(Test Bank) Serway Physics Test Bank With Solutions

Serway, 6

Physics

Test Bank

6th Edition

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1 CHAPTER OUTLINE 1.1 Standards of Length, Mass, and Time1.2 Matter and Model-Building 1.3 Density and Atomic Mass 1.4 Dimensional Analysis 1.5 Conversion of Units 1.6 Estimates and Order-of- Magnitude Calculations 1.7 Significant Figures 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 inorder to use the density of water as a standard. Q1.3 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 (b) and (d). 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. Q1.7 If I were a runner, I might walk or run 10 1 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 200 miles per day on vacation. Q1.8 On February 7, 2001, I am 55 years and 39 days old. 55365 25 139 20 12886 400 117 4 1 0 1 099 yr d yr d d s d s s..~F HGI KJ+= FHGIKJ=× . Many college students are just approaching 1 Gs. Q1.9 Zero digits. An order-of-magnitude calculation is accurate only within a factor of 10. Q1.10 The mass of the forty-six chapter textbook is on the order of 100 kg. Q1.11 With one datum known to one significant digit, we have 80 million yr + 24 yr = 80 million yr. 12 Physics and Measurement SOLUTIONS TO PROBLEMS Section 1.1 Standards of Length, Mass, and Time No problems in this section Section 1.2 Matter and Model-Building P1.1 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, LL Ldiag=+22. Thus, since the atoms are separated by a distance L=02 0 0. n m , the diagonal planes are separated by 1 201 4 122LL+= . n m . Section 1.3 Density and Atomic Mass *P1.2 Modeling the Earth as a sphere, we find its volume as 4 34 36 37 10 1 08 1036321 3ππr=×= × .. m m ej . Its density is then ρ==× ×=×m V59 8 1 0 10 8 1 055 2 1 024 21 333 . .. kg m kg m . This value is intermediate between the tabulated densities of aluminum and iron. Typical rocks have densities around 2 000 to 30 0 03 kg m . The average density of the Earth is significantly higher, so higher-density material must be down below the surface. P1.3 With V=base area heightaf bg Vr h=π2ej and ρ=m V, we have ρ ππ ρ==F HGI KJ =×m rh229 431 19 5 39 010 1 21 5 1 0 kg mm mm mm m kg m3 3.. ..af af *P1.4 Let V represent the volume of the model, the same in ρ=m V for both. Then ρiron kg=93 5. V and ρgoldgold=m V. Next, ρ ρgold irongold kg=m 93 5. and mgold33 3 kg19.3 10 kg / m kg / m kg =× ×F HGI KJ= 93 5 78 6 1 023 03. ... P1.5 VV V r roi=−= −4 323 13πej ρ=m V, so mV rrrr == FHGIKJ−=− ρρ ππρ4 34 323 1323 13ejej.Chapter 1 3 P1.6 For either sphere the volume is Vr=4 33π and the mass is mV r==ρρ π4 33. We divide this equation for the larger sphere by the same equation for the smaller: m mr rr r s ssA A A== =ρπ ρπ434353 33 3. Then rrsA== = 5 4 50 1 71 7 693... cm cm af . P1.7 Use 1 u . g=×−16 6 1 024. (a) For He, m0244 00 6 64 10=× F HGI KJ=×−.. u1.66 10 g 1 ug-24 . (b) For Fe, m02355 9 9 29 10=× F HGI KJ=×−.. u1.66 10 g 1 ug-24 . (c) For Pb, m024 2220716 6 1 034 4 1 0 =× F HGI KJ=×− − u g 1 ug ... *P1.8 (a) The mass of any sample is the number of atoms in the sample times the mass m0 of one atom: mN m=0. The first assertion is that the mass of one aluminum atom is m027 262 7 0 2 7 0 1 6 61 0 1 4 4 81 0==× × = ×−−.. . . u u k g u k g . Then the mass of 6 02 1023.