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FinalExamSolution - PHYSICS 9B FINAL EXAM December 11, 2009...

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Unformatted text preview: PHYSICS 9B FINAL EXAM December 11, 2009 I certify by my signature below that I will abide by the UC Davis Code of Academic Conduct. This includes • not copying from anyone else’s exam • not letting any other student copy from my exam • not discussing this exam with any student who has not yet taken it, nor providing any information, written or oral, that might get to a student who has not yet taken it. Anyone suspected of cheating will be automatically reported to Student Judicial Affairs. LAST NAME: (P R I N T) FIRST NAME: (P R I N T) STUDENT ID: (LAST 4 DIGITS) Signature: Problem 1. (2 points) An ice cube floats in a glass with an orange juice. When the ice melts will the orange juice level A. Rise B. Fall C. Remain unchanged Solution: The key is to realize that the density of orange juice (OJ) is larger than the density of water. Let’s denote volume of OJ without the ice cube as Voj. When the ice cube is immersed, it is partially under the orange juice, because its weight should be compensated by the buoyant force. Let’s denote the full volume of ice as Vice and its volume under the orange juice as V’. B= ρojV’g = ρiceViceg which produces V’=Viceρice/ρoj as a part of the ice volume under the OJ. Therefore the OJ level (when the ice floats) is Voj+V’. When the ice melts completely it adds the volume Vadd=mice/ρw=Viceρice/ρw and the volume level is determined by Voj +Vadd. Since ρoj>ρw we conclude that Vadd>V’ therefore the OJ level should raise. Answer A. Problem 2. (2 points) A sound source is moving towards standing listener with speed v but a strong wind of the same speed v is blowing in the direction opposite to the source motion. Compared to the source frequency the listener will hear the sound of A. B. C. D. Zero frequency Lower (but finite) frequency The same frequency Higher frequency. Solution: Let’s consider a coordinate system associated with the wind. Then, the source is moving with speed 2v and the listener is moving with the speed v in the same direction. fL=(vsound-vL)/λfront=(vsound-vL)/ (vsound-vS)fS=(vsound-v)/(vsound-2v)fS >fS. The answer is D. Problem 3. (2 points) Consider two lenses, (first is divergent and second is convergent) with focal distances equal to f. which are placed at distance f from each other. The object is located as shown. The image is A. Erect, located to the left of both lenses. B. Erect, located in the middle of both lenses. C. Erect, located to the right of both lenses. D. Inverted, located to the left of both lenses. E. Inverted, located in the middle of both lenses. F. Inverted, located to the right of both lenses. 1 Object 2 f f f Solution. Consider two rays, one directed towards the center of the first lens and another one towards its focus We obtain inverted image on the right, answer F. Object Image Problem 4. (2 points) Consider a horizontal plane of thin film with a thickness t. This film is located in between air and water (see sketch). Light is directed from the air downward through the film and into the water, perpendicular to the surfaces. The index of refraction of the film n1. The index of refraction of water is n2. The wavelength of the incident wave in the air is λ. The smallest non-zero thickness tmin leading to no reflection is A. tmin = B. tmin = C. tmin = λ 2 λ λ 2n1 2n2 D. tmin = λ E. tmin = F. tmin = λ n1 λ n2 Solution Problem 5. (2 points) Why do radio waves diffract around buildings, while light waves do not? A. Radio waves travel much slower than light waves B. Radio waves are electromagnetic waves while light waves are not. C. Radio waves have a much longer wavelengths than light waves. Solution: The wavelengths of AM radio waves are hundreds of meters, much larger than the size of buildings, so they easily diffract around buildings. Light, with wavelengths a tiny fraction of a centimeter, show no