College physics serway, vuille 10th edition

Instructor Solutions Manual College Physics ! TENTH EDITION ! ! ! ! RAYMOND A SERWAY Emeritus, James Madison University ! CHRIS VUILLE Embry-Riddle Aeronautical University
 ! ! ! ! ! ! ! ! ! Prepared by ! Vahe Peroomian & John Gordon ! ! ! ! Australia • Brazil • Japan • Korea • Mexico • Singapore • Spain • United Kingdom • United States www.elsolucionario.org Introduction ANSWERS TO WARM-UP EXERCISES (a) The number given, 568 017, has six significant figures, which we will retain in converting the number to scientific nota5 tion Moving the decimal five spaces to the left gives us the answer, 5.680 17 × 10 (b) The number given, 0.000 309, has three significant figures, which we will retain in converting the number to scientific –4 notation Moving the decimal four spaces to the right gives us the answer, 3.09 × 10 We first collect terms, then simplify: [ M ][L]2 [T ] [ M ][L]2 [T ]2 [ M ][L] [T ] = = [ L] [T ] [T ] [T ]3 [L] As we will see in Chapter 6, these are the units for momentum Examining the expression shows that the units of meters and seconds squared (s2) appear in both the numerator and the denominator, and therefore cancel out We combine the numbers and units separately, squaring the last term before doing so: 7.00 m s2 1.00 km 1.00 × 103 m 1.00 1.00 × 103 = (7.00) = 25.2 3600 1.00 m s km m s2 km The required conversion can be carried out in one step: h = (2.00 m ) 60.0s 1.00 1.00 cubitus = 4.49 cubiti 0.445 m The area of the house in square feet (1 420 ft2) contains significant figures Our answer will therefore also contain three significant figures Also note that the conversion from feet to meters is squared to account for the ft units in which the area is originally given ( A = 420 ft ) 1.00 m 3.281 ft = 131.909 m = 132 m Using a calculator to multiply the length by the width gives a raw answer of 783 m2 This answer must be rounded to contain the same number of significant figures as the least accurate factor in the product The least accurate factor is the length, which contains significant figures, since the trailing zero is not significant (see Section 1.6) The correct answer for the area of the airstrip is 6.80 × 10 m2 Adding the three numbers with a calculator gives 21.4 + 15 + 17.17 + 4.003 = 57.573 However, this answer must be rounded to contain the same number of significant figures as the least accurate number in the sum, which is 15, with two significant figures The correct answer is therefore 58 The given Cartesian coordinates are x = –5.00 and y = 12.00 The least accurate of these coordinates contains significant figures, so we will express our answer in three significant figures The specified point, (–5.00, 12.00), is in the second quadrant since x < and y > To find the polar coordinates (r, θ ) of this point, we use r = x + y = (5.00)2 + (12.00) = 13.0 and θ = tan −1 y 12.00 = tan −1 = –67.3° x –5.00 Since the point is in the second quadrant, we add 180° to this angle to obtain θ = −67.3° + 180° = 113° The polar coordinates of the point are therefore (13.0, 113°) Refer to ANS FIG The height of the tree is described by the tangent of the 26° angle, or tan 26° = h 45 m from which we obtain h = ( 45 m) tan 26° = 22 m ANS FIG ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS Atomic clocks are based on the electromagnetic waves that atoms emit Also, pulsars are highly regular astronomical clocks (a) (b) (c) Let us assume the atoms are solid spheres of diameter 10−10 m Then, the volume of each atom is of the order of 10−30 m3 .) Therefore, since , the number of atoms in the cm3 solid is on (More precisely, volume = the order of atoms A more precise calculation would require knowledge of the density of the solid and the mass of each atom However, our estimate agrees with the more precise calculation to within a factor of 10 Realistically, the only lengths you might be able to verify are the length of a football field and the length of a housefly The only time intervals subject to verification would be the length of a day and the time between normal heartbeats 10 In the metric system, units differ by powers of ten, so it’s very easy and accurate to convert from one unit to another 12 Both answers (d) and (e) could be physically meaningful Answers (a), (b), and (c) must be meaningless since quantities can be added or subtracted only if they have the same dimensions ANSWERS TO EVEN NUMBERED PROBLEMS (a) (b) L All three equations are dimensionally