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UW-Madison PHYSICS 107 - Answer Key - Chapters 6 & 8 Assignments

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107 Answer Key - Chapters 6 & 8 assignmentsConceptual: Ch.6: 36, Ch.8: 8, 12, 30, 34Problems: Ch.6: 14, 18, Ch.8: 2, 10October 8, 20051 Chapter 61.1 Conceptual Exercises36.) “Figure 6.16 is a graph of a roller coaster’s height above the groundversus the length of track it covers. The coaster is powered up to its highpoint at 100 m from the starting point. From the high point, the coastercoasts freely all the way to the end. Assume that the coaster starts from restat the high point and encounters no friction or air resistance. Between 200m and the finish, whe re is it moving slowest? Fastest?”To find where the car will be moving the slowest or fastest, we ne edto use conservation of energy. Since the car is barely moving at the highpoint, all of its energy is potential energy. Conservation of energy says toset the initial energy equal to the energy of the point you’re interested in:Efinal=12mass ∗ (speed)2+ mass ∗ g ∗ height. If we want the speed as largeas possible, we want the height at that point to be as small as possible - thatmeans more of its energy is kinetic energy. The smallest height we can geton the Figure is height = 0 at 600 m.If we want the s peed to be as small as possible, we want the he ight tobe large - that means that more of its energy is potential energy, not kineticenergy. The largest height we can get (not including the starting point) isan elevation of 30 m at a distance of 500 m.Note that in both parts of the question we did not care about whathappened before or after the fastest or slowest point, i.e. big/small hill, flat,start or end of the ride. Only the height matters - not how it got there!11.2 Problems14). “Jack, who has a mass of 30 kg and weighs 300 newtons, sits in a child’sswing. You pull the swing back so that it is 2 m above its low point, andrelease it. What form of energy, and how much energy, does Jack have whenhe is pulled back and held at rest?”If Jack is at rest, then he cannot have kinetic energy. Since the weight ofJack is displaced a certain height (and someone had to do work to put Jackthere), Jack has gravitational potential energy. If we measure our heightsfrom the low point of the swing, then Jack has gravE = (weight) ∗ (height)= 300 N ∗ 2 m = 600 J.18.) “You do a pullup, lifting yourself by 0.5 m in 2 s. If your weight is600 N, how much work did you do, and what was your power output duringlifting?”Work = (Force)*(distance) = 600 N ∗ 0.5 m = 300 J.Power =W ork donetimeelapsed=300 J2 s= 150 W atts.2 Chapter 82.1 Conceptual Exercises8.) “What happens to the energy of the two waves in Figure 8.9 when theyinterfere destructively, as shown in the second of the three sketches? Did theenergy vanish?”Because the pulse is a transverse wave, the velocities of the small pieces ofstring are in the vertical direction (but the pulse itself travels horizontally).The leading edge of a pulse pushes the string up, while the trailing edgepushes the string down. When the waves overlap, the amplitudes cancel, butthe velocities do not! On one side, part of the string is moving up due to theleading edge of one pulse and the trailing edge of the other, and on the otherend, part is moving down due to the trailing edge of one and leading edgeof the other. This means that the string has some velocity, even whe n thereis complete destructive interference. A non-zero velocity means a non-zerokinetic energy, thus the energy is stored in the kinetic energy of the string.12.) “You shine two flashlights on a wall. Why don’t you see an inter-ference pattern?”You don’t see an interference pattern because the light from the flashlightisn’t coherent. Coherent light is light that is synchronized; the flashlight2light is not coherent because the thermal motion that generates the lightis random. Synchronized light is necessary to view the interference pattern,which can be obtained from a laser. You would also need just one laser, sincethere wouldn’t be the necessary synchronization between two.30.) “According to Figure 8.24, what are the atomic numbers of carbonand helium? Roughly how much more massive is the carbon atom than thehelium atom?”The atomic number of an element is the number of protons in the nucleus.Counting the blue circles in the Figure, theatomic number of carbon is 6,and the atomic number of helium is 2. Since an elec tron’s mass is much lessthan a proton’s, we can compare the mass of carbon and helium by compar-ing the size of the nucleus - carbon is 3 times more massive than helium ,counting protons alone. It turns out, though, that the mass of a neutron(white circle) is about the same as a proton’s; comparing the total numberof protons and neutrons for carbon (12) and helium (4), gives us the sameresult: carbon is 3 times more massive than helium.34.) “Making Estimates About how many atoms thick is a sheet ofpaper?”An atom is about 10−10m/atom across. Our textbook looks roughly4cm thick, and has about 600 pages. So a sheet of paper in this book is0.04m/600 = 6.67X10−5m thick. Since this is just an estimate, let’s call it10−4m. We can get 10−4m/(10−10m/atom) = 106atoms in this thickness, soa sheet of paper is about 106, or 1,000,000, atoms thick!2.2 Problems2.) “Typical AM radio wavelengths are about as long as a football field,while typical FM radio wavelengths are about a meter long. Which one thenhas the highest (largest) frequency? If an AM wavelength is 100 times longerthan an FM radio wavelength, then how do the frequencies compare?”The relation between wavelength and frequency is given by speed =(wavelength)∗(frequency), or dividing by wavelength, frequency = (speed)/(wavelength).If the wavelength of AM radio waves is longer than FM, then the frequency ofFM radio waves will be higher than AM, because there is a smaller numberin the denominator for FM waves.If we increase the wavelength by 100 times, then we need to divide the3frequency by 100 to make the equation true:frequencyF M=speedwavelengthF M⇒frequencyF M100=speed100 ∗ wavelengthF M,(1)where the speed for both waves is the same (speed of light). Thus, thefrequency for AM radio waves is smaller by a factor of 100.10.) “Find the wavelength of an AM radio wave whose frequency is 1000kHz”Once again we use the equation speed = (wavelength) ∗ (frequency),inverting it this time to get wavelength = (speed)/(frequency). Using thespeed of light as 3 ∗ 108m/s, and noting that 1000 kHz = 1, 000, 000 Hz


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UW-Madison PHYSICS 107 - Answer Key - Chapters 6 & 8 Assignments

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