appreciable diffraction around buildings. Answer C. Problem 6. (2 points) You have four samples of ideal gas, each of which contains the same number of moles of gas, has the same initial temperature, volume and pressure. You expand each sample to twice of its initial volume. Sample 1 is a monoatomic gas expanded isothermally. Sample 2 is a monoatomic gas expanded adiabatically. Sample 3 is a diatomic gas expanded isothermally, Sample 4 is a diatomic gas expanded adiabatically. Rank the four samples in order from highest to lowest value of the final pressure. A. 2 then 4 then 1 and 3 tie B. 1 and 3 tie then 4 then 2 C. 2 and 4 tie then 1 then 3 D. 1 then 3 then 4 then 2 E. 2 then 4 then 3 then 1 Solution. Samples 1 and 3 are expanded isothermally therefore PV=const. The volume of each sample increases twice of its initial value, so the final pressure is one half of the initial pressure: P1=P3=P/2 By contrast, samples 2 and 4 are expanded adiabatically, so PVγ=const and the pressure decreases by a factor of 2γ. For monoatomic gases γ=CP/CV=2.5/1.5=5/3 therefore its final pressure is 25/3=3.17 times smaller than the initial pressure: P2=P/3.17 For diatomic gases γ=CP/CV=3.5/2.5=7/5 so its final pressure is 27/5=2.64 times smaller than the initial pressure: P4=P/2.64. We obtain (P1=P3)>P4>P2. Answer B. Problem 7. (2 points) The PV-diagrams below show four types of cyclic processes done by an ideal gas. Rank them according to the total work done by the gas greatest first. 1 2 3 4 P P2 P1 V1 V2 V P P2 P1 V1 V2 V P P2 P1 V1 V2 V P P2 P1 V1 V2 V A. W3>W2>W1>W4 B. W1>W3>W4>W2 C. W1>W2>W3>W4 D. W3>W1>W4>W2 E. W4>W1>W2>W3 Solution. The work done by the gas during a cyclic process is the insider area on PV diagram. It is largest for process 3, then 1, then 4, then 2. Answer D. Problem 8. (2 points) Which statements (if any) contradict with the second law of thermodynamics? A. A cyclic process can transfer heat from a colder place to a hotter place with an input of mechanical work. B. Heat supplied to working substance of a heat engine cannot be completely converted into work. C. An internal combustion engine can be made more efficient than a Carnot engine operating between the same temperatures by adjusting its compression ratio. D. Mechanical energy can be converted to thermal energy and vice versa. Solution. Statement A is correct, that is the principle of refrigerator. statement B is correct, that is the principle of heat engine, statement C is incorrect, a Carnot engine is a model of highest efficiency. statement D is incorrect. Answers C and D. Problem 9. (3 points) The earth does not have a uniform density; it is most dense at its center and least dense at its surface. An approximation of its density is ρ(r)=A-Br where A=12,700 kg/m3, B=1.5 x 10-3 kg/m4, and r is the distance away from the center. Find the mass of the earth within this model assuming that the radius of the earth R=6.37 x 106 m. Solution. Since the density is not uniform we need to take the integral of the density over the earth volume assuming it is a sphere. Since the density is spherically symmetric we obtain R R 4 3 M =∫ ρ dV = ∫ ρ ( r )4π r 2dr = 4π ∫ ( A − Br ) r 2 dr = π R 3 ( A − BR ) Sphere 0 0 3 4 4 3 = 3.1415*[6.37 x106 ]3 (1.27 x104 − 1.5 x10−3 * 6.37 x106 ) = 5.99 x1024 kg. 3 4 Problem 10. (3 points) A triangular wave pulse on a taut string travels in the positive x direction with speed v. The tension in the string is F and the linear mass density of the string is μ. At t=0, the shape of the pulse is given by 0 for x < − L ⎧ ⎪ h( L + x ) / L for − L < x < 0 ⎪ y ( x,0) = ⎨ ⎪ h( L − x ) / L for 0 < x < L ⎪ 0 for x > + L ⎩ a) Draw the pulse at t=0. b) Determine the wave function y(x,t) at all times t. c) Find the instantaneous power P(x,t) of the wave. Show that it is zero everywhere except for –L<(x-vt)<L where it is a constant. Solution. a) The pulse looks like this y h -L +L x b) The wave function at any time is obtained by replacing x by x-vt 0 for x − vt < − L ⎧ ⎪ h( L + x − vt ) / L