incorrect (b) Ft = p (b) 22.7 (a) (a) 10 (a) (b) 12 (a) (b) 14 (a) 22.6 797 (b) (c) (c) 1.1 (c) 16 18 22.6 is more reliable (a) (b) 17.66 www.elsolucionario.org (c) (d) 20 22 24 26 28 ~ 30 (b) (c) 32 (a) 34 (a) ~ (c) The very large mass of prokaryotes implies they are important to the biosphere They are responsible for fixing carbon, producing oxygen, and breaking up pollutants, among many other biological roles Humans depend on them! 36 2.2 m 38 8.1 cm (b) 40 42 2.33 m 44 (a) 1.50 m 46 8.60 m 48 (a) and (b) (b) 2.60 m (c) (d) 50 52 (a) 54 Assumes population of 300 million, average of can/week per person, and 0.5 oz per can (a) (b) ~ (b) 56 (a) (b) 58 (a) (b) 60 (a) 62 ~ 500 yr (b) (c) ~ (c) 1.03 h 6.6 × 104 times Assumes lost ball per hitter, 10 hitters per inning, innings per game, and 81 games per year PROBLEM SOLUTIONS and recognizing that 2π is a dimensionless constant, we have 1.1 Substituting dimensions into the given equation or Thus, the 1.2 (a) (b) From x = Bt2, we find that If Thus, B has units of , then But the sine of an angle is a dimensionless ratio Therefore, 1.3 (a) The units of volume, area, and height are: , , and We then observe that or is Thus, the equation where (b) where 1.4 (a) In the equation , while Thus, the equation is , (b) In (c) In the equation is also but Hence, this equation is , we see that , while 1.5 From the universal gravitation law, the constant G is Its units are then Therefore, this equation www.elsolucionario.org 1.6 (a) Solving for the momentum, p, gives where the numeral is a dimensionless constant Di- mensional analysis gives the units of momentum as: Therefore, in the SI system, the units of momentum are (b) Note that the units of force are or Then, observe that From this, it follows that force multiplied by time is proportional to momentum: (See the impulse–momentum , which says that a constant force F multiplied by a duration of time ∆t equals the theorem in Chapter 6, change in momentum, ∆p.) 1.7 1.8 (a) Computing without rounding the intermediate result yields to three significant figures (b) Rounding the intermediate result to three significant figures yields Then, we obtain to three significant figures (c) because rounding in part (b) was carried out too soon 1.9 (a) has (b) has (c) has (d) with the uncertainty in the tenths position has The two zeros were originally included only to position the decimal 1.10 (a) Rounded to significant figures: (b) Rounded to significant figures: (c) Rounded to significant figures: the width , and the height 1.11 Observe that the length Thus, any product of these quantities should contain significant figures (a) (b) (c) all contain significant figures (d) In the rounding process, small amounts are either added to or subtracted from an answer to satisfy the rules of signify cant figures For a given rounding, different small adjustments are made, introducing a certain amount of randomness in the last significant digit of the final answer 1.12 (a) Recognize that the last term in the brackets is insignificant in comparison to the other two Thus, we have (b) 1.13 The least accurate dimension of the box has two significant figures Thus, the volume (product of the three dimensions) will contain only two significant figures 1.14 (a) The sum is rounded to because 756 in the terms to be added has no positions beyond the decimal must be rounded to (b) because has only two significant figures (c) must be rounded to because 5.620 has only four significant figures 1.15 The answer is limited to one significant figure because of the accuracy to which the conversion from fathoms to feet is given 1.16 giving 1.17 1.18 (a) (b) (c) (d) In (a), the answer is limited to three significant figures because of the accuracy of the original data value, 348 miles In (b), (c), and (d), the answers are limited to four significant figures because of the accuracy to which the kilometers-to-feet conversion factor is given www.elsolucionario.org 1.19 1.20 1.21 (a) (b) (c) 1.22 This means that the proteins are assembled at a rate of many layers of atoms each second! 