s .
h ton 60 mm 60 s 1 mi 2 1 609 m = 1.609 km; thus, to go from mph to km/ h, multiply by 1.609. (a) 1 mi/h= 1.609 km/h (b) 55 mi/h= 88.5 km/h (c) 65 mi/h=104.6 km/h. Thus, A22 2 16.1 km/h . Chapter 1 9 P1 29 (a) 6 X 1012 $ 1 h 1 day 1 yr  190 ears
' 1 000 $/s 3600s 24 h 365 days y (b) The circumference of the Earth at the equator is 271(6378 X 103 In) 2 4.01 X 107 In. The length of one dollar bill is 0.155 In so that the length of 6 trillion bills is 9.30 X 1011 In. Thus, the
6 trillion dollars would encircle the Earth 9.30 X 1011 m 7 = 2.32.><104 times .
4.01><0 In 1.99 1030 k
P1.30 N 2 ms“ 2 X g 2 atoms —27 1.19 X 1057 atoms
matom 1.67 X 10’ kg 73 3
P1.31 V=At so tzlzwz 1.51>< 10*4 m (or 151 ,uIn) A 25.0 m2 1 [(13.0 acres)(43 560 ftZ/acre)] P132 V — Bh —
3 3 = 9.08 X 107 £13, 72 3
V = (9.08 X 107 PEWMJ 1 ft3
FIG. P132 (481 ft) 01‘ = 2.57x106 m3 P133 Pg (2.50 tons/block)(2.00><106 blocks)(20001b/ton)= 1.00><1010 lbs >*P1.34 The area covered by water is Z
Aw = 0,7031%1h = (0.70)(472R§mh) (0.70)(47z)(6.37>< 106 m) = 3.6 X 1014 m2 . The average depth of the water is
d = (2.3 Iniles)(1 609 In/l mile) = 3.7 X 103 m.
The volume of the water is V = Awd = (3.6 X 1014 m2)(3.7 X 103 m) = 1.3 X 1018 m3 and the mass is m:pV:(1000 kg/m3)(1.3><1018 m3): 1.3x 1021 kg . 10 Physics and Measurement d atom, scale J 2 (2.40 X 10715 : 6.79 X 1073 ft, or P135 a d = d
( ) nucleus, scale nucleus, real[ d X 10,10 In atom, real dnudeus, scale = (6.79 X 10’3 ft)(304.8 mm/l ft): 2.07 mm
r3 3
b Vatom _ 4”?” _ ratom 3 _ dam 3 _ 1.06x10’10 m
( ) Vnucleus 4ﬂrgudeus rnucleus dnucleus X 10715 m 3 = 8.62 X 1013 times as large between distance factor 1.4 X 109 m >*P136 scale distance real scale ’3
=( ]= (4.0 X 1013 km)(m] = 200 km P137 The scale factor used in the "dinner plate” model is 0.25 In S 2 5—
1.0 X 10 lightyears = 2.5 X 10’6 m/lightyears . The distance to Andromeda in the scale model will be Dscale = DactuaIS = (2.0 X 106 lightyears)(2.5 X 10’6 m/lightyears) = 5.0 m . A 4,,,2 h r 2 (6.37x106 m)(100 cm/m) Z (a) Earth : liart : Earth 2 8 Z AMOOr1 477 rMoon rMoon 1.74 X 10 cm 4771'sa 3 6 3
(b) VEarth _ 3m _ ,Emh _ (6.37 X 10 m)(100 cm/m) _ 49.1 VMoon 4ﬂr§°m rMoon X 108 CHI P139 To balance, 1tie 2 mAl or pFeVFe = pAlVAl p (4)777 32p (2)777 3
Fe 3 Fe Al 3 Al 1/ 3 7 86 1/ 3
7A1 : TFe[pFe] = (2.00 cm)['—j = 286 CH1 
M 2.70 Chapter 1 11 P140 The mass of each sphere is 471p r 3
mAl = PAlVAl = $
and
47w J as
mFe : pFeVFe : L
3
Setting these masses equal,
477PA17A13 477PF 7F 3 PF
—=—e e and rAl =rFes e
3 3 PM
Section 1.6 Estimates and Orderof—Magnitude Calculations P1.41 Model the room as a rectangular solid with dimensions 4 In by 4 In by 3 In, and each pingpong ball as a sphere of diameter 0.038 In. The volume of the room is 4 X 4 X 3 = 48 In3 , while the volume of one ball is
3
4%[0'0328 1“) = 2.87 X 10*5 m3.