for − L < x − vt < 0 ⎪ y ( x, t ) = ⎨ ⎪ h( L − x + vt ) / L for 0 < x − vt < L ⎪ 0 for x − vt > + L ⎩ c) The instantaneous power is given by −F * 0 * 0 = 0 for x − vt < − L ⎧ ⎪ − F ( h / L)( − vh / L) = Fv (h / L) 2 for − L < x − vt < 0 ∂y ( x, t ) ∂y ( x, t ) ⎪ P ( x, t ) = − F =⎨ 2 ∂x ∂t ⎪ − F ( −h / L)( vh / L) = Fv ( h / L) for 0 < x − vt < L ⎪ for x − vt > + L −F * 0 * 0 = 0 ⎩ We see that P(x,t) is zero everywhere except for –L<(x-vt)<L where it is given by Fv(h/L)2. Problem 11. (3 points) You are looking at your own reflection in a shiny silvered Christmas tree ornament 0.75 m away. The diameter of the ornament is 7.2 cm. Your height is approximately 1.7 m. Although not drawn to scale the figure below illustrates the setting. In the following use paraxial approximation. 0.75 m a) Draw a principal ray diagram showing the position of your image. State whether the image is virtual or real and whether it is direct or inverted. b) At what distance from the surface of the ornament the image appears? c) What is the height of your image in the ornament? Solution. a) Here we obtain a virtual direct image b) Note that the focal distance for convex mirrors is negative. Here f=-d/4=-7.2/4=-1.8 cm. The distance from the surface (in cm) is s’=(1/f-1/s)-1=(-1/1.8-1/75)=-1.76 cm. c) The lateral magnification is s’/s=1.76/75=0.0234. Therefore, the image height is 0.0234*170=3.978 cm. Problem 12. (3 points) In a single slit diffraction experiment the wavelength of the incident light is 520 nm, the slit width is 8 μm and the distance between the slit and the screen is 8.4 m. In the following use small angle approximation tanθ= sinθ=θ. a) Find the position y = y1 of the first intensity minimum. b) Find the position y = y2 of the second intensity maximum (approximate it by point between first and second intensity minima). c) If the intensity of the first (central) maximum is 0.5 W/m2, estimate the intensity at the second intensity maximum. y2 Solution a) For single slit diffraction, destructive interference occurs when, a sinθ =mλ. The small angle approximation gives us y1/L = tanθ = sinθ = mλ/a or for m=1 y1 = Lλ/a = 8.4 * 520 * 10-9/ (8 * 10-6) = 546 * 10-3 = 0.546 m. b) The second minimum can be found at the position 2y1=1.092 m, therefore the second intensity maximum is approximately located at y2=0.819 m. c) The intensity is approximately given by I0 ⎡ sin[π a sin θ / λ ⎤ ⎡ sin[3π / 2) ⎤ 2 I = I0 ⎢ ⎥ = I 0 ⎢ 3π / 2 ⎥ = 9π 2 / 4 = 0.022 W / m ⎣ π a sin θ / λ ⎦ ⎣ ⎦ 2 2 Problem 13. (3 points) Evaporation of sweat is an important mechanism for temperature control in some warmblooded organisms. a) What amount of heat is lost by a 70 kg man when his body is cooled off by 1 C? The specific heat of typical human body is 3480 J/kg*K. b) What mass of water must evaporate from the skin of this man to cool his body by 1 C. The heat of vaporization of water at body temperature is 2.42 x 106 J /kg*K. c) What volume of water must the man drink to replenish the evaporated water? Compare to the volume of a soft-drink can (335 cm3) The density of water is 1000 kg/m3. Solution. a) The heat lost by the man Q= cmanmmanΔT=70*3480*(-1)=-243600 J. The minus sign reflects the fact that the heat flows away from the man. b) The heat flowing out of the man when he cools off his body by 1 C is equal to the heat needed to evaporate the water from the body. Therefore -cmanmmanΔT=mwaterLv. So we have mwater= -cmanmmanΔT/Lv=70*3480*(-1)/[ 2.42 x 106]=0.101 kg. c) The volume of water V=m/ρ=0.101/1000=1.01x10-4 m3=101 cm3. This is about 28 % of volume of soft-drink can Problem 14. (3 points) The surface of the sun has a temperature of about 5800K and consists largely of hydrogen atoms. a) Find the rms speed of a hydrogen atom at this temperature. b) Find the escape speed of a hydrogen atom from the sun (that is the minimal speed needed for a projectile to leave the surface to infinite distance). c) Find the temperature of the sun at which appreciable amount of hydrogen would start leaving its