1.23 1.24 1.25 1.26 (Where L = length of one side of the cube.) 1.27 Thus, and 1.28 We estimate that the length of a step for an average person is about 18 inches, or roughly 0.5 m Then, an estimate for the number of steps required to travel a distance equal to the circumference of the Earth would be or 1.29 We assume an average respiration rate of about 10 breaths/minute and a typical life span of 70 years Then, an estimate of the number of breaths an average person would take in a lifetime is or 1.30 We assume that the average person catches a cold twice a year and is sick an average of days (or week) each time Thus, on average, each person is sick for weeks out of each year (52 weeks) The probability that a particular person will be sick at any given time equals the percentage of time that person is sick, or The population of the Earth is approximately billion The number of people expected to have a cold on any given day is then 1.31 (a) Assume that a typical intestinal tract has a length of about m and average diameter of cm The estimated total intestinal volume is then The approximate volume occupied by a single bacterium is If it is assumed that bacteria occupy one hundredth of the total intestinal volume, the estimate of the number of microorganisms in the human intestinal tract is (b) The large value of the number of bacteria estimated to exist in the intestinal tract means that they are probably not dangerous Intestinal bacteria help digest food and provide important nutrients Humans and bacteria enjoy a mutually beneficial symbiotic relationship 1.32 (a) (b) Consider your body to be a cylinder having a radius of about inches (or 0.15 m) and a height of about 1.5 meters Then, its volume is (c) The estimate of the number of cells in the body is then 1.33 A reasonable guess for the diameter of a tire might be ft, with a circumference ( = distance travels per revo- www.elsolucionario.org (e) 30.9 at a price of $130 or less per kilogram of uranium The total energy required for one year is The number of fission events needed will be and the mass of this number of 30.10 atoms is (a) At a concentration of the mass of uranium dissolved in the oceans covering is two-thirds of Earth’s surface to an average depth of or (b) makes up 0.7% of all uranium, so Fissionable collected and caused to undergo fission, with the release of about ergy supply is and at a consumption rate of or (c) If we assume all of the is per event, the potential en- the time this could supply the world’s energy needs is The uranium comes from dissolving rock and minerals Rivers carry such solutes into the oceans, but the Earth’s supply of uranium is not renewable However, if breeder reactors are used, the current ocean supply can last about a half-million years 492 30.11 (a) (b) (c) The total energy released in this pair of fusion reactions is 30.12 The energy released in the reaction 30.13 The energy released in the reaction 30.14 (a) (b) (c) (d) (e) (f) 30.15 is is or With the deuteron and triton at rest, the total momentum before the reaction is zero To conserve momentum, the neutron and the alpha particle must move in opposite directions with equal magnitude momenta after the Neglecting relativistic effects, we use the classical relationship between momentum and reaction, or kinetic energy, and write or To conserve energy, it is necessary that the kinetic energies of the reaction products satisfy the relation Then, using the result from above, we have or the kinetic energy of the emerging neutron must be 493 www.elsolucionario.org 30.16 30.17 (a) The energy released in the reaction is (b) The proton and the boron nucleus both have positive charges but must come very close to one another in order for fusion to occur Thus, they must have sufficient kinetic energy to overcome the repulsive Coulomb force one exerts on the other Note that pair production cannot occur in a vacuum It must occur in the presence of a massive particle which can absorb at least some of the momentum of the photon and allow total linear momentum to be conserved When a particle-antiparticle pair is produced by a photon having the minimum possible frequency, and hence minimum possible energy, the nearby massive particle absorbs all the momentum of the photon, allowing both components of the particle-antiparticle pair to be left at rest In such an event, the total kinetic energy afterwards is essentially zero, and the photon need only supply the total rest energy of the pair produced The minimum photon energy required to produce a proton-antiproton pair is minimum photon frequency is Thus, the and 30.18 The total kinetic energy after the pair production is The kinetic energy of the antiproton is then 30.19 The total rest energy of the is converted into energy of the photons Since the total momentum was zero before the decay, the two photons must go in opposite directions