. 48 6 . .
Therefore, one can ﬁt about W ~ 10 p1ngpong balls 1n the room.
. X As an aside, the actual number is smaller than this because there will be a lot of space in the
room that cannot be covered by balls. In fact, even in the best arrangement, the socalled "best packing fraction” is éﬂﬁ = 0.74 so that at least 26% of the space will be empty. Therefore, the above estimate reduces to 1.67 X 106 X 0.740 ~ 106 . P1.42 A reasonable guess for the diameter of a tire might be 2.5 ft, with a circumference of about 8 ft. Thus,
the tire would make (50 000 mi)(5 280 ft /mi)(1 rev/8 ft): 3 X 107 rev ~ 107 rev . P1.43 In order to reasonably carry on photosynthesis, we might expect a blade of grass to require at least % 'n2 = 43 X 10’5 ft2 . Since 1 acre 2 43 560 ftZ, the number of blades of grass to be expected on a quarteracre plot of land is about total area _ (0.25 acre)(43 560 ft2 /acre) 11  5 Z  2.5 X 107 blades ~ 107 blades .
area per blade 43 x 10’ ft /blade 12 Physics and Measurement P1.44 A typical raindrop is spherical and might have a radius of about 0.1 inch. Its volume is then approximately 4 X 10’3 in3 . Since 1 acre 2 43 560 ftZ, the volume of water required to cover it to a
depth of 1 inch is Z . Z
(1 «mm inch) = (1 144 m
1 acre 1 ftz Jz63>< 106 ms. The number of raindrops required is n _ volume of water required _ 6.3 X 1036 in: _ 1.6 X 109 ~ 109
volume of a s1ngle drop 4 x 10’ in >*P1.45 Assume the tub measures 1.3 m by 0.5 m by 0.3 m. Onehalf of its volume is then
V = (0.5)(1.3 m)(0.5 m)(0.3 m) = 0.10 m3 . The mass of this volume of water is mmter = pmrvz (1 000 kg/m3)(0.10 m3) = 100 kg ~10Z kg . Pennies are now mostly zinc, but consider copper pennies filling 50% of the volume of the tub. The
mass of copper required is mmpper = mePerv = (8 920 kg/m3)(0.10 m3) = 892 kg ~ 103 kg . P1.46 The typical person probably drinks 2 to 3 soft drinks daily. Perhaps half of these were in aluminum
cans. Thus, we will estimate 1 aluminum can disposal per person per day. In the U.S. there are ~250
million people, and 365 days in a year, so (250x 106 cans/day)(365 days/year); 1011 cans are thrown away or recycled each year. Guessing that each can weighs around 1/10 of an ounce, we estimate this represents (1011 cans)(0.1 oz/can)(1 lb/16 oz)(1 ton/2 0001b)z 3.1 X 105 tons/year. ~ 105 tons P1.47 Assume: Total population 2 107 ; one out of every 100 people has a piano; one tuner can serve about
1 000 pianos (about 4 per day for 250 weekdays, assuming each piano is tuned once per year).
Therefore, 1 .
