Page Physics 207 – Lecture 30Physics 207: Lecture 30, Pg 1Lecture 30Goals:Goals:••Wrap up chapter 20. Wrap up chapter 20. ••Review for the final.Review for the final.• Final exam on Friday, Dec 23, at 10:05 am in BASCOM 272.Final will cover Chapters 1-20 (excluding Chapter 13).Semi-cumulative.• HW 12 due tomorrow night.Physics 207: Lecture 30, Pg 2The Doppler effect l The frequency of the wave that is observed depends on the relative speed between the observer and the source. vsobserverPage Physics 207 – Lecture 30Physics 207: Lecture 30, Pg 3The Doppler effect f=f0/(1-vs/v)l Approaching source:l Receding source:f=f0/(1+vs/v)Physics 207: Lecture 30, Pg 4lThe figure shows a snapshot graph D(x, t = 2 s) taken at t = 2 s of a pulse traveling to the left along a string at a speed of 2.0 m/s. Draw the history graph D(x = −2 m, t) of the wave at the position x = −2 m.WavesPage Physics 207 – Lecture 30Physics 207: Lecture 30, Pg 5l History Graph:23647time (sec)2-25D(x = −2 m, t) Physics 207: Lecture 30, Pg 6lA concert loudspeaker emits 35 W of sound power. A small microphone with an area of 1 cm2 is 50 m away from the speaker.lWhat is the sound intensity at the position of the microphone?lHow much sound energy impinges on the microphone each second?Page Physics 207 – Lecture 30Physics 207: Lecture 30, Pg 7RI =Psource4πR2Psource=35 WR=50 ml The power hitting the microphone is:Pmicrophone= I AmicrophonePhysics 207: Lecture 30, Pg 8Intensity of soundsl If we were asked to calculate the intensity level in decibels:β=10log10II0 I0: threshold of human hearingI0=10-12 W/m2Page Physics 207 – Lecture 30Physics 207: Lecture 30, Pg 9l Suppose that we measure intensity of a sound wave at two places and found them to be different by 3 dB. By which factor, do the intensities differ?β1=10log10I1I0 β2=10log10I2I0 β1−β2=10log10I1I2 = 3I1I2=100.3= 2Physics 207: Lecture 30, Pg 10Enginesl For the engine shown below, find, Wout, QHand the thermal efficiency. Assume ideal monatomic gas.Pi4PiVi2ViQ=90JQ=25JPage Physics 207 – Lecture 30Physics 207: Lecture 30, Pg 11l First, use the ideal gas law to find temperaturesPi4PiVi2ViQ=90JQ=25JTi4Ti8Ti2Til From the right branch, we have:nCV∆T=90 Jn(3R/2)6Ti=90JnRTi=10JPhysics 207: Lecture 30, Pg 12l Work output is the area enclosed by the curve:Pi4PiVi2ViQ=90JQ=25JTi4Ti8Ti2TiWout=area=3PiVi=3nRTi=30JPage Physics 207 – Lecture 30Physics 207: Lecture 30, Pg 13lFrom energy conservation:Wout=QH-QCWout=30JQC=115J=0.2lThe thermal efficiency is:QH=145JPhysics 207: Lecture 30, Pg 14The Carnot EnginelAll real engines are less efficient than the Carnot engine because they operate irreversibly due to the path and friction as they complete a cycle in a brief time period.llCarnot showed that the Carnot showed that the thermal efficiency of a thermal efficiency of a Carnot engine is:Carnot engine is:hotcoldcycleCarnot TT1−=ηPage Physics 207 – Lecture 30Physics 207: Lecture 30, Pg 15lFor which reservoir temperatures would you expect to construct a more efficient engine?A) Tcold=10o C, Thot=20o CB) Tcold=10o C, Thot=800o CC) Tcold=750o C, Thot=800o CPhysics 207: Lecture 30, Pg 16Kinetic theoryl A monatomic gas is compressed isothermally to 1/8 of its original volume. l Do each of the following quantities change? If so, does the quantity increase or decrease, and by what factor? If not, why not?a. The temperatureb. The rms speed vrmsc. The mean free path d. The molar heat capacity CVPage Physics 207 – Lecture 30Physics 207: Lecture 30, Pg 17λ=14 2πN /V( )r2N/V: particles per unit volumelMean free path is the average distance particle moves between collisions:lThe average translational kinetic energy is:avg=(1/2) mvrms2=(3/2) kBTCV=3R/2 (monatomic gas) lThe specific heat for a monatomic gas is:Physics 207: Lecture 30, Pg 18Simple Harmonic Motionl A Hooke’s Law spring is on a horizontal frictionless surface is stretched 2.0 m from its equilibrium position. An object with mass m is initially attached to the spring however, at equilibrium position a lump of clay with mass 2m is dropped onto the object. The clay sticks. What is the new amplitude?kmkm-220(Xeq)2m2mPage Physics 207 – Lecture 30Physics 207: Lecture 30, Pg 19½ k A2=½ m vmax2vmax= Al The speed when the mass reaches the equilibrium position:l The speed after clay sticks can be found using momentum conservation: m vmax=(m+2m)vnewvnew=vmax/3l The new amplitude can be found using energy conservation:½ (m+2m)vnew2=½ k Anew2Anew=A/√3Physics 207: Lecture 30, Pg 20FluidslWhat happens with two fluids? lConsider a U tube containing liquids of density ρ1and ρ2 as shown:Compare the densities of the liquids:(A) ρ1< ρ2(B) ρ1= ρ2(C) ρ1> ρ2ρρρρ1ρρρρ2Page Physics 207 – Lecture 30Physics 207: Lecture 30, Pg 21Fluidsl What happens with two fluids?? l Consider a U tube containing liquids of density ρ1and ρ2 as shown:l At the red arrow the pressure must be the same on either side. ρ1x = ρ2yv Compare the densities of the liquids:(A) ρ1< ρ2(B) ρ1= ρ2(C) ρρρρ1>
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