Resolução Física resnick - halliday- krane - 5ta ed vol 1, Exercícios de Física

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PAUL STANLEY INSTRUCTOR*S SOLUTIONS MANUAL to accompany Physics FIFTH EDITION Volume 1 DAVID HALLIDAY ROBERT RESNICK KENNETH KRANE INSTRUCTOR'S SOLUTIONS MANUAL PAUL STANLEY Beloit College to accompany Volume One Physics Fifth Edition DAVID HALLIDAY University of Pittsburgh ROBERT RESNICK Rensselaer Polytechnic Institute KENNETH KRANE Oregon State University (3) JOHN WILEY & SONS, INC. A Note To The Instructor. The solutions here are somewhat brief, as they are designed for the instructor, not for the student. Check with the publishers before electronically posting any part of these solutions; website, ftp, or server access must be restricted to your students. 1 have been somewhat casual about subscripts whenever it is obvious that a problem is one dimensional, or that the choice of the coordinate system is irrelevant to the numerical solution. Although this does not change the validity of the answer, it will sometimes obfuscate the approach 1 viewed by a novice. There are some traditional formula, such as vi= vê, +20çã, which are not used in the text. The worked solutions use only material from the text, so there may be times when the solution here seems unnecessarily convoluted and drawn out. Yes, I know an easier approach existed. But if it was not in the text, I did not use it here. I also tried to avoid reinventing the wheel. There are some exercises and problems in the text which build upon previous exercises and problems. Instead of rederiving expressions, I simply refer you to the previous solution. 1 adopt a different approach for rounding of significant figures than previous authors; in partie- ular, I usually round intermediate answers. As such, some of my answers will differ from those in the back of the book. Exercises and Problems which are enclosed in a box also appear in the Student's Solution Manual with considerably more detail and, when appropriate, include discussion on any physical implications ofthe answer. These student solutions carefully discuss the steps required for solving problems, point out the relevant equation numbers, or even specify where in the text additional information can be found. When two almost equivalent methods of solution exist, often both are presented. You are encouraged to refer students to the Student's Solution Manual for these exercises and problems. However, the material from the Student's Solution Manual must not be copied. Paul Stanley Beloit College stanleyOclunet.edu E1-1 (a) Megaphones; (b) Microphones; (c) Decacards (Deck of Cards); (d) Gigalows (Gigolos); (e) Terabulis (Terribles); (f) Decimates; (g) Centipedes; (h) Nanonanettes (7); (1) Picoboos (Peek-a- Boo); (j) Attoboys ('atta boy); (k) Two Hectowithits (To Heck With It); (1) Two Kilomockingbirds (To Kill A Mockingbird, or Tequila Mockingbird). E1-2 (a) $36,000/52 week = $692/week. (b) $10, 000, 000/(20 x 12 month) = 841, 700/month. (c) 30 x 10º/8 = 3.75 x 10º. Multiply out the factors which make up a century. 365 de) (x hours) /60 as) 1 year 1 day 1 hour 1 century = 100 years ( This gives 5.256 x 107 minutes in a century, so a microcentury is 52.56 minutes. The percentage difierence from Fermi's approximation is (2.56 min)/(50 min) x 100% or 5.12%. E1-4 (3000 mi)/(3 hr) = 1000 mi/timezone-hour. There are 24 time-zones, so the circumference is approximately 24 x 1000 mi = 24, 000 miles. E1-5 Actual number of seconds in a year is 24 hr 60 min (365.25 days) (E) ( Thr ) ( The percentage error of the approximation is then =3.1558 x 10's. 31416» 107 x 107 s =—0,45 % E1-6 (a) 1078 seconds per shake means 10º shakes per second. There are 365 days) /24 hr) /60 min 05). ; ( 1 year ) (E) ( Thr ) (1 = 3.1536 x 10fa/year. This means there are more