× atoms is mN m== × × × = =− 023 266 02 10 4 48 10 0 027 0 27 0.. . . kg kg g . Thus the first assertion implies the second. Reasoning in reverse, the second assertion can be written mN m=0. 0 027 0 6 02 1023 0 .. kg=× m, so m0 2326 00 2 7 60 2 1 044 8 1 0 = ×=×− . .. kg kg , in agreement with the first assertion. (b) The general equation mN m=0 applied to one mole of any substance gives MN M g u= , where M is the numerical value of the atomic mass. It divides out exactly for all substances, giving 1 000 000 0 10 1 660 540 2 1032 7.. ×= ×−− kg kg N . With eight-digit data, we can be quite sure of the result to seven digits. For one mole the number of atoms is N=F HGI KJ=×−+ 1 16 6 05 4 0210 6 022 137 1032 7 2 3 ... (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 H O2 is 2 1 008 0 15 999 18 0 ...bg+= u u. Then the molar mass is 18 0. g. (d) For CO2 we have 12 011 2 15 999 44 0... g g g+=bg as the mass of one mole.4 Physics and Measurement P1.9 Mass of gold abraded: ∆m=−==F HGI KJ=×−3 80 3 35 0 45 0 45145 1 04... . . g g g g kg 10 g kg3 bg . Each atom has mass m027 25197 19716 6 1 0 132 7 1 0 ==× F HGI KJ=×− − u u kg u kg... Now, ∆∆mN m=0, and the number of atoms missing is ∆∆Nm m==× ×=×− − 04 2521 45 1 0 32 7 1 013 8 1 0. .. kg kg atoms . The rate of loss is ∆ ∆ ∆ ∆N t N t=× FHGIKJFHGIKJFHGIKJFHGIKJ =×13 8 1 0 501 11 1 87 2 1 021 11. .. atoms yr yr 365.25 d d 24 h h 60 min min 60 s atoms s P1.10 (a) mL== × = × =×−− −ρ33 6316 197 86 5 00 10 9 83 10 9 83 10. g c m c m g k g ej e j .. . (b) Nm m==× ×=×− −019 277 98 3 1 0 55 9 1 66 1010 6 1 0. ... kg u k g 1 uatomsej P1.11 (a) The cross-sectional area is A=+ =×−2 0 150 0 010 0 340 0 010 64 0 1 03.. .. .. m m m m m2a fa fa fa f . The volume of the beam is VA L== × = ×−−6 4 01 0 1 5 0 9 6 01 033.. . m m m23ej af . Thus, its mass is mV== × × =−ρ 7 56 10 9 60 10 72 633.. . kg / m m kg33ej e j . FIG. P1.11 (b) The mass of one typical atom is m027 2655 916 6 1 0 192 8 1 0 =× F HGI KJ=×− −... u kg u kg af . Now mN m=0 and the number of atoms is Nm m== ×=×− 02626 72 6 92 8 1 078 2 1 0. .. kg kg atoms .Chapter 1 5 P1.12 (a) The mass of one molecule is m027 2618 016 6 1 029 9 1 0 =× F HGI KJ=×− −... u kg 1 u kg . The number of molecules in the pail is Nm mpail kg 2.99 kgmolecules == ×=×− 02625 12 0 1040 2 1 0... (b) Suppose that enough time has elapsed for thorough mixing of the hydrosphere. NNm Mboth pailpail total25(4.02 10 molecules) kg kg=F HGI KJ=× ×F HGI KJ12 0 13 2 1 021. ., or Nboth molecules =×36 5 1 04. . Section 1.4 Dimensional Analysis P1.13 The term x has dimensions of L, a has dimensions of LT−2, and t has dimensions of T. Therefore, the equation xk a tmn= has dimensions of LL T T=−2ej afmn or L T L T10 2=− mn m. The powers of L and T must be the same on each side of the equation. Therefore, LL1=m