INSTRUCTOR'S SOLUTIONS MANUAL FOR SERWAY AND JEWETT'S PHYSICS FOR SCIENTISTS AND ENGINEERS SIXTH EDITION Ralph V McGrew Broome Community College James A Currie Weston High School Australia • Canada • Mexico • Singapore • Spain • United Kingdom • United States Physics and Measurement CHAPTER OUTLINE 1.1 1.2 1.3 1.4 1.5 1.6 1.7 ANSWERS TO QUESTIONS Standards of Length, Mass, and Time Matter and Model-Building Density and Atomic Mass Dimensional Analysis Conversion of Units Estimates and Order-ofMagnitude Calculations Significant Figures 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 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: chimpanzee = 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 miles per day Since I am a college professor, I walk about 10 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 55 yr F 365.25 d I + 39 d = 20 128 dFG 86 400 s IJ = 1.74 × 10 GH yr JK H 1d K s ~ 10 s Many college students are just approaching 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 10 kg Q1.11 With one datum known to one significant digit, we have 80 million yr + 24 yr = 80 million yr 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 Matter and Model-Building 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 = L2 + L2 Thus, since the atoms are separated by a distance L = 0.200 nm , the diagonal planes are separated by Section 1.3 *P1.2 L + L2 = 0.141 nm Density and Atomic Mass Modeling the Earth as a sphere, we find its volume as 4 π r = π 6.37 × 10 m 3 e j = 1.08 × 10 21 m Its m 5.98 × 10 24 kg = = 5.52 × 10 kg m3 This value is intermediate between the V 1.08 × 10 21 m tabulated densities of aluminum and iron Typical rocks have densities around 000 to 000 kg m3 The average density of the Earth is significantly higher, so higher-density material must be down below the surface density is then ρ = P1.3 a fb g e j With V = base area height V = π r h and ρ = ρ= a fa ρ = 2.15 × 10 kg m 3 m for both Then ρ iron = 9.35 kg V and V 19.3 × 10 kg / m3 = 23.0 kg = 9.35 kg 7.86 × 10 kg / m3 Let V represent the volume of the model, the same in ρ = ρ gold = P1.5 F 10 mm I f GH m JK kg m = π r h π 19.5 mm 39.0 mm *P1.4 m , we have V m gold V V = Vo − Vi = ρ= Next, ρ gold ρ iron π r23 − r13 e = m gold 9.35 kg and m gold F GH j FG IJ e H K e 4π ρ r23 − r13 m , so m = ρV = ρ π r23 − r13 = V 3 j j I JK Chapter P1.6 4 π r and the mass is m = ρV = ρ π r We divide this equation 3 for the larger sphere by the same equation for the smaller: For either sphere the volume is V = m A ρ 4π rA3 rA3 = = = m s ρ 4π rs3 rs3 a f Then rA = rs = 4.50 cm 1.71 = 7.69 cm P1.7 *P1.8 Use u = 1.66 × 10 −24 g F 1.66 × 10 GH u F 1.66 × 10 = 55.9 uG H 1u F 1.66 × 10 = 207 uG H 1u -24 I = 6.64 × 10 JK gI JK = 9.29 × 10 gI JK = 3.44 × 10 g −24 g −23 g −22 g (a) For He, m = 4.00 u (b) For Fe, m (c) For Pb, m (a) The mass of any sample is the number of atoms in the sample times the mass m of one atom: m = Nm The first assertion is that the mass of one aluminum atom is -24 −24 m = 27.0 u = 27.0 u × 1.66 × 10 −27 kg u = 4.48 × 10 −26 kg Then the mass of 6.02 × 10 23 atoms is m = Nm = 6.02 × 10 23 × 4.48 × 10 −26 kg = 0.027 kg = 27.0 g Thus the first assertion implies the second Reasoning in reverse, the second assertion can be written m = Nm 0.027 kg = 6.02 × 10 23 m , so m = 0.027 kg 6.02 × 10 23 = 4.48 × 10 −26 kg , in agreement with the first assertion (b) The general equation m = Nm applied to one mole of any substance gives M g = NM u , where M is the numerical value of the atomic mass It divides out exactly for all substances, giving 1.000 000 × 10 −3 kg = N 1.660 540 × 10 −27 kg 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 I 10 GH 1.660 540 JK −3 + 27 = 6.022 137 × 10 23 (c) The atomic mass of hydrogen is 1.008 u and that of oxygen is 15.999 u The mass of one molecule of H O is 1.008 + 15.999 u = 18.0 u Then the molar mass is 18.0 g (d) For CO we have 12.011 g + 15.999 g = 44.0 g as the mass of one mole b g b g Physics and Measurement P1.9 b gFGH 101 kgg IJK = 4.5 × 10 kg I JK = 3.27 × 10 kg Mass of gold abraded: ∆m = 3.80 g − 3.35 g = 0.45 g = 0.45 g F 1.66 × 10 GH u −27 Each atom has mass m = 197 u = 197 u −4 kg −25 Now, ∆m = ∆N m , and the number of atoms missing is ∆N = ∆m = m0 4.5 × 10 −4 kg 3.27 × 10 −25 kg = 1.38 × 