# tuners ~ mi ﬂ (107 people): 100 .
1 000 p1anos 100 people Chapter 1 13 Section 1.7 Significant Figures >"P148 METHOD ONE
We treat the best value with its uncertainty as a binomial (2.1.3 i0.2) cm (9.8 $0.1) cm, A = [21.3(9.8) i 21.3(0.1) i0.2(9.8) i (0.2)(0.1)] cmZ . The first term gives the best value of the area. The cross terms add together to give the uncertainty
and the fourth term is negligible. A = 2.09 cm2 i4 cm2 METHOD TWO
We add the fractional uncertainties in the data. A = (21.3 cm)(9.8 cm) :[£+ﬂj = 209 cm2 i2% = 209 cm2 :4 cm2
21.3 9.8
P1.49 (a) 7172 = 72(105 m 50.2 m)2 = 72 [(10.5 m)2 :2(10.5 m)(0.2 m) +(0.2 m)2] = 346 m2 i13 m2 (b) 2771’ = 277(105 m i 0.2 m): 66.0 mi1.3 m P1.50 (a) 3 (b) 4 (c) 3 (d) 2
121,51 1’ = (6.50 : 0.20) cm = (6.50 i020) X 10’2 m
m = (1.85 : 0.02) kg
_ m
p 8w
also,
5 5 35
_P 2 _m + _’ _
p m r In other words, the percentages of uncertainty are cumulative. Therefore, 5_p _ 0.02 + 3(0.20)
p 1.85 6.50 = 0.103 , p=i= 1.61x103 kg/m3 (%)7Z(6.5 X 10’2 m)3 and pi5p= (1.61i0.17)x103 kg/m3 =(1.6i0.2)x103 kg/ms. 14 Physics and Measurement P1.52 (a) 756??
37.2? 0.83 + 2.5? 796.53 = 797 (b) 0.003 2(2 5.1.) X 356.3(4 5.1.) = 1.140 16 = (2 5.1.) 1.1 (c) 5.620(4 s.f.) >< 74> 4 s.f.) = 17.656 = (4 s.f.) 17.66 >*P1.53 We work to nine significant digits: 365.242199d '
1yr1yr [2411)[60 mum 60,5 )— 31556926.0s .
1yr 1d 1h 1m1n P154 The distance around is 38.44 m + 19.5 m + 38.44 m + 19.5 In 2 115.88 In , but this answer must be
rounded to 115.9 In because the distance 19.5 In carries information to only one place past the decimal. 115.9 In P1.55 V = 2V1 + 2V2 = 2(1/1 + V2) 7 19.0 m —» V (170 +10m+10m)(10m)(009m) 170m3 = . m . . . . = . V1 = (10.0 m)(1.0 m)(0.090 m) = 0.900 m3 Z 10l0 m g
2 3 3 3 l V = 2(170 m +0.900 m )= 5.2 m FIG. P1.55 % = 0'12 m = 0.0063
31 19.0 m 6"” = 0'01 m = 0.010 W — 0.006 : 0.010 1 0.011 — 0.027 — 3% ml 1.0 m V
i = 0'1 cm = 0.011
151 9.0 cm Additional Problems
P1.56 It is desired to find the distance x such that x _1000m
100m x (i.e., such that x is the same multiple of 100 m as the multiple that 1 000 m is of x). Thus, it is seen that
x2 = (100 m)(1 000 m) = 1.00 X 105 m2 and therefore x=\/1.00><105 m2 = 316 m . Chapter 1 15 >9P1.57 Consider one cubic meter of gold. Its mass from Table 1.5 is 19 300 kg. One atom of gold has mass 1.66 10’27 k
m0 = (197 u)[#] = 3.27 X 10’25 kg.
1 u
50, the number of atoms in the cube is
19 300 kg = —%= 5.90 X 1028.
3.27 X 10 kg The imagined cubical volume of each atom is 1m3 = W = 1.69 X 10*29 m3.
. >< d3 50 P1.58 Atotal =(N)(Admp)=[vt°tal](Admp) Vt? (472%) V 47!?
drop 3
76 3
AW _[3Vtota1)_ 3 30.0 x 10 75m _ 450 m2
1’ 2.00 X 10 m P1.59 One month is 1 mo = (30 day)(2.4 h/day)(3 600 s/h) = 2.592 X 106 s.