and m=1. Likewise, equating terms in T, we see that nm−2 must equal 0. Thus, n=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. Expression (i) has dimension L L L2122ej/=, so this must be area (c). Expression (ii) has dimension L, so it is (a). Expression (iii) has dimension L L L23ej=, so it is (b). Thus, (a) ii; (b) iii, (c) i== = .6 Physics and Measurement P1.15 (a) This is incorrect since the units of ax are ms22, while the units of v are ms . (b) This is correct since the units of y are m, and coskxaf is dimensionless if k is in m−1. *P1.16 (a) aF m∝∑ or akF m=∑ represents the proportionality of acceleration to resultant force and the inverse proportionality of acceleration to mass. If k has no dimensions, we have akF m= , L T1F M2= , FML T2=⋅. (b) In units, ML Tkg m s22⋅=⋅, so 11 newton kg m s2=⋅ . P1.17 Inserting the proper units for everything except G, kg m skg m2LNMOQP=G2 2. Multiply both sides by m2 and divide by kg2; the units of G are m kg s3 2⋅. Section 1.5 Conversion of Units *P1.18 Each of the four walls has area 8 00 12 0 96 0 .. . ft ft ft2 af af = . Together, they have area 49 601 32 835 722 .. ft m . f t m2ejFHGIKJ= . P1.19 Apply the following conversion factors: 12 5 4 in cm=., 1 8 6 4 0 0 d s= , 100 1 cm m= , and 10 19 nm m= 1 3225 4 1 0 1 0 91 929 in day c mi n mc m n mm 86 400 s day nm s FHGIKJ =−. .bg ej e j. This means the proteins are assembled at a rate of many layers of atoms each second! *P1.20 85 0 85 000 2 5413 9 1 03 4.... in in m 1 in m33 3= FHGIKJ=×−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 m m m2≈=30 50 1 500af af . Categorize : We model the lot as a perfect rectangle to use Area = Length × Width. Use the conversion: 1 m 3.281 ft = . Analyze : AL W== FHGIKJFHGIKJ =× 1001 32 8 11501 32 8 113 9 1 03 ft m ft ft m ft= 13 9 0 m m22 af af.... 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 . m3 a fa fa f96 0 1 03 V=× = ×9 60 10 3 39 1035 3. m 3 . 2 8 f t 1 m f t3 3 bg . (b) The mass of the air is mV== × = ×ρair33 kg m 9.60 10 m . kg1 20 1 15 1034.ej e j . The student must look up weight in the index to find Fm gg== × = × 1.15 10 kg 9.80 m s 1.13 10 N42 5ej e j . Converting to pounds, Fg=× =×1 13 10 2 54 1054. N 1 lb 4.45 N lb ej bg . . P1.23 (a) Seven minutes is 420 seconds, so the rate is r== ×− 30 0 42071 4 1 02 .. gal s gal s . (b) Converting gallons first to liters, then to m3, r r=×F HGI KJF HGI KJ =×−− −71 4 1 037 8 6 1 0 27 0 1 023 4.. .. gal s L 1 gal m 1 L ms3 3ej (c) At that rate, to fill a 1-m3 tank would take t= ×F HGI KJF HGI KJ=−1 27 0 1 0110 34 m ms h 36 0 0 h3 3...8 Physics and Measurement *P1.24 (a) Length of Mammoth Cave = FHGIKJ== × = × 34816 0 9 1560 5 60 10 5 60 1057 mi km mi km m cm... . (b) Height of Ribbon Falls = FHGIKJ== = × 16 1 20 1491 m 0 491 4 91 104 ft.304 8 m ft km cm.. . (c) Height of Denali = FHGIKJ== × = × 20 3200 16 6 19 10 6 19 1035 ft.304 8 m ft.19 km m cm .. . (d) Depth of King’s Canyon = FHGIKJ== × = × 82 0 00 12 2 50 10 2 50 1035 ft.304 8 m ft.50 km m cm .. . P1.25 From Table 1.5, the density