shakes in a second. (b) Humans have existed for a fraction of 10º yeara/1010 years = 1072. That fraction of a day is 10-4 (24 hr) (5 min ( 60 8 hr fis) =864s. We'lt assume, for convenience only, that the runner with the longer time ran exactly one mile. Let the speed of the runner with the shorter time be given by v:, and call the distance actually ran by this runner dy. Then w = di/t;. Similarly, vz = do /t2 for the other runner, and da = 1 mile. We want to know when vw > vz. Substitute our expressions for speed, and get di/t > do/ta. Rearrange, and di/da > tr/ta or di/da > 0.99937. Then dy > 0.99937 mile x (5280 feet/1 mile) or di > 5276.7 feet is the condition that the first runner was indeed faster. The first track can be no more than 3.3 feet too short to guarantee that the first runner was faster. area of the base is the area of a semicircle. Then V=4h= (Gr) h. The volume is 14)(2000 x 1000 m)2(3000 m) = 1:88 x 1018 im? 4 em 3 a O 885 1018 q? ne | 100 mA gn - Im E1-16 The volume is (77x 10tm?)(26m) = 2.00x 107 mê. This is equivalent to (2.00x 107 mº)(10-? km/m)? = 0.02 km? The volume of Antarctica is approximated by the area of the base time the height; the ELI7T (2) C=2xr = 27(6.37 x 108 km) = 4.00 x 104 km. (b) A = 4772 = 47(6.37 x 10º km)? = 5.10 x 108 km. (c) V = $7(6.37 x 103 km)? = 1.08 x 1012 mê. E1-18 The conversions: squirrel, 19 km/hr(1000 m/km) /(3600 s/hr) = 5.3m/s; rabbit, 30 knots(1.688ft/s/knot)(0.3048 m/ft) = 15m/s; snail, 0.030 mi/hr(1609 m/mi)/(3600 s/hr) = 0.013 m/s; spider, 1.8 ft/s(0.3048 m/f) = 0.55m/s; cheetah, 1.9 km /min(1000 m/km)/(60 s/min) = 32m/s; human, 1000 em/s/(100 cm/m) = 10m/s; fox, 1100 m/min/(60 s/min) = 18m/s; lion, 1900 km/day(1000 m/km)/(86, 400 s/day) = 22m/s. The order is snail, spider, squirrel, human, rabbit, fox, lion, cheetah. raspar light o (Sem o E ) 100 year aSsty e x 108 m/s) x (1 year) 1 mi “3600 s/ Ad century which is equal to 0.00286 light-year /century. E1-20 Start with the British units inverted, mê -2 ; gal (a) (peço =) ( mi ) = 784102 pm, 30.0 mi gal in' 1.609 km E1-21 (b) À light-year is (3.00 x 108 km/s) (E ) ( Zá tar DT = 12 Thr 7 =) (365 days) = 9.46 x 10? km. A parsec is (5 ru) = 3:09 x 1088 kem. Em rad 1.50 x 108 km / 360º 5 001” 2x rad One light-year is the distance traveled by light in one year, or (3 x 10º m/s) x (1 year). (a) (1.5 x 108 km) /(3.09 x 1018 km /pe = 4.85 x 1078 pc. (1.5 x 10º km)/(9.46 x 1022 km/1y) = 1.59 x 1075 ly. El-22 First find the “logarithmic average” by logdy = 5 (log(2 x 1028) + log(1 x 10-15), Flog (2 x 10% x 1 x 1075), Flog? x 104 = log (v2x 10) . Solve, and day = 450 km. The number of atoms is given by (1 kg)/(1.00783 x 1.661 x 1072” kg), or 5.974 x 1028 atoms. El-24 (a) (2x 10+ 16)u(1:661 x 10-27kg) = 3.0 x 10-28kg. (b) (1.4 x 1021kg) /(3.0 x 10-28kg) = 4.7 x 104º molecules. El-25 The coflee in Paris costs $18.00 per kilogram, or 0.4536 kg 818.00 kg”! ( didndb ia + ) = Seo ib-! , a) “16 1h Tt is cheaper to buy coffee in New York (at least according to the physics textbook, that is.) E1-26 The room volume is (21 x 13 x 12)ft*(0.3048 m/f)? = 92.8m?. The mass contained in the room is (92.8m2)(1.21 kg/m?) = 112kg. Ei-27| One mole of sugar cubes would have a volume of Na x 1.0 cmê, where Na is the Avogadro constant. Since the volume of a cube is equal to the length cubed, V = 23, then | = Na em =84x10" em. E1-28 The number of seconds in a week is 60 x 60 x 24 x 7 = 6.05 x 105. The “weight” loss per second is then (0.23kg)/(6.05 x 1025) = 3.80 x 107! mg/s. 9| The definition of the meter was wavelengths per meter; the question asks for meters per wavelength, so we want to tale the reciprocal. The definition is accurate to 9 figures, so the reciprocal should be written as 1/1,650, 763.73 = 6.05780211 x 10-"m = 605.78024 nm. E1-30 (a) 37.76 + 0.132 = 37.89. (b) 16.264 — 16.26325 = 0.001 The