10 21 atoms The rate of loss is ∆N ∆t ∆N ∆t P1.10 P1.11 = FG H yr 1.38 × 10 21 atoms 365.25 d 50 yr IJ FG d IJ FG h IJ FG IJ K H 24 h K H 60 K H 60 s K = 8.72 × 10 11 atoms s e je j (a) m = ρ L3 = 7.86 g cm 5.00 × 10 −6 cm (b) N= (a) The cross-sectional area is = 9.83 × 10 −16 g = 9.83 × 10 −19 kg 9.83 × 10 −19 kg m = = 1.06 × 10 atoms m 55.9 u 1.66 × 10 −27 kg u e j a f a fa fa A = 0.150 m 0.010 m + 0.340 m 0.010 m = 6.40 × 10 −3 m f The volume of the beam is ja e f V = AL = 6.40 × 10 −3 m 1.50 m = 9.60 × 10 −3 m3 Thus, its mass is e FIG P1.11 je9.60 × 10 m j = 72.6 kg F 1.66 × 10 kg I = 9.28 × 10 The mass of one typical atom is m = a55.9 ufG H u JK m = ρV = 7.56 × 10 kg / m (b) −3 3 −27 m = Nm and the number of atoms is N = −26 kg Now 72.6 kg m = = 7.82 × 10 26 atoms −26 m 9.28 × 10 kg Chapter P1.12 (a) F 1.66 × 10 GH u −27 The mass of one molecule is m = 18.0 u kg I = 2.99 × 10 JK −26 kg The number of molecules in the pail is N pail = (b) 1.20 kg m = = 4.02 × 10 25 molecules m 2.99 × 10 −26 kg Suppose that enough time has elapsed for thorough mixing of the hydrosphere N both = N pail F m I = (4.02 × 10 GH M JK pail 25 F 1.20 kg I , GH 1.32 × 10 kg JK molecules) total 21 or N both = 3.65 × 10 molecules Section 1.4 P1.13 Dimensional Analysis The term x has dimensions of L, a has dimensions of LT −2 , and t has dimensions of T Therefore, the equation x = ka m t n has dimensions of e L = LT −2 j aTf m n or L1 T = Lm T n − m The powers of L and T must be the same on each side of the equation Therefore, L1 = Lm and m = Likewise, equating terms in T, we see that n − 2m must equal Thus, n = 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 e j Expression (i) has dimension L L2 1/2 = L2 , so this must be area (c) Expression (ii) has dimension L, so it is (a) Expression (iii) has dimension L L2 = L3 , so it is (b) Thus, (a) = ii; (b) = iii, (c) = i e j Physics and Measurement P1.15 *P1.16 (a) This is incorrect since the units of ax are m s , while the units of v are m s (b) This is correct since the units of y are m, and cos kx is dimensionless if k is in m −1 (a) a f a∝ ∑F or a = k ∑F represents the proportionality of acceleration to resultant force and m m the inverse proportionality of acceleration to mass If k has no dimensions, we have a = k (b) P1.17 In units, M ⋅L T = kg ⋅ m s2 F F L M ⋅L , F = , =1 m T M T2 , so newton = kg ⋅ m s Inserting the proper units for everything except G, LM kg m OP = G kg Ns Q m 2 Multiply both sides by m Section 1.5 *P1.18 m3 and divide by kg ; the units of G are kg ⋅ s Conversion of Units a fa f Each of the four walls has area 8.00 ft 12.0 ft = 96.0 ft Together, they have area e 96.0 ft P1.19 2 jFGH 3.128mft IJK = 35.7 m Apply the following conversion factors: in = 2.54 cm , d = 86 400 s , 100 cm = m , and 10 nm = m FG H 32 IJ b2.54 cm inge10 m cmje10 K 86 400 s day −2 in day j= nm m 9.19 nm s This means the proteins are assembled at a rate of many layers of atoms each second! *P1.20 8.50 in = 8.50 in FG 0.025 m IJ H in K = 1.39 × 10 −4 m Chapter P1.21 Conceptualize: We must calculate the area and convert units Since a meter is about feet, we should expect the area to be about A ≈ 30 m 50 m = 500 m a fa f Categorize: We model the lot as a perfect rectangle to use Area = Length × Width Use the conversion: m = 3.281 ft a Analyze: A = LW = 100 ft 1m I F m IJ = 390 m 150 ft fG f FGH 3.281 a J K H 3.281 ft K ft = 1.39 × 10 m Finalize: Our calculated result agrees reasonably well with our initial estimate and has the proper units of m Unit conversion is a common technique that is applied to many problems P1.22 (a) a fa fa f V = 40.0 m 20.0 m 12.0 m = 9.60 × 10 m 3 b g V = 9.60 × 10 m 3.28 ft m (b) = 3.39 × 10 ft The mass of the air is e je j m = ρ air V = 1.20 kg m3 9.60 × 10 m3 = 1.15 × 10 kg The student must look up weight in the index to find e je j Fg = mg = 1.15 × 10 kg 9.80 m s = 1.13 × 10 N Converting to pounds, jb e g Fg = 1.13 × 10 N lb 4.45 N = 2.54 × 10 lb P1.23 (a) Seven minutes is 420 seconds, so the rate is r= (b) 30.0 gal = 7.14 × 10 −2 gal s 420 s Converting gallons first to liters, then to m3 , e r = 7.14 × 10 −2 gal s jFGH 3.1786galL IJK FGH 10 Lm IJK −3 r = 2.70 × 10 −4 m3 s (c) At that rate, to fill a 1-m3 tank would take t= F 1m GH 2.70 × 10 −4 m IF h I = G J s JK H 