of lead is 11 3 1 04. k g m3× , 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 ρ=m V. We must convert to SI units in the calculation. ρ=F HGI KJFHGIKJ=×23 94 21 01 10 0 0100 111 4 1 033 4 . . g cm kg g cm m. k g m3 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, thedifference in the two values is probably due to measurement uncertainty and should not be aconcern. One important common-sense 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. P1.26 It is often useful to remember that the 1 600-m 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 11 64016 0 940 5 1 02 3 acre mi acres m mi m2 2 afF HGI KJFHGIKJ=×. . *P1.27 The weight flow rate is 12 0 020 0 0 11667 ton h lb ton h 60 min min 60 s lb s FHGIKJFHGIKJFHGIKJ= . P1.28 1 1 609 1 609 mi m km== . ; thus, to go from mph to km h, multiply by 1.609. (a) 11 6 0 9 mi h km h=. (b) 55 88 5 mi h km h= . (c) 65 104 6 mi h km h= . . Thus, ∆v=16 1. k mh .Chapter 1 9 P1.29 (a)61 0 1 1119012×F HGI KJF HGI KJFHGIKJF HGI KJ= $ 10 0 0 $ s h 36 0 0 s day 24 h yr 365 days years (b) The circumference of the Earth at the equator is 2 6 378 10 4 01 1037π..×= × m m ej . The length of one dollar bill is 0.155 m so that the length of 6 trillion bills is 9 30 1011.× m. Thus, the 6 trillion dollars would encircle the Earth 93 0 1 023 2 1 011 4 ..× ×=× m 4.01 0 m times7. P1.30 Nm matomsSun atom kg 1.67 kg atoms ==× ×=×−19 9 1 0 1011 9 1 030 2757 .. P1.31 VA t= so tV A==×=×− − 37 8 1 0 25 015 1 1 0 1 5 13 4 . .. m m m or m3 2µ bg P1.32 VB h== =×1 313 0 43 560 3481 90 8 1 07. ., acres ft acre ft ft2 3a fejaf or V=×× F HGI KJ =×− 90 8 1 028 3 1 0 1 25 7 1 072 6.. . ft m ft m33 3 3ejBh Bh FIG. P1.32 P1.33 Fg=× = ×2 50 2 00 10 2 000 1 00 1061 0.. . tons block blocks lb ton lbs bg ej bg *P1.34 The area covered by water is AA Rw== = × = ×07 0 36 1 021 4. 0.70 4 0.70 4 6.37 10 m . mEarth Earth622 af ej af a f ej ππ . The average depth of the water is d== ×2.3 miles 1 609 m l mile . maf bg 37 1 03. The volume of the water is VA dw== × × = × 3 6 10 3 7 10 1 3 1014 2 3 18 3. m . m . mej e j and the mass is mV== × =×ρ 10 0 0 13 1 0 13 1 031 8 3 2 1 kg m . m kg ej e j . .10 Physics and Measurement P1.35 (a) ddd dnucleus, scale nucleus, real atom, scale atom, real m ft 1.06 10 m ft =F HGI KJ=× ×FHGIKJ=×− −−24 0 1 030067 9 1 015 103..ej , or dnucleus, scale ft mm 1 ft mm =× =−67 9 1 0 3 0 48 20 73...ej bg (b)V Vr rd dr ratom nucleusatom nucleusatom nucleusatom nucleus m m times as large==FHGIKJ=FHGIKJ=× ×F HGI KJ =×− −4 3 4 33310 153 133 310 6 1 0 24 0 1 0 86 2 1 0π π. . . *P1.36 scale distance betweenreal distancescale factor km m m km =FHGIKJFHGIKJ=×× ×F HGI KJ=− 40 1 070 1 0 14 1 0200133 9.. .ej P1.37 The scale factor used in the “dinner plate” model is S= ×=×− 02 5 10 1 021 056 . . m lightyears.5 m lightyears . The distance to Andromeda in the scale model will be DD Sscale actual66 2.0 10 lightyears 2.5 10 m lightyears m == × × =−ej e j 50.. P1.38 (a)A Ar rr rEarth MoonEarth Moon2Earth Moon m c m m cm==FHGIKJ=× ×F HGGI KJJ=4 463 7 1 0 1 0 0 17 4 1 013 422 