surface. Boltzmann constant is 1.38x10-23 J/K. Mass of hydrogen atom is 1.67x10-27 kg. The gravitational constant is 6.673 x10-11 N*m2/kg2. Mass of the sun is 1.99x1030 kg Radius of the sun is 6.96x108 m. Solution. a) Kinetic energy of hydrogen atoms determines the temperature of the sun: mvrms2/2=3/2kT so that vrms=(3kT/m)0.5=(3*1.38x10-23*5800/1.67x10-27)0.5=1.2 x 104 m/s. b) To find the escape speed, we use that the total energy of any body at the surface of the sun which is made of the potential energy U=-G*mbody*Msun/Rsun and kintetic energy K=mvesc2/2 should be equal to zero when the body reaches infinite distance from the sun (that is both potential and kinetic energies of the body are zero at infinity). As a result, mvesc2/2-G*mbody*Msun/Rsun=0 and vesc=(2GMsun/Rsun)0.5=(2*6.673 x10-11*1.99x1030 /6.96x108)0.5=6.18x105 m/s. which is much larger than the rms speed of hydrogen atoms at the surface of the sun. c) For appreciable amount of hydrogen то escape, the rms speed has to be equal to escape speed which gives us the temperature T=mvesc2/3k=1.67x10-27*(6.18x105)2/(3*1.38x10-23)=1.54x107 K This is much larger than the actual temperature of the sun’s surface. Problem 15. (3 points) A cylinder with a frictionless movable piston contains a quantity of helium gas. Initially the gas is at a pressure of 1.00 x 105 Pa, has a temperature of 300 K and occupies a volume of 1.5 L. The gas then undergoes two processes. In the first, the gas is heated and the piston is allowed to move to keep the temperature equal to 300 K. This continues until the pressure reaches 2.5 x 104 Pa. In the second process, the gas is compressed at constant pressure until it returns to its original volume of 1.5 L. Assume that the gas may be treated as ideal. a) In a PV diagram show both processes b) Find the volume of the gas at the end of the first process, and find the temperature and the pressure at the end of the second process. c) Find the total work done by the gas during both processes Solution. a) The PV diagram is represented by an isotherm and by an isobar: P 1 3 2 V b) At constant temperature PV=const. which results V2=V1(P1/P2)=1.5(1.00 x 105/2.5 x 104)=6 L. c) The work done in the first process W12=nRTln(V2/V1)=P1V1ln(P1/P2)= 1.00 x 105*(1.5x10-3)ln(1.00 x 105/2.5 x 104)=208 J. The work done during the second process W23=P2(V3-V2)=P2(V1-V2)=P2V1(1-P1/P2)= 2.5 x 104*1.5x10-3(1-1.00 x 105/2.5 x 104)=-113 J The total work is Wtot= W12+W23=208-113=95 J. Problem 16. (3 points) A typical coal fired power plant generates 1000 MW of usable power at an overall thermal efficiency of 40%. a) What is the rate of heat input to the plant and at what rate is heat ejected into the cool reservoir which is the nearby river? b) The plant burns anthracite coal, which has the heat of combustion of 2.65 x 107 J/kg. How much coal does the plant use per day, if it operates continuously. c) The river’s temperature is 18 C before it reaches the power plant and 18.5 C after it receives plant’s waste heat. Calculate the river’s flow rate in cubic meters per second if the specific heat of water is 4190 J/kg and the density of water is 1000 kg/m3. Solution. a) Since the thermal efficiency coefficient e=W/QH=(W/t)/(QH/t)=P/PH we obtain the rate of heat input PH=P/e=1000/0.4=2500 MW. Similarly, the heat that is lost is connected to the work as follows |QC|=|QH|-|W| therefore per unit time: PC=PH-P=2500-1000=1500 MW. b) The heat input in one day is PH times number of seconds per day: QH=PH*24*3600=2500*106*24*3600=2.16x1014 J This heat is obtained by burning so that QH=Lcomb*mcoal We obtain mcoal=QH/Lcomb=2.16x1014/2.65 x 107=8.15x106 kg. c) The heat input to the river each second is given by QC/t=PC=1500 MW which cools off some mass of water. According to the equation QC=mwcwΔT. Therefore each second the amount of mass cooled off (mw/t)=PC/(cwΔT)=1.5*109/4190/0.5=7.16x 105 kg/s. The river flow rate is volume per unit time: (mw/t)/ρ= 7.16x 105/103=716 m3/s. ...
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