600 K 1.03 h Physics and Measurement *P1.24 (a) (b) (c) (d) P1.25 FG 1.609 km IJ = 560 km = 5.60 × 10 m = 5.60 × 10 cm H mi K F 0.304 m IJ = 491 m = 0.491 km = 4.91 × 10 cm Height of Ribbon Falls = 612 ftG H ft K F 0.304 m IJ = 6.19 km = 6.19 × 10 m = 6.19 × 10 cm Height of Denali = 20 320 ftG H ft K F 0.304 m IJ = 2.50 km = 2.50 × 10 m = 2.50 × 10 cm Depth of King’s Canyon = 200 ftG H ft K Length of Mammoth Cave = 348 mi 5 From Table 1.5, the density of lead is 1.13 × 10 kg m , 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 = ρ= 23.94 g 2.10 cm F kg I FG 100 cm IJ GH 000 g JK H m K m We must convert to SI units in the calculation V = 1.14 × 10 kg 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, the difference in the two values is probably due to measurement uncertainty and should not be a concern One important common-sense check on density values is that objects which sink in water must have a density greater than g cm , and objects that float must be less dense than water P1.26 It is often useful to remember that the 600-m race at track and field events is approximately mile in length To be precise, there are 609 meters in a mile Thus, acre is equal in area to mI a1 acrefFGH 6401 miacres IJK FGH 609 J mi K *P1.27 P1.28 The weight flow rate is 200 FG H ton 000 lb h ton = 4.05 × 10 m IJ FG h IJ FG IJ = K H 60 K H 60 s K 667 lb s mi = 609 m = 1.609 km ; thus, to go from mph to km h , multiply by 1.609 (a) mi h = 1.609 km h (b) 55 mi h = 88.5 km h (c) 65 mi h = 104.6 km h Thus, ∆v = 16.1 km h 632 *P46.28 Particle Physics and Cosmology (a) Let Emin be the minimum total energy of the bombarding particle that is needed to induce the reaction At this energy the product particles all move with the same velocity The product particles are then equivalent to a single particle having mass equal to the total mass of the product particles, moving with the same velocity as each product particle By conservation of energy: em c j + b p c g Emin + m c = 2 p = p1 By conservation of momentum: b g = bp cg e 2 = Emin − m1 c Substitute (2) in (1): Emin + m c = ∴ p3 c (1) j (2) em c j 2 e + Emin − m1 c j Square both sides: e + 2Emin m c + m c Emin ∴ Emin em = j j = em c j 2 e + Emin − m1 c j − m12 − m 22 c 2m ∴ K = Emin − m1 c em = j − m12 − m 22 − 2m1 m c 2m = b m 32 − m1 + m g 2m Refer to Table 46.2 for the particle masses (b) K = a 938.3 f e 938.3 MeV c b497.7 + 115.6g = (c) K (d) K = a MeV c − 938.3 a f 938.3 + 135 2 K MeV c a MeV c − 139.6 + 938.3 a f f = 5.63 GeV MeV c 2 938.3 MeV c a MeV c − 938.3 a f f MeV c 2 938.3 MeV c LMe91.2 × 10 j − a938.3 + 938.3f =N 2a938.3f MeV c (e) j f 2 MeV c OP Q= = 768 MeV = 280 MeV 4.43 TeV c2 633 Chapter 46 Section 46.8 Finding Patterns in the Particles Section 46.9 Quarks Section 46.10 Multicolored Quarks Section 46.11 The Standard Model P46.29 (a) The number of protons F 6.02 × 10 molecules I FG 10 protons IJ = 3.34 × 10 JK H molecule K GH 18.0 g F 6.02 × 10 molecules I FG neutrons IJ = 2.68 × 10 = 000 g G JK H molecule K 18.0 g H 23 N p = 000 g and there are Nn 23 (b) 26 neutrons e j e 2.68 × 10 j + 3.34 × 10 3.34 × 10 26 + 2.68 × 10 26 = 9.36 × 10 26 up quarks 26 and there are protons 3.34 × 10 26 electrons So there are for electric neutrality The up quarks have number 26 26 = 8.70 × 10 26 down quarks Model yourself as 65 kg of water Then you contain: e j ~ 10 65 e9.36 × 10 j ~ 10 65 e8.70 × 10 j ~ 10 65 3.34 × 10 26 28 electrons 26 29 up quarks 26 29 down quarks Only these fundamental particles form your body You have no strangeness, charm, topness or bottomness P46.30 (a) strangeness baryon number charge proton e u 1/3 2e/3 u 1/3 2e/3 d 1/3 –e/3 total e strangeness baryon number charge neutron u 1/3 2e/3 d 1/3 –e/3 d 1/3 –e/3 total (b) P46.31 Quark composition of proton = uud and of neutron = udd Thus, if we neglect binding energies, we may write mp = m u + md and m n = m u + md (2) Solving simultaneously, we find mu = 1 m p − m n = 938 MeV c − 939.6 meV c = 312 MeV c 3 e (1) j e and from either (1) or (2), md = 314 MeV c j 634 P46.32 Particle Physics and Cosmology d 1/3 –e/3 –1/3 e/3 total 0 strangeness baryon number charge Λ0 –1 u 1/3 2e/3 d 1/3 –e/3 s –1 1/3 –e/3 (b) P46.33 (a) (c) (d) total –1 π − + p → K + Λ0 In terms of constituent quarks: (b) s strangeness baryon number