6 82 π π. ..ej bg (b)V Vr rr rEarth MoonEarth Moon3 3 Earth Moonm c m m cm==FHGIKJ=× ×F HGGI KJJ=4 3 4 36 83 363 7 1 0 1 0 0 17 4 1 049 1π π. ..ej bg P1.39 To balance, mmFe Al= or ρρFe Fe Al AlVV= ρπ ρπ ρ ρFe Fe Al Al Al FeFe Al cm cm4 34 3 20 078 6 27 028 633 13 13FHGIKJ= FHGIKJ =FHGIKJ= FHGIKJ=rr rr/ / .. ... afChapter 1 11 P1.40 The mass of each sphere is mVr Al Al AlAl Al==ρπρ4 33 and mVr Fe Fe FeFe Fe==ρπρ4 33 . Setting these masses equal, 4 34 333πρ πρAl Al Fe Ferr= and rrAl FeFe Al=ρ ρ3 . Section 1.6 Estimates and Order-of-Magnitude Calculations P1.41 Model the room as a rectangular solid with dimensions 4 m by 4 m by 3 m, and each ping-pong ball as a sphere of diameter 0.038 m. The volume of the room is 4434 8××= m3, while the volume of one ball is 4 300 3 828 7 1 03 5 π .. m 2 m3 FHGIKJ=×−. Therefore, one can fit about 48 28 7 1 01056 .~ ×− ping-pong balls in the room. 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 so-called “best packing fraction” is 1 620 7 4π=. so that at least 26% of the space will be empty. Therefore, the above estimate reduces to 16 7 1 0 07 4 0 1 066.. ~×× . 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 5 280 1 3 107 mi ft mi rev 8 ft rev ~ 10 rev7 bg b g b g =× . P1.43 In order to reasonably carry on photosynthesis, we might expect a blade of grass to require at least 1 16 in 43 10 ft25 2=×−. Since 1 acre 43 560 ft2= , the number of blades of grass to be expected on a quarter-acre plot of land is about n== ×=×−total area area per blade acre ft acre ft blade2.5 10 blades blades2 27 02 5 4 35 6 0 43 101057. ~af ej.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 41 03×− in3. Since 1 acre 43 560 ft2= , the volume of water required to cover it to a depth of 1 inch is 1 acre 1 inch 1 acre in ft 1 acre in ft6.3 10 in63 af a f a f =⋅F HGI KJF HGI KJ≈×43 560 144 12 2 2. The number of raindrops required is n==× ×=×−volume of water required volume of a single dropin in. 63 1 0 41 01 6 10 1063 3399 .~. *P1.45 Assume the tub measures 1.3 m by 0.5 m by 0.3 m. One-half of its volume is then V==05 13 05 03 01 0.. . . .a f af af af m m m m3. The mass of this volume of water is mVwater water33 kg m m kg kg == =ρ 1 000 0 10 100 102ej e j .~ . Pennies are now mostly zinc, but consider copper pennies filling 50% of the volume of the tub. The mass of copper required is mVcopper copper33 kg m m kg kg == =ρ 8 920 0 10 892 103ej e j .~ . 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 ~250million people, and 365 days in a year, so 250 10 365 1061 1×≅ cans day days year cans ej bg are thrown away or recycled each year. Guessing that each can weighs around 1 10 of an ounce, we estimate this represents 1 0 01 1 1 31 1 011 5 cans oz can lb 16 oz ton 2 000 lb tons year ej bg b g b g.. ≈× . ~105 tons P1.47 Assume: Total population =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, # t u n e r s ~1 tuner 10 0 0 p i a n o s1 piano 100 peoplepeople F HGI KJF HGI KJ = ()10 1007.Chapter 1 13 Section 1.7 Significant Figures *P1.48 METHOD ONE We treat the best value with its uncertainty as a binomial 21 3 0 2 9 8 0 1 .. ..