easiest approach is to first solve Darcy's Law for K, and then substitute the known for the other quantities. Then VL . (mê) (m) K= AHE has units of ESTO) which can be simplified to m/s. P1-3 (a) The circumference of the Earth is approximately 40,000 km; 0.5 seconds of an arc is 0.5/(60 x 60 x 360) = 3.9 x 107? of a circumference, so the north-south error is (3.9 x 10-7)(4 x 10"m) = +15.6m. This is a range of 31 m. (b) The east-west range is smaller, because the distance measured along a latitude is smaller than the circumference by a factor of the cosine of the latitude. Then the range is 31 cos 43.6º = 22m. (c) The tanker is in Lake Ontario, some 20 km off the coast of Hamlin? Pi-4 Your position is determined by the time it takes for your longitude to rotate ”underneath” the sun (in fact, that's the way longitude was measured originally as in 5 hours west of the Azores...) the rate the sun sweep over at equator is 25,000 miles/86,400 s = 0.29 miles /second. 'The correction factor because of latitude is the cosine of the latitude, so the sun sweeps overhead near England at approximately 0.19 mi/s. Consequently a 30 mile accuracy requires an error in time of no more than (30 mi)/(0.19 mi/s) = 158 seconds. Trip takes about 6 months, so clock accuracy needs to be within (158s)/(180 day) = 1.2 sec- onds/day. (b) Same, except 0.5 miles accuracy requires 2.6 s accuracy, so clock needs to be within 0.007 s/day! P1-5 Let B be breaths/minute while sleeping. Each breath takes in (1.43 g/L)(0.3 L) = 0.429 g; and lets out (1.96 g/L)(0.3 L) = 0.288 g. The net loss is 0.141 g. Multiply by the number of breaths, (8 hr)(60 min. /hr) (0.141 8) = B(67.68 g). FI take a short nap, and count my breaths, then finish the problem. I'm back now, and I found my breaths to be 8/minute. So I lose 541 g/night, or about 1 pound. P1-6 The mass of the water is (1000kg/mº)(5700m$) = 5.7 x 109kg, The rate that water leaks lrains out is (5.7 x 10 kg) 2 hr)(3600 s/hr) = 132 a Let the radius of the grain be given by rg. Then the surface area of the grain is A, = arg, and the volume is given by V = (4/3)xr;. H N grains of sand have a total surface area equal to that of a cube 1 m on a edge, then NA, —6 m?, The total volume V of this number of grains of sand is NV,. Eliminate N from these two expressions and get (6 m?) Ag Then V. = (2 m2)(50 x 10% m) = 1 x 104 mê. The mass of a volume V, is given by a 2600 kg 4 m3 = 1x 10“ m ( T mê ) 0.26 kg. — (ômêro VK= NV, = 3 Va P1-8 For a cylinder V = nr2h, and A = 2772? + 27rh. We want to minimize A with respect to changes in r, so dA d 2 v Fa & (2m +2mr53) V dr 2 Set this equal to zero; then V = 2xr3. Notice-that À = 2r in this expression. P1-9 (a) The volume per particle is (9.27 x 10-6kg)/(7870 kg/m?) = 1.178 x 10-2m?, The radius of the corresponding sphere is — a/MLITBX 1020) E = 1.41 x 10710m., Double this, and the spacing is 282 pm. (b) The volume per particle is (3.82 x 10-8kg) /(1013 kg/m?) = 3.77 x 10Pmº. The radius of the corresponding sphere is a [BTT X 10 Bd) = 2.08 x 10710m. 4m Double this, and the spacing is 416 pm. P1-10 (a) The area of the plate is (8.43 cm)(5.12 cm) = 43.2 cm2, (b) (3.14)(3.7 em)? = 43 cm?, The stated angle is measured from the east-west axis, counter clockwise from east. So O is measured against the north-south axis, with north being positive; À is measured against east-west with east being positive. Since her individual steps are displacement vectors which are only north-south or east-west, she must eventually take enough north-south steps to equal 1.96 km, and enough east-west steps to equal 2.80 km. Any individual step can only be along one or the other direction, so the minimum total will be 4.76 km. E2-6 Let F = 124] km and F; = (72.62 + 31.45) km. Then the ship needs to travel AF=F E = (51.4i+431.4) km. Ship needs to travel 51.42 431.42 km = 60.2 km in a direction 6 = tan”!