with equal magnitude momenta (and hence equal energies) Thus, the rest energy of the is split equally between the two photons, giving for each photon 494 and 30.20 Observe that the given reactions involve only mesons and baryons With no leptons before or after the reactions, we not have to consider the conservation laws concerning the various lepton numbers All interactions always conserve both charge and baryon numbers The strong interaction also conserves strangeness Conservation of strangeness may be violated by the weak interaction but never by more than one unit With these facts in mind consider the given interactions: total before total after Charge 0 +1 –1 Baryon number 0 0 Strangeness +1 +1 0 This reaction conserves both charge and baryon number but does violate strangeness by one unit Thus, it but not other interactions total before total after Charge 0 +1 –1 Baryon number +1 +1 0 Strangeness –1 –1 0 This reaction via any interaction 495 www.elsolucionario.org 30.21 Reaction Conservation Law Violated (a) (b) (c) (d) (e) 30.22 Reaction Conservation Law Violated (a) (b) (c) (d) (e) 496 30.23 Reaction Conservation Law Cannot Occur (a) (b) All conservation laws observed May occur via the Strong Interaction (c) Strangeness violated by unit All other conservation laws observed Can occur via Weak Interaction, but not by Electromagnetic or Strong Interactions (d) Strangeness violated by unit All other conservation laws observed Can occur via Weak Interaction, but not by Electromagnetic or Strong Interactions (e) 30.24 Conclusion All conservation laws observed (a) Baryon number, B: Charge, Q: Baryon number, B: Charge, Q: (b) Strangeness is conserved in the first reaction: Strangeness, S: 497 Allowed via all interactions, but photons are the mediator of the electromagnetic interaction and the lifetime of the is consistent with decay by that interaction www.elsolucionario.org The second reaction does not conserve strangeness: Strangeness, S: The second reaction (c) were also produced in the second reaction, giving If then strangeness would no longer be violated: Strangeness, S: +1 Because the total mass of the product particles in this reaction would be greater than that in the first reaction [see part (a)], the total incident energy of the reacting particles 30.25 (a) Strangeness, S: (b) Strangeness, S: (c) Strangeness, S: (d) Strangeness, S: (e) Strangeness, S: (f) Strangeness, S: 498 30.26 30.27 proton u u D total strangeness 0 0 Baryon number 1 charge e e neutron u d D total strangeness 0 0 Baryon number 1 charge 0 The number of water molecules in one liter ( ) of water is Each molecule contains 10 protons, 10 electrons, and neutrons Thus, there are and Each proton contains up quarks and down quark, while each neutron has up quark and down quarks Therefore, there are and 499 www.elsolucionario.org 30.28 Particle d total strangeness 1 baryon number 0 charge 0 Particle strangeness 30.29 30.30 u d 0 S total baryon number 1 charge 0 Compare the given quark states to the entries in Table 30.4: (a) (b) (c) (d) (a) This is the (b) This is the 500 30.31 The reaction is or on the quark level, The right side has plus the quark composition of the unThe left side has a net known particle To conserve the total number of each type of quark, the composition of the unknown particle Therefore, the unknown particle must be a must be 30.32 30.33 (a) is forbidden by (b) is forbidden by both (c) is forbidden by To the reaction for nuclei, and we add three electrons to both sides to obtain Then we use the masses of the neutral atoms from Appendix B of the textbook to compute 30.34 For the particle reaction, the lepton numbers before the event are values must be conserved by the reaction so one of the emerging neutrinos must have has The emerging particles are and 30.35 (a) and These while the other Since this occurs via the strong interaction, all conservation rules must be observed Baryon Number: so X is not a baryon Strangeness: or X has Charge: or X has Lepton Numbers: or X has Of the particles in Table 30.2, the only non-baryon with Thus, this is an elastic scattering process is the positive kaon, The weak interaction observes all conservation rules except strangeness, and 501 www.elsolucionario.org (b) Baryon Number: or X has Strangeness: or X has Charge: or X has Lepton Numbers: or X has The particle must be a neutral baryon with strangeness Thus, it is the (c) Baryon Number: so X is not a baryon Strangeness: or X has Charge: or X has Lepton Numbers: or X has or X has or X has The particle must be a neutral non-baryon with strangeness the neutral pion, 30.36 and This is Assuming a head-on collision, the total momentum is zero both before and after the reaction Therefore, since the proton and the pion are at rest after the reaction, particle X must also be left at rest Particle X must be a neutral baryon in order to conserve charge and baryon number in the reaction Conservation of energy requires that the rest energy this particle be or Particle X is 502 30.37 If a neutron starts with kinetic energy and loses one-half of its kinetic energy in each collision with a moderator atom, its kinetic energy after n collisions will be The average kinetic energy associated with particles in a gas at temperature 10 of the textbook) is (see Chapter Thus, the number of collisions the neutron must make before it reaches the energy associated with a room temperature gas is or 30.38 (a) The number of deuterons in one kilogram of deuterium is Each occurrence of the reaction consumes two deuterons and releases of energy The total energy available from the one kilogram of deuterium is then 30.39 (b) At a rate of eight cents per kilowatt-hour, the value of this energy is (c) Deuterium makes up four-twentieths or one-fifth of the mass of a heavy water molecule Thus, five kilograms of heavy water are necessary to obtain one kilogram of deuterium The cost for this water (d) Whether it would be cost-effective depends on how much it cost to fuse the deuterium and how much net energy was produced If the cost is nine-tenths of the value of the energy produced, each kilogram of deuterium would still yield a profit of (a) The number of moles in 1.0 gal of water is 503 www.elsolucionario.org so the number of hydrogen atoms (2 per water molecule) will be and one of every 500 of these contains a deuteron Thus, the number of deuterons contained in 1.0 gal of water is and the available energy is 30.40 (b) At a consumption rate of the time that this could supply a person’s energy needs is (a) The number of water molecules in the oceans is Since there are hydrogen atoms per water molecule, and the fraction of all hydrogen atoms which conthe number of deuterons in the oceans is tain deuterons is The Each fusion event consumes deuterons, so the number of fusion events possible is releases 3.27 MeV per event, so the total energy that could be released is reaction (b) One hundred times the current world electric power consumption is At this rate, the time the energy computed in part (a) would last is 504 30.41 Conserving energy in the decay with the initially at rest gives or [1] Since the total momentum was zero before the decay, conservation of momentum requires the muon and antiThe relativistic relation beneutrino to recoil in opposite directions with equal magnitude momenta, or or The same retween total energy and momentum of a particle gives for the antineutrino: lation applied to the muon gives Since we must have this may be rewritten as and using the fact that , we have Rearranging and factoring then gives and [2] Substituting Equation [1] into [2] gives or [3] and Subtracting Equation [3] from Equation [1] yields 30.42 The reaction is at the quark level There is This reaction conserves the net number of each type quark For the reaction or there is a net This reaction does not conserve the net number of each type quark 30.43 To compute the energy released in each occurrence of the reaction 505 www.elsolucionario.org we add two electrons to each side to produce neutral atoms and obtain Then, recalling that the neutrino and the photon both have zero rest mass, and using the neutral atomic masses from Appendix B in the textbook gives Each occurrence of this reaction consumes four protons Thus, the energy released per proton consumed is Therefore, the rate at which the Sun must be fusing protons to provide its power output is 30.44 With the kaon initially at rest, the total momentum was zero before the decay Thus, to conserve momentum, the two pions must go in opposite directions and have equal magnitudes of momentum after the decay Since the pions have equal mass, this means they must have equal speeds and hence, equal energies The rest energy of the kaon is then split equally between the two pions, and the energy of each is The total energy of one of the pions is related to its rest energy by fore, the speed of each of the pions after the decay will be 506 where There-