±± af a f cm cm , A=± ± ±2 1398 2 1301 0298 02 01.. .. .. . .a f a f a fa fa f cm2. 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=±209 422 cm cm . METHOD TWO We add the fractional uncertainties in the data. A=± + FHGIKJ=± =± 21 3 9 802 21 301 98209 2% 209 422 2... .. . cm cm cm cm cm a fa f P1.49 (a)ππ πr2 2 22 2210 5 0 2 10 5 2 10 5 0 2 0 2 346 13=± =± + =±. m . m m m m m m maf (. ) (. ) ( . )( . ) (b) 22 1 0 5 0 2 6 6 0 1 3ππr=± = ± . m . m m maf .. P1.50 (a) 3 (b) 4 (c) 3 (d) 2 P1.51 r m m r=± =±× =± =−6 50 0 20 6 50 0 20 10 18 5 00 22 4 33.. c m .. m .. k gaf af af chρ π also, δρ ρδδ=+m mr r3. In other words, the percentages of uncertainty are cumulative. Therefore, δρ ρ=+ =00 2 18 5302 0 65 001 0 3. .. ..a f, ρ π= ×=× −18 5 65 1 016 1 1 0 4 32333 . ..ch ej m kg m and ρδ ρ±= ± × =±× 1 61 0 17 10 1 6 0 2 1033.. . .af a f kg m kg m33.14 Physics and Measurement P1.52 (a) 756.?? 37.2? 0.83+ 2.5? 796./5/3 =797 (b) 0 003 2 356 3 1 140 16 1 1. 2 s.f. . 4 s.f. . 2 s.f. af af af×= = . (c) 5.620 4 s.f. >4 s.f. 17.656= 4 s.f. 17.66 af a f af×=π *P1.53 We work to nine significant digits: 11365 242 199 24 60 6031 556 926 0 yr yr d 1 yr h 1 d min 1 h s 1 min s =F HGI KJFHGIKJFHGIKJFHGIKJ=... P1.54 The distance around is 38.44 m 19.5 m 38.44 m 19.5 m 115.88 m++ += , but this answer must be rounded to 115.9 m because the distance 19.5 m carries information to only one place past the decimal. 115 9. m P1.55 VV V V V V V V=+= + =+ + = == =+=22 2 1 7 01 01 0 1 0 0 0 9 1 7 0 10 0 1 0 0 090 0 900 217 0 09 0 0 5212 1 2 1 2 3bg af a f a faf a f a f ej... . . . ... . .. . m m m m m m m m m m m m m3 3 33 δ δ δδA A1 1 1 1 1 101 200 0 6 3 00 100 1 0 0100 1 10 006 0 010 0 011 0 027 3%== ====U V||| W|||=++==.. .. ...... m 19.0 m m 1.0 m cm 9.0 cmw w t tV V FIG. P1.55 Additional Problems P1.56 It is desired to find the distance x such that x x 10010 0 0 m m= (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 x25100 1 000 1 00 10== × m m m2 af bg . and therefore x=× =10 0 1 0 3 1 65. m m2.Chapter 1 15 *P1.57 Consider one cubic meter of gold. Its mass from Table 1.5 is 19 300 kg. One atom of gold has mass m027 25197 3 27 10=× F HGI KJ=×− − u1.66 10 kg 1 u kg af .. So, the number of atoms in the cube is N= ×=×−19 30059 0 1 02528 kg 3.27 10 kg.. The imagined cubical volume of each atom is d3 2829 1 59 0 1 016 9 1 0 = ×=×− m m3 3 ... So d=×−25 7 1 010. m . P1.58 AN AV VAVr rtotal droptotal dropdroptotal 4==F HGI KJ=F HGGI KJJaf ej ej e jππ3 324 AV rtotaltotal3 2 m m m =FHGIKJ=× ×F HGI KJ=− −3330 0 10 20 0 1 045 06 5. .. P1.59 One month is 1 30 24 3 600 2 592 106 mo day h day s h s== ×bg b g b g .. Applying units to the equation, Vt t=+15 0 00 0 80 02.. Mft mo Mft mo33 2ej e j . Since 11 06 Mft ft33= , Vt t=× + ×1 5 01 0 0 0 0 8 0 01 066 2.. ft mo ft mo33 2ej e j . Converting months to seconds, Vt t=× ×+× ×1 5 01 0 0 0 0 8 0 01 066 22 .. ft mo 2.592 10 s mo ft mo 2.592 10 s mo3 632 6ej. Thus, Vt t ft ft s ft s33 3 2[] . .=+ ×−05 7 9 11 9 1 092ej e j .