(31.4/51.4) = 31.4º west of north. (a) In unit vector notation we need only add the components; a+5 = (5i+3)+(-3i+2)) = CHA =2+5. (b) If we define E = & + B and write the magnitude of E as c, then c= /2+&=vVZ+5= 5.39. The direction is given by tan 9 = c,/c» which gives an angle of 68.2º, measured counterclock- wise from the positive x-axis. E28 ()d+b=(4-Di+(c3+Dj+(1+4)k=3i-2+5k. (b)ã-B=(4--Di+(-3-Dj+(L- 4)k = 6] — 4) ak. (c) Resrrange, andE=b-&orb-d=(-1-49i+(1--3j+(4-1)k=—5i+4]+3k. E2-9 (a) The magnitude of dis 202 +(-3.0)! = 5.0; the direction is 6 = tan !(-3.0/4.0) = 323º, (b) The magnitude of B is 6.027 8/0º = 10.0; the direction is 9 = tan”1(6.0/8.0) = 36.9º. 'c) The resultant vector is & + b = (40+6.0)]+ (-30 4 8.0). The magnitude of & + 6 is 10.02 (5.0)? = 11.2; the direction is 8 = tan *(5.0/10.0) = 26.6º. d) The resultant vector is &— B = (40 — 6.0)i+ (-3.0 — 8.0)). The magnitude of à — b is (2.02 + (110) = 11.2; the direction is 9 = tan-1(-11.0/ — 2.0) = 260º. e) The resultant vector is b— & = (6.0 — 4.0) + (8.0 — —3.0)). The magnitude of b — & is Vade + (10) = 11.2; the direction is 6 = tan(11.0/2.0) = 79.7º. E2-10 (a) Find components of ã; a, = (12.7) cos(28.2º) = 11.2, ay = (12.7) sin(28.2º) = Find components of D; be = (12.7) cos(133º) = —8.66, by = (12.7) sin(133º) = 9.29. Then F=ã+6=(11.2-8.66)] + (6.00 + 9.29)] = 2.54i + 15.29). (b) The magnitude of F is 2.542 + 15.29? = 15.5. (c) The angle is 8 = tan-2(15.29/2.54) = 80.6º. E2-11 Consider the figure below. 10 E2-12 Consider the figure below. Our axes will be chosen so that Í points toward 3 O'clock and j points toward 12 O'clock. (a) The two relevant positions are Fi = (11.3 em)i and f; = (11.3 cm)j. Then AF = WE = (113 cm) - (11.3 em. (b) The two relevant positions are now É; = (11.3 cm)j and F = (-11.3 em)j. Then AF = HR = (118emj-(-11.3 cmj (22.6 emj. (c) , . The two relevant positions are now Fi = (—11.3 em)j and E = (—11.3 em). Then AF = ER = (-113cmj-(-113 em = (0cmj. E2-14 (a) The components of Fj are rio = (4.13) cos(225º) = —2.92m Tiy = (4.13m)sin(225º) = —2.92m. 1 (a) Evaluate F when t=2s. F = (Om/ — (5 m/)gi+(6m) (7 m/s (2 m/8*(2 5)? — (5 m/s)2 si + [(6 m) — (7 m/s*)(2 s)* [(16 m) — (10 má +[(6 m) — (112 m)j [6 m)Ji + [-(106-m)]3. (b) Evaluate: so emp (m/s (7 m/s)Aeh dê K6 m/s — (5 m/s)i+I-(28 m/s). Into this last expression we now evaluate V(t = 2 s) and get 7 = (6m/5)2s)? — (5 m/s)li+|-(28 m/s'(2 s/ 24 m/s) — (5 m/s) + [=(224 m/s) [18 m/s)li+ [-(224 m/s)li, for the velocity Y when t=2s. (c) Evaluate py h | ] (6 m/s*jxtji + [-(28 m/s)32] [12 m/6i + (-(84 m/8" JP). Into this last expression we now evaluate &(t = 25) and get ã = [12m/)2 s)i+i-(8 m/s!)(2 224 24 m/s? já + [-(336 m/s"). E2-18 (a) Let ui point north, j point east, and É point up. The displacement is (8.71 + 9.7) + 2.9k) km. The average velocity is found by dividing each term by 3.4 hr; then Fay = (2.614 2.95 + 0.85) km/he. The magnitude of the average velocity is 2.62 + 2.92 + 0.85? km/hr = 4,0 km/hr. (b) The horizontal velocity has a magnitude of 2.62 4 2.9? km/hr = 3.9 km/hr. The angle with the horizontal is given by 6 = tan” !(0.85/8.9) = 13º. E2-19 (a) The derivative of the velocity is à= [(60m/s?) — (8.0m/s)gi so the acceleration at t = 3s is &= (—18.0m/s?)i. (b) The acceleration is zero when (6.0m/s2) — (8.0m/s?)t = 0, or t = 0.75s. (c) The velocity is never zero; there is no way to “cancel” out the y component. (d) The speed equals 10m/s when 10 = vi +82, or vs = +6.0m/s. This happens when (6.0m/s2) — (8.0m/sº)t = +6.0m/s, or when t = Os. 13 E2-20 Hv is constant then so is v? = v2 +v2. Take the derivative; d d Disto + oyo = Uvas + ya). But if the value is constant the derivative is zero. Let the actual flight time, as measured by the passengers, be T. There is some time diference between the two