Applying units to the equation, V=(1.50 Mft3/mo)t+(0.008 00 Mft3/moz)tz. Since 1 Mft3 :106 £13,
V = (1.50 X 106 £13 /mo)t +(0.008 00 X 106 £13 /moz)t2. Converting months to seconds, _ 1.50 X 106 113/5101f ‘ 0.008 00 X 106 113/11102 t2 V
2.592x106 s/mO (2.592x106 s/mo)Z Thus, V[ft3]=(0.579ft3‘/s)t+(1.19x10’9 £13/s2)tZ . 16 Physics and Measurement 2.4.6° P1.61 2711’ = 15.0 In
7 = 2.39 In 2 tan 55.0o E
1,
h = (2.39 m)tan(55.0°) = 3.41 In FIG. P1.61 >*P1.62 Let d represent the diameter of the coin and h its thickness. The mass of the gold is Zﬂdz
4 m=pV=pAt=p[ +71%}? where t is the thickness of the plating. (2.41)Z
4 m = 19.3[272 + 7r(2.41)(0.178)](0.18 X 104) = 0.003 64 grams
cost = 0.003 64 grams >< $10/gram = $0.036 4 = 3.64 cents This is negligible compared to $4.98.
P1.63 The actual number of seconds in a year is (86 400 s/day)(365.25 day/yr) = 31 557 600 s/yr. The percent error in the approximation is ‘(ﬂ X 107 s/yr) —(31 557 600 s/yr)‘ ><100%= 0.449% .
31557600 s/yr Chapter 1 17
P1.64 (a) [V] = L3, [A] = L2, [h] = L [V] = [A][h] L3 = LZL 2 L3. Thus, the equation is dimensionally correct. (b) chﬁnder =72R2h=(7zR2)h=Ah,where 14:7sz
Vrectangular object 2 [wk 2 (£w)h 2 Ah , where A = [w P1.65 (a) The speed of rise may be found from _ (Vol rate of flow) _ 16.5 cm3/s _ 22 DZ 2 0.529 cm/s .
(Area: H4 ) —”(6'32 cm)
(b) Likewise, at a 1.35 cm diameter,
16.5 3
222%: 11.5 cm/s .
I 4
P1.66 (a) 1 cubic meter of water has a mass m = pV = (1.00 X 10’3 kg /cm3)(1.00 m3)(102 cm/m)3 = 1 000 kg (b) As a rough calculation, we treat each item as if it were 100% water. cell: m = pV = p[§ﬂR3) = p[%ﬂD3) = (1 000 kg/m3)[%7rj(1.0 X 10’6 m)3 = 5.2 X 10’16 kg kidney: m = pV = p[§ﬂR3) = (1.00 X 10’3 kg/cm3 {37001.0 cm)3 = 0.27 kg fly: m = pgozhj =(1 X 10*3 kg /cm3)(3(2.0 mm)2(4.0 mm)(10’1 cm/mm)3 = 1.3 X 10’5 kg _ (108 cars)(104 mi/yr) _ 10 VZOmPg —W—5.0X10 (108 cars)(104 mi/yr) 10
V = — = 4.0 10 l
25 mpg X ga Fuel saved 2 V25 mpg — V20 mpg = 1.0 x 1010 gal/yr 18 P1.68 P1.69 P1.70 P1.71 Physics and Measurement v _ 5.00 furlongs 220 yd 0.914 4 m 1 fortn1ght [ 1 day) 1 hr _ 8.32 X 104 m/S
fortn1ght 1 furlong 1 yd 14 days 2.4 hrs 3 600 s
This speed is almost 1 mm/s; so we might guess the creature was a snail, or perhaps a sloth. The volume of the galaxy is
2
nr2t=n(1021 m) (1019 m)~1061 m3.