charge K0 0 (a) ud + uud → ds + uds up quarks: −1 + → + 1, or 1→1 down quarks: + → + 1, or 2→2 strange quarks: + → −1 + , or 0→0 π + + p → K+ + Σ+ du + uud → us + uus up quarks: + → + 2, or 3→3 down quarks: −1 + → + , or 0→0 strange quarks: + → −1 + , or 0→0 K − + p → K + + K + Ω− us + uud → us + ds + sss up quarks: −1 + → + + , or 1→1 down quarks: +1→ +1+0, or 1→1 strange quarks: + → −1 − + , or 1→1 p + p → K0 + p + π + + ? uud + uud → ds + uud + ud + ? The quark combination of ? must be such as to balance the last equation for up, down, and strange quarks up quarks: 2+ = 0+ 2+1+? (has u quark) down quarks: 1+1 =1+1−1+? (has d quark) strange quarks: + = −1 + + + ? (has s quark) quark composition = uds = Λ0 or Σ P46.34 In the first reaction, π − + p → K + Λ0 , the quarks in the particles are: ud + uud → ds + uds There is a net of up quark both before and after the reaction, a net of down quarks both before and after, and a net of zero strange quarks both before and after Thus, the reaction conserves the net number of each type of quark In the second reaction, π − + p → K + n , the quarks in the particles are: ud + uud → ds + uds In this case, there is a net of up and down quarks before the reaction but a net of up, down, and anti-strange quark after the reaction Thus, the reaction does not conserve the net number of each type of quark Chapter 46 P46.35 Σ0 + p → Σ+ + γ + X dds + uud → uds + + ? The left side has a net 3d, 2u and 1s The right-hand side has 1d, 1u, and 1s leaving 2d and 1u missing The unknown particle is a neutron, udd Baryon and strangeness numbers are conserved P46.36 P46.37 Compare the given quark states to the entries in Tables 46.4 and 46.5: (a) suu = Σ + (b) ud = π − (c) sd = K (d) ssd = Ξ − (a) uud : (b) udd : Section 46.12 P46.38 FG eIJ + FG − eIJ + FG eIJ = − e This is the antiproton H K H K H3 K F I F1 I F1 I charge = G − eJ + G eJ + G eJ = This is the antineutron H K H3 K H3 K charge = − The Cosmic Connection Section 39.4 says fobserver = fsource + va c − va c The velocity of approach, v a , is the negative of the velocity of mutual recession: v a = − v c Then, P46.39 (a) λ′ c λ λ′ = λ 1+ v c 1− v c 1.18 = 1+v c = 1.381 1−v c 1+ v = HR : 1−v c 1+v c and 510 nm = 434 nm v v = 1.381 − 1.381 c c v = 0.160 c (b) = or 2.38 λ′ = λ 1+ v c 1− v c 1+v c 1−v c v = 0.381 c v = 0.160 c = 4.80 × 10 m s R= 4.80 × 10 m s v = = 2.82 × 10 ly H 17 × 10 −3 m s ⋅ ly 635 636 P46.40 Particle Physics and Cosmology (a) λ n′ = λ n 1+ v= (b) P46.41 P46.42 a 1+v c = Z+1 1−v c f f − FGH vc IJK aZ + 1f F Z + 2Z I cG H Z + 2Z + JK a v = Z+1 c a FG v IJ eZ H cK f j + Z + = Z2 + Z 2 F GH v c Z + 2Z = H H Z + 2Z + I JK e1.7 × 10 H= v = HR −2 ms j ly 1+ v c = 590 1.000 113 = 590.07 nm 1− v c b g e j λ′ = λ e j λ ′ = 590 + 0.011 33 = 597 nm − 0.011 33 e j λ ′ = 590 + 0.113 = 661 nm − 0.113 (a) v 2.00 × 10 ly = 3.4 × 10 m s (b) v 2.00 × 10 ly = 3.4 × 10 m s (c) v 2.00 × 10 ly = 3.4 × 10 m s (a) Wien’s law: λ maxT = 2.898 × 10 −3 m ⋅ K Thus, λ max = (b) *P46.43 R= 1+ v c = Z + λn 1− v c 2.898 × 10 −3 m ⋅ K 2.898 × 10 −3 m ⋅ K = = 1.06 × 10 −3 m = 1.06 mm T 2.73 K This is a microwave We suppose that the fireball of the Big Bang is a black body ja e I = eσT = (1) 5.67 × 10 −8 W m ⋅ K 2.73 K f = 3.15 × 10 −6 W m As a bonus, we can find the current power of direct radiation from the Big Bang in the section of the universe observable to us If it is fifteen billion years old, the fireball is a perfect sphere of radius fifteen billion light years, centered at the point halfway between your eyes: e ja fe P = IA = I ( 4π r ) = 3.15 × 10 −6 W m 4π 15 × 10 ly P = 7.98 × 10 47 W F I j GH ×110ly yrm s JK e3.156 × 10 s yr j Chapter 46 P46.44 637 The density of the Universe is ρ = 1.20 ρ c = 1.20 F 3H I GH 8π G JK Consider a remote galaxy at distance r The mass interior to the sphere below it is M=ρ FG π r IJ = 1.20FG 3H IJ FG π r IJ = 0.600 H r H K H 8π G K H K G 3 both now and in the future when it has slowed to rest from its current speed v = H r The energy of this galaxy-sphere system is constant as the galaxy moves to apogee distance R: GmM GmM =0− mv − r R −0.100 = −0.600 r R I JK F GH F GH so Gm 0.600 H r Gm 0.600 H r =0− mH r − r G R G so R = 6.00r I JK The Universe will expand by a factor of 6.00 from its current dimensions P46.45 (a) k B T ≈ 2m p c so (b) (a) T≈ kB = a f F 1.60 × 10 J I G J J K j H MeV K ~ 10 13 K f F 1.60 × 10 J I j GH MeV JK ~ 10 10 K −13 938.3 MeV e1.38 × 10 −23 a −13 0.511 MeV 2m e c = kB 1.38 × 10 −23 J K e The Hubble constant is defined in v = HR The distance R between any two far-separated objects opens at constant speed according to R = v t Then the time t since the Big Bang is found from v = H vt (b) 2m p c k B T ≈ 2m e c so *P46.46 T≈ 1= Ht F GH t= I JK H × 10 m s 1 = = 1.76 × 10 10 yr = 17.6 billion years H 17 × 10 −3 m s ⋅ ly ly yr 638 *P46.47 Particle Physics and Cosmology (a) Consider a sphere around us of radius R large compared to the size of galaxy clusters If the matter M inside the sphere has the critical density, then a galaxy of mass m at the surface of the sphere is moving just at escape speed v according to GMm = mv − R K +Ug = The energy of the galaxy-sphere system is conserved, so this equation is true throughout the dR Then history of the Universe after the Big Bang, where v = dt F dR I GH d t JK R3 / 32 = GM R R = GM t 0 *P46.48 R3 2 = GM z T dt R 2GM R 2R v so T= Now Hubble’s law says v = HR So T= e R dR = GM 2GM =v R From above, T= R 3/2 = GM T R T T= (b) z dR = R −1 / 2 GM dt 17 × 10 −3 R = HR H F × 10 m s I = G J m s ⋅ ly j H ly yr K 1.18 × 10 10 yr = 11.8 billion years In our frame of reference, Hubble’s law is exemplified by v = HR and v = HR From these we may form the equations − v = − HR and v − v = H R − R These equations express Hubble’s b g b g law as seen by the observer in the first galaxy cluster, as she looks at us to find − v = H − R and as she looks at cluster two to find v − v = H R − R b Section 46.13 P46.49 g Problems and Perspectives =G = c3 e1.055 × 10 − 34 (a) L= (b) This time is given as T = je J ⋅ s 6.67 × 10 −11 N ⋅ m kg e3.00 × 10 ms j Yes 1.61 × 10 −35 m L 1.61 × 10 − 35 m = = 5.38 × 10 −44 s , c 3.00 × 10 m s which is approximately equal to the ultra-hot epoch (c) j= Chapter 46 Additional Problems P46.50 We find the number N of neutrinos: a f e 10 46 J = N MeV = N × 1.60 × 10 −13 J j N = 1.0 × 10 58 neutrinos The intensity at our location is N N 1.0 × 10 58 = = A 4π r 4π 1.7 × 10 ly e F G j GH e3.00 × 10 ly je m s 3.16 × 10 I J sj JK = 3.1 × 10 14 m −2 The number passing through a body presenting 000 cm = 0.50 m FG 3.1 × 10 H is then m2 IJ e0.50 m j = 1.5 × 10 K 14 ~ 10 14 or *P46.51 14 A photon travels the distance from the Large Magellanic Cloud to us in 170 000 years The hypothetical massive neutrino travels the same distance in 170 000 years plus 10 seconds: b g b c 170 000 yr = v 170 000 yr + 10 s 170 000 yr v = = c 170 000 yr + 10 s + 10 s { g e1.7 × 10 yrje3.156 × 10 s yr = j} 1 + 1.86 × 10 −12 For the neutrino we want to evaluate mc in E = γ mc : v2 mc = = E − = 10 MeV − γ c + 1.86 × 10 −12 e mc ≈ 10 MeV e 1.86 × 10 −12 e1 + 1.86 × 10 j − e1 + 1.86 × 10 j −12 E j = 10 MeV j = 10 MeVe1.93 × 10 j = 19 eV −6 Then the upper limit on the mass is m= m= P46.52 19 eV c2 19 eV c F GH 931.5 × 10u eV c I = 2.1 × 10 JK −8 u (a) π − + p → Σ+ + π is forbidden by charge conservation (b) µ− → π − +νe is forbidden by energy conservation (c) p→π+ +π+ +π− is forbidden by baryon number conservation −12 639 640 P46.53 Particle Physics and Cosmology The total energy in neutrinos emitted per second by the Sun is: a0.4fLMN4π e1.5 × 10 j OPQW = 1.1 × 10 11 23 W Over 10 years, the Sun emits 3.6 × 10 39 J in neutrinos This represents an annihilated mass m c = 3.6 × 10 39 J m = 4.0 × 10 22 kg About part in 50 000 000 of the Sun’s mass, over 10 years, has been lost to neutrinos P46.54 p+p→ p+π+ + X We suppose the protons each have 70.4 MeV of kinetic energy From conservation of momentum for the collision, particle X has zero momentum and thus zero kinetic energy Conservation of system energy then requires e j e M p c + Mπ c + M X c = M p c + K p + M p c + K p a j f M X c = M p c + 2K p − Mπ c = 938.3 MeV + 70.4 MeV − 139.6 MeV = 939.5 MeV X must be a neutral baryon of rest energy 939.5 MeV Thus X is a neutron *P46.55 (a) If 2N particles are annihilated, the energy released is Nmc The resulting photon E Nmc = Nmc Since the momentum of the system is conserved, the momentum is p = = c c rocket will have momentum 2Nmc directed opposite the photon momentum p = Nmc (b) Consider a particle that is