cities, call it AT =: Namulevu time - Los Angeles time. The AT will be positive if Namulevu is east of Los Angeles. The time in Los Angeles can then be found from the time in Namulevu by subtracting AT. The actual time of flight from Los Angeles to Namulevu is then the difference between when the plane lands (LA times) and when the plane takes off (LA time): T (18:50 — AT) — (12:50) 6:00— AT, where we have written times in 24 hour format to avoid the AM/PM issue. The return fight time can be found from T (18:50) — (1:50 — AT) = I700+47T, where we have again changed to LA time for the purpose of the calculation. (b) Now we just need to solve the two equations and two unknowns. 17:00+4T = 600-AT 2AT = 6:00 17:00 AT = —5:30. Since this is a negative number, Namulevu is located west of Los Angeles. (a) T=6:00- AT =11:30, or eleven and a half hours. (c) The distance traveled by the plane is given by d = vt = (520 mi/hr)(11.5 br) = 5980 mi. We'll draw a circle around Los Angeles with a radius of 5980 mi, and then we look for where it intersects with longitudes that would belong to a time zone AT away from Los Angeles. Since the Earth rotates once every 24 hours and there are 360 longitude degrees, then each hour corresponds to 15 longitude degrees, and then Namulevu must be located approximately 15º x 5.5 = 83º west of Los Angeles, or at about longitude 160 east. The location on the globe is then latitude 5º, im the vicinity of Vanuatu. When this exercise was originally typeset the times for the outbound and the inbound flights were inadvertently switched. I suppose that we could blame this on the airlines; nonetheless, when the answers were prepared for the back of the book the reversed numbers put Namulevu cast of Los Angeles. That would put it in either the North Atlantic or Brazil. E2-22 There is & three hour time zone difference. So the flight is seven hours long, but it takes 3 hr 51 min for the sun to travel same distance. Look for when the sunset distance has caught up with plane: daunset = planes Vsunset(t — 1:35) = Uplanet, (E 1:35)/8551 = 4/7:00, so t = 3:31 into flight. 14 w =d/ty ando =do/ta. vs = d/t, where d total distance and £ is the total time. The total distance is dy + do = 2d. The total time t is just the sum of t, and ta, so 2d; tita 2d d/w + dafva 2 1704 1/0 Take the reciprocal of both sides to get a simpler looking expression 2 1 1 Vavo Ut Then the average speed is 48 km/hr. E2-30 (a) Average speed is total distance divided by total time. Then (240 Fe) + ÃO ft) “0 Ta A/AD Rs) (240 RO 75) 7 ts (b) Same approach, but different information given, so — (OS) eo) + (00 6)00 Ro) pop May (60 5) + (605) E2-31 The distance traveled is the total area under the curve. The “curve” has four regions: (1) a triangle from O to 2 s; (II) a rectangle from 2 to 10 s; (HI) a trapezoid from 10 to 12 s; and (IV) a rectangle from 12 to 16 s. The area underneath the curve is the sum of the areas of the four regions. d= 52 s)(8m/s) + (8.0s)(8m/s) + S(25X8m/s +4m/s) + (4.0s)(4m/s) = 100m. E2-32 The acceleration is the slope of a velocity-time curve, — (8m/s) = (4m/s) “="Qos)- (ão) = 2m/e. The initial velocity is Y; = (18 m/s)i, the final velocity is Y; = (—30 m/s)i. The average acceleration is then . . . um DE MEO (30 moi (18 m/o)i a — ACO A 245 which gives day = (—20.0 m/s?)i. 16 E2-34 Consider the figure below. 0 RE NE a -s E2-35 (a) Upto 4 v, > 0 and is constant. From A to B v, is decreasing, but still positive. From BtoCw=0. FomCtoDv, O and is decreasing. From Ato Bv, =0. From Bto O vz > O and is increasing. From C' to D w, > O and is constant. (b) No. Constant acceleration would appear as (part of) a parabola; but it would be chailenging to distinguish between a parabola and an almost parabola. Consider the figure below. PS Nes Pass ngpraçaãoo E pole q E2-38 Consider the figure below. w”