If the distance between stars is 4 x 1016 m, then there is one star in a volume on the order of (4x 1016 m)3 ~1050 m3. 1061 ms
The number of stars is about er 1011 stars .
10 m /star The density of each material is p = ﬂ = m 4m V ﬂrzh : ﬂDZl’Z. 4(51.5 g) Al: p Z  2.75 g3 The tabulated value [2.70 g3 ] is 2% smaller.
7z(2..52. cm) (3.75 cm) cm cm
4(56.3 g) g g .
Cu: pz—Zz 9.36 3 The tabulated value 8.92. 3 1s 5% smaller.
7r(1.2.3 cm) (5.06 cm) Cm cm
4 94.4
Brass: p = $ = 8.91 g3
7r(1.54 cm) (5.69 cm) cm
4 69.1
Sn: p ( 2 g)  7.68 g3
7r(1.75 cm) (3.74 cm) Cm
4(216.1 g) g g .
Fe: p Z  7.88 3 The tabulated value 7.86 3 1s 0.3% smaller.
7r(1.89 cm) (9.77 cm) cm cm (a) (3 600 s/hr)(24 hr/day)(365.25 days/yr): 3.16 X 107 s/yr 3
(b) me = gnﬁ = $745.00 X 10’7 m) = 5.24 X 10’19 m3
chbe _ 1 m3 = 1.91 X 1018 micrometeorites me _ 5.2.4>< 10491113 18  
This would take 191 X 10 mlcrometeorltes  6.05 X 1010 yr . 3.16 X 107 micrometeorites/yr Chapter 1 19 ‘ NSWERS TO EVEN PROBLEMS P1.2 5.52 X 103 kg/m3 , between the densities P134 1.3 X 1021 kg of aluminum and iron, and greater than the densities of surface rocks. P136 200 km
P14 230 kg P138 (a) 13.4; (b) 49.1
P16 7.69 cm 1/3 P1.40 7A1 = rFe[pFe]
, PAI P1.8 (a) and (b) see the solutlon, NA = 6.022 137 X 1023 ; (c) 18.0 g; 7 (CD440 g P1.42 ~10 rev 9 .
P110 (a) 9.83 X 10*16 g; (b) 1.06 X 107 atoms P144 10 mmdrops
25 P1.46 ~1011 cans; ~105 tons P1.12 (a) 4.02. X 10 molecules; (b) 3.65 x 104 molecules P148 (209 i 4) cm2 P1.14 (a) ii; (b) iii; (c)i P150 (a) 3; (b) 4; (C) 3; (d) 2 P116 (a) 1vif;(b)1newton=1kgm/sz P152 (a) 797; (b) 11; (c) 17.66 P1.54 115.9 m
P118 35.7 m2 P156 316 m
P1.20 1.39 X 10’4 m3 P158 4.50 m2
P122 (a) 3.39 X 105 £13; (b) 2.54>< 1041b
P1.60 see the solution; 2.4.6° P1.24 (a) 560 km = 5.60 X 105 m = 5.60 X 107 cm;
(b) 491 m =0.491 km: 4.91 X 104 cm;
(c) 6.19 km = 6.19 X 103 m = 6.19 X 105 cm;
(cl) 2.50 km = 2.50 X 103 m = 2.50 X 105 cm P1.62 3.64 cents ; no P1.64 see the solution P166 (a) 1 000 kg; (b) 5.2 X 10*16 kg; 0.27 kg ; P126 4.05 103 2
X m 1.3x10’5 kg P1.28 (a) 1 mi/h=1.609 km/h; (b) 88.5 km/h;
(c) 16.1 km/h P168 8.32x10*4 m/s; asnail P130 1.19 X 1057 atoms P1.70 see the solutlon P132 2.57 X 106 m3 ...
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This note was uploaded on 04/07/2008 for the course PHY 2400 taught by Professor Donaldshaw during the Fall '08 term at Villanova.
 Fall '08
 DonaldShaw
 mechanics

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