annihilated and gives up its rest energy mc to another particle which also has initial rest energy mc (but no momentum initially) e j Thus e 2mc j = p c + emc j E = p c + mc 2 2 2 2 Where p is the momentum the second particle acquires as a result of the annihilation of the e j first particle Thus mc 2 e j ,p = p c + mc 2 e j So p = = mc 2 mc N N protons and antiprotons) Thus the total 2 momentum acquired by the ejected particles is 3Nmc , and this momentum is imparted to the rocket This process is repeated N times (annihilate p = Nmc (c) Method (a) produces greater speed since Nmc > Nmc 641 Chapter 46 P46.56 (a) ∆E∆t ≈ = , and ∆t = r 1.4 × 10 −15 m = = 4.7 × 10 −24 s c × 10 m s ∆E ≈ = 1.055 × 10 −34 J ⋅ s MeV = = 2.3 × 10 −11 J = 1.4 × 10 MeV ∆t 4.7 × 10 −24 s 1.60 × 10 −13 J m= (b) P46.57 IJ K jFGH e ∆E ≈ 1.4 × 10 MeV c ~ 10 MeV c c2 From Table 46.2, mπ c = 139.6 MeV a pi - meson m Λ c = 1115.6 MeV Λ0 → p + π − m p c = 938.3 MeV mπ c = 139.6 MeV The difference between starting rest energy and final rest energy is the kinetic energy of the products K p + K π = 37.7 MeV and p p = pπ = p Applying conservation of relativistic energy to the decay process, we have LM a938.3f N OP LM a139.6f Q N + p c − 938.3 + OP Q + p c − 139.6 = 37.7 MeV Solving the algebra yields pπ c = p p c = 100.4 MeV Then, Kp = Kπ = P46.58 em c j + a100.4f a139.6f + a100.4f p 2 − m p c = 5.35 MeV 2 − 139.6 = 32.3 MeV By relativistic energy conservation in the reaction, By relativistic momentum conservation for the system, Eγ c = X= Subtracting (2) from (1), me c = − 3X 1− X and X = so Eγ = 4m e c = 2.04 MeV − v2 c2 3m e v − v2 c2 Eγ Dividing (2) by (1), Solving, = 3m e c Eγ + m e c = Eγ + m e c = 3m e c 1− X2 (1) (2) v c − 3m e c X 1− X2 642 P46.59 Particle Physics and Cosmology jb e ga f Momentum of proton is qBr = 1.60 × 10 −19 C 0.250 kg C ⋅ s 1.33 m p p = 5.32 × 10 −20 kg ⋅ m s c p p = 1.60 × 10 −11 kg ⋅ m s = 1.60 × 10 −11 J = 99.8 MeV Therefore, p p = 99.8 MeV c The total energy of the proton is Ep = E02 + cp b g = a983.3f + a99.8f 2 = 944 MeV For pion, the momentum qBr is the same (as it must be from conservation of momentum in a 2particle decay) pπ = 99.8 MeV c E0π = 139.6 MeV b g Eπ = E02 + cp = a139.6f + a99.8f 2 = 172 MeV Etotal after = Etotal before = Rest energy Thus, Rest Energy of unknown particle = 944 MeV + 172 MeV = 116 MeV (This is a Λ0 particle!) Mass = 116 MeV c P46.60 Σ → Λ0 + γ From Table 46.2, m Σ = 192.5 MeV c m Λ = 1115.6 MeV c and Conservation of energy in the decay requires e j E0, Σ = Eo , Λ + K Λ + Eγ F GH 192.5 MeV = 115.6 MeV + or I JK pΛ + Eγ 2m Λ System momentum conservation gives p Λ = pγ , so the last result may be written as F p I 192.5 MeV = G 115.6 MeV + J +E 2m K H F p c I 192.5 MeV = G 115.6 MeV + J +E 2m c K H γ γ Λ γ or 2 Λ m Λ c = 115.6 MeV Recognizing that P46.61 192.5 MeV = 115.6 MeV + Solving this quadratic equation, Eγ = 74.4 MeV γ pγ c = Eγ and we now have b Eγ 2 115.6 MeV g +E γ p+p→ p+ n+π+ The total momentum is zero before the reaction Thus, all three particles present after the reaction may be at rest and still conserve system momentum This will be the case when the incident protons have minimum kinetic energy Under these conditions, conservation of energy for the reaction gives e j m p c + K p = m p c + mn c + mπ c so the kinetic energy of each of the incident protons is Kp = mn c + mπ c − m p c 2 = a939.6 + 139.6 − 938.3f MeV = 70.4 MeV Chapter 46 P46.62 π − → µ− +νµ : From the conservation laws for the decay, mπ c = 139.6 MeV = Eµ + Eν P46.63 [1] d i + a105.7 MeVf = bp cg + a105.7 MeVf − E = a105.7 MeV f 2 and p µ = pν , Eν = pν c : Eµ2 = p µ c or Eµ2 Since Eµ + Eν = 139.6 MeV and dE µ ν 2 ν i a id a105.7 MeVf = Eµ − Eν Subtracting [3] from [1], 2Eν = 59.6 MeV [2] [1] + Eν Eµ − Eν = 105.7 MeV then f [2] 139.6 MeV = 80.0 and [3] Eν = 29.8 MeV The expression e − E k BT dE gives the fraction of the photons that have energy between E and E + dE The fraction that have energy between E and infinity is z z ∞ E ∞ z z ∞ e − E k BT dE = e − E k BT dE E ∞ b e − E k BT − dE k B T b e − E k BT − dE k B T g g = e − E k BT e ∞ E − E k BT ∞ = e − E k BT 0 We require T when this fraction has a value of 0.0100 (i.e., 1.00%) and E = 1.00 eV = 1.60 × 10 −19 J Thus, 0.010 = e b g or ln 0.010 = − P46.64 643 (a) e − 1.60 × 10 −19 J 1.60 × 10 −19 J e1.38 × 10 −23 j j e1.38 ×10 JKT =− −23 j JK T 1.16 × 10 K giving T = 2.52 × 10 K T This diagram represents the annihilation of an electron and an antielectron From charge and lepton-number conservation at either vertex, the γ γ exchanged particle must be an electron, e − (b) This is the tough one A neutrino collides with a neutron, changing it into a proton with release of a muon This is a weak interaction The exchanged particle has charge +e and is a W + (a) (b) FIG P46.64 644 P46.65 Particle Physics and Cosmology (a) The mediator of this weak interaction is a Z boson (b) The Feynman diagram shows a down quark and its antiparticle annihilating each other They can produce a particle carrying energy, momentum, and angular momentum, but zero charge, zero baryon number, and, it may be, no color charge In this case the product particle is a photon FIG P46.65 For conservation of both energy and momentum in the collision we would expect two photons; but momentum need not be strictly conserved, according to the uncertainty principle, if the photon travels a sufficiently short distance before producing another matterantimatter pair of particles, as shown in Figure P46.65 Depending on the color charges of the d and d quarks, the ephemeral particle could also be a gluon , as suggested in the discussion of Figure 46.14(b) *P46.66 (a) At threshold, we consider a photon and a proton colliding head-on to produce a proton and a pion at rest, according to p + γ → p + π Energy conservation gives mp c 2 1−u c + Eγ = m p c + mπ c m pu Momentum conservation gives 1−u c − Eγ c = Combining the equations, we have mp c 1−u c + b mp c u = m p c + mπ c c 1−u c 938.3 MeV + u c b1 − u cgb1 + u cg (b) so u = 0.134 c and Eγ = 127 MeV g = 938.3 MeV + 135.0 MeV λ maxT = 2.898 mm ⋅ K λ max = (c) 2.898 mm ⋅ K = 1.06 mm 2.73 K Eγ = hf = hc λ = continued on next page 240 eV ⋅ 10 −9 m 1.06 × 10 -3 m = 1.17 × 10 −3 eV Chapter 46 (d) 645 u′ = 0.134 and the c photon is moving to the left with hf ′ = 1.27 × 10 eV In the unprimed frame, In the primed reference frame, the proton is moving to the right at hf = 1.17 × 10 −3 eV Using the Doppler effect equation from Section 39.4, we have for the speed of the primed frame 1+ν c 1.17 × 10 −3 −ν c 1.27 × 10 = v = − 1.71 × 10 −22 c Then the speed of the proton is given by 0.134 + − 1.71 × 10 −22 u u′ c + ν c = = = − 1.30 × 10 −22 −22 c + u ′ν c + 0.134 − 1.71 × 10 e j And the energy of the proton is mp c 1−u c = 938.3 MeV e − − 1.30 × 10 j −22 = 6.19 × 10 10 × 938.3 × 10 eV = 5.81 × 10 19 eV ANSWERS TO EVEN PROBLEMS P46.2 2.27 × 10 23 Hz ; 1.32 fm P46.4 ν µ and ν e P46.6 ~ 10 −16 m P46.8 ~ 10 −23 s P46.10 νµ P46.12 (a) electron lepton number and muon lepton number; (b) charge; (c) baryon number; (d) baryon number; (e) charge P46.14 the second violates conservation of baryon number P46.16 0.828 c P46.18 (a) see the solution; 469 MeV for both; (b) 469 MeV ; c (c) 0.999 999 4c P46.20 see the solution P46.22 (a) electron lepton number and muon lepton number; (b) electron lepton number; (c) strangeness and charge; (d) baryon number; (e) strangeness P46.24 see the solution P46.26 686 MeV 200 MeV and ; c c (b) 627 MeV c ; (c) 244 MeV , 130 MeV , 370 MeV ; (a) (d) 190 MeV c , 0.500 c P46.28 (a) see the solution; (b) 5.63 GeV ; (c) 768 MeV ; (d) 280 MeV ; (e) 4.43 TeV P46.30 see the solution P46.32 see the solution P46.34 see the solution P46.36 (a) Σ + ; (b) π − ; (c) K ; (d) Ξ − P46.38 see the solution 646 P46.40 Particle Physics and Cosmology (a) v = c F Z + 2Z I ; (b) c F Z + 2Z I GH Z + 2Z + JK H GH Z + 2Z + JK 2 P46.56 (a) ~ 10 MeV c ; (b) a pi - meson P46.58 2.04 MeV P46.60 74.4 MeV P46.62 29.8 MeV P46.64 (a) electron-position annihilation; e − ; (b) a neutrino collides with a neutron, producing a proton and a muon; W + P46.66 (a) 127 MeV ; (b) 1.06 mm; (c) 1.17 meV ; (d) 5.81 × 10 19 eV P46.42 (a) 1.06 mm; (b) microwave P46.44 6.00 P46.46 (a) see the solution; (b) 17.6 Gyr P46.48 see the solution P46.50 ~ 10 14 P46.52 (a) charge; (b) energy; (c) baryon number P46.54 neutron [...]... a= (b) Maximum positive acceleration is at t = 3 s, and is approximately 2 m s 2 (c) a = 0 , at t = 6 s , and also for t > 10 s (d) Maximum negative acceleration is at t = 8 s, and is approximately −1.5 m s 2 Section 2.4 P2.18 ∆v 16.0 m s − 10.0 m s = = 6.00 m s 2 ∆t 3.00 s − 2.00 s (c) d (6.00 − 2.00)= 6.00 m s 2 (This includes both t = 2.00 s and t = 3.00 s ) dt ∆v 8.00 m s = = 1.3 m s 2 ∆t 6.00... Since 1 acre = 43 560 ft 2 , the number of blades of grass to be expected on a 16 quarter-acre plot of land is about n= a fe j 0.25 acre 43 560 ft 2 acre total area = = 2.5 × 10 7 blades ~ 10 7 blades area per blade 43 × 10 −5 ft 2 blade 12 P1.44 Physics and Measurement A typical raindrop is spherical and might have a radius of about 0.1 inch Its volume is then approximately 4 × 10 −3 in 3 Since 1 acre... also, δ ρ δ m 3δ r = + ρ m r In other words, the percentages of uncertainty are cumulative Therefore, a f δ ρ 0.02 3 0.20 = + = 0.103 , 6.50 ρ 1.85 ρ= and a 1.85 c hπ e6.5 × 10 4 3 f −2 j m 3 = 1.61 × 10 3 kg m 3 a f ρ ± δ ρ = 1.61 ± 0.17 × 10 3 kg m3 = 1.6 ± 0.2 × 10 3 kg m3 2 14 P1.52 *P1.53 Physics and Measurement (a) 756.?? 37.2? 0.83 + 2.5? 796.5/ 3/ = 797 (b) 0.003 2 2 s.f × 356.3 4 s.f = 1.140... 25 mi gal Fuel saved = V25 mpg − V20 mpg = 1.0 × 10 10 gal yr jFGH π4 IJK a2.0 mmf a4.0 mmfe10 2 −1 j cm mm 3 17 18 P1.68 P1.69 Physics and Measurement F GH IF JK GH IF JK GH IF JK GH I FG JK H IJ FG KH I JK furlongs 220 yd 0.914 4 m 1 fortnight 1 day 1 hr = 8.32 × 10 −4 m s fortnight 1 furlong 1 yd 14 days 24 hrs 3 600 s This speed is almost 1 mm/s; so we might guess the creature was a snail, or perhaps... 2.3 2.4 2.5 2.6 2.7 Position, Velocity, and Speed Instantaneous Velocity and Speed Acceleration Motion Diagrams One-Dimensional Motion with Constant Acceleration Freely Falling Objects Kinematic Equations Derived from Calculus ANSWERS TO QUESTIONS Q2.1 If I count 5.0 s between lightning and thunder, the sound has traveled 331 m s 5.0 s = 1.7 km The transit time for the light is smaller by b ga f 3.00... 8−0 t 2 − t1 x = 10t 2 : For af xamf ts = 2.0 2.1 3.0 = 44.1 90 40 (a) v= ∆x 50 m = = 50.0 m s ∆t 1.0 s (b) v= ∆x 4.1 m = = 41.0 m s ∆t 0.1 s IJ FG 10 mm IJ = KH 1 m K 3 5 × 10 8 yr 23 24 P2.5 Motion in One Dimension (a) Let d represent the distance between A and B Let t1 be the time for which the walker has d the higher speed in 5.00 m s = Let t 2 represent the longer time for the return trip in t1... ft s 2 ( 25 2 + 24.0 + 23.8 + 23.0) s Acceleration is the slope of the graph of v vs t a (m/s2) 2.0 For 0 < t < 5.00 s, a = 0 1.6 For 15.0 s < t < 20.0 s , a = 0 For 5.0 s < t < 15.0 s , a = a= v f − vi t f − ti 1.0 8.00 − (−8.00) 15.0 − 5.00 0.0 = 1.60 m s 2 t (s) 0 5 a= v f − vi t f − ti (i) For 5.00 s < t < 15.0 s , ti = 5.00 s , vi = −8.00 m s , t f = 15.0 s v f = 8.00 m s a= (ii) x = 2.00... of each sphere is m Al = ρ Al VAl = 4π ρ Al rAl 3 3 m Fe = ρ FeVFe = 4π ρ Fe rFe 3 3 and Setting these masses equal, 4π ρ Al rAl 3 4π ρ Fe rFe 3 ρ = and rAl = rFe 3 Fe 3 3 ρ Al Section 1.6 P1.41 Estimates and Order-of-Magnitude Calculations 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 4 ×... We must find separately the time t1 for speeding up and the time t 2 for coasting: x f − xi = x f − xi = 1 1 v xf + v xi t1 : 40 m = 3 × 10 6 m s + 0 t1 2 2 t1 = 2.67 × 10 −5 s d i e j 1 1 v xf + v xi t 2 : 60 m = 3 × 10 6 m s + 3 × 10 6 m s t 2 2 2 t 2 = 2.00 × 10 −5 s d i e j total time = 4.67 ×10−5 s *P2.35 (a) Along the time axis of the graph shown, let i = 0 and f = t m Then v xf = v xi + a x... c = 0 + am tm am = (b) vc tm The displacement between 0 and t m is x f − xi = v xi t + 1 vc 2 1 1 axt 2 = 0 + t m = v c tm 2 tm 2 2 The displacement between t m and t 0 is x f − xi = v xi t + a f 1 a x t 2 = v c t0 − tm + 0 2 The total displacement is ∆x = FG H 1 1 v c t m + v c t 0 − v c t m = v c t 0 − tm 2 2 IJ K (c) For constant v c and t 0 , ∆x is minimized by maximizing t m to t m = t 0 ...INSTRUCTOR'S SOLUTIONS MANUAL FOR SERWAY AND JEWETT'S PHYSICS FOR SCIENTISTS AND ENGINEERS SIXTH EDITION Ralph V McGrew Broome Community College James... • United States Physics and Measurement CHAPTER OUTLINE 1.1 1.2 1.3 1.4 1.5 1.6 1.7 ANSWERS TO QUESTIONS Standards of Length, Mass, and Time Matter and Model-Building Density and Atomic Mass... mile”, and “a dollar”, were measured to threedigit precision We have interpreted “up in the sky” as referring to the free fall time, not to the launch and landing times Both the takeoff and landing