1 / 20

Lecture 30

Lecture 30. Goals:. Review for the final. Final exam on Monday, Dec 20, at 5:05 pm, at Sterling 1310, Graham 19. HW 11 due tonight. Waves. The figure shows a snapshot graph D ( x , t = 2 s) taken at

afram
Download Presentation

Lecture 30

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Lecture 30 Goals: • Review for the final. • Final exam on Monday, Dec 20, at 5:05 pm, at Sterling 1310, Graham 19. • HW 11 due tonight.

  2. Waves • The 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.

  3. History Graph: 2 -2 3 4 6 7 2 5 time (sec)

  4. A concert loudspeaker emits 35 W of sound power. A small microphone with an area of 1 cm2 is 50 m away from the speaker. • What is the sound intensity at the position of the microphone? • How much sound energy impinges on the microphone each second?

  5. Psource=35 W R=50 m • The power hitting the microphone is: R Pmicrophone= I Amicrophone

  6. Intensity of sounds • If we were asked to calculate the intensity level in decibels: I0: threshold of human hearing I0=10-12 W/m2

  7. 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?

  8. Engines • For the engine shown below, find, Wout, QH and the thermal efficiency. Assume ideal monatomic gas. 4Pi Q=90J Pi Q=25J Vi 2Vi

  9. First, use the ideal gas law to find temperatures 8Ti 4Ti 4Pi Q=90J 2Ti Pi Ti Q=25J 2Vi Vi • From the right branch, we have: nCVΔT=90 J n(3R/2)6Ti=90J nRTi=10J

  10. Work output is the area enclosed by the curve: 8Ti 4Ti 4Pi Q=90J 2Ti Pi Ti Q=25J 2Vi Vi Wout=area=3PiVi=3nRTi=30J

  11. From energy conservation: Wout=QH-QC Wout=30J QC=115J QH=145J • The thermal efficiency is: η=0.2

  12. The Carnot Engine • Carnot showed that the thermal efficiency of a Carnot engine is: • All 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.

  13. For which reservoir temperatures would you expect to construct a more efficient engine? A) Tcold=10o C, Thot=20o C B) Tcold=10o C, Thot=800o C C) Tcold=750o C, Thot=800o C

  14. Kinetic theory • A monatomic gas is compressed isothermally to 1/8 of its original volume. • 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 temperature b. The rms speed vrms c. The mean free path d. The molar heat capacity CV

  15. The average translational kinetic energy is: εavg=(1/2) mvrms2=(3/2) kBT • Mean free path is the average distance particle moves between collisions: N/V: particles per unit volume • The specific heat for a monatomic gas is: CV=3R/2 (monatomic gas)

  16. The average kinetic energy of the molecules of an ideal gas at 10°C has the value K1. At what temperature T1 (in degrees Celsius) will the average kinetic energy of the same gas be twice this value, 2K1? (A) T1 = 20°C (B) T1 = 293°C (C) T1 = 100°C • Suppose that at some temperature we have oxygen molecules moving around at an average speed of 500 m/s. What would be the average speed of hydrogen molecules at the same temperature? (A) 100 m/s (B) 250 m/s (C) 500 m/s (D) 1000 m/s (E) 2000 m/s

  17. Simple Harmonic Motion • 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? 2m k m 2m k m -2 0(≡Xeq) 2

  18. The speed when the mass reaches the equilibrium position: ½ k A2=½ m vmax2 vmax=ωA • The speed after clay sticks can be found using momentum conservation: m vmax=(m+2m)vnew vnew=vmax/3 • The new amplitude can be found using energy conservation: ½ (m+2m)vnew2=½ k Anew2 Anew=A/√3

  19. (A)r1 < r2 (B)r1 = r2 (C)r1>r2 Fluids • What happens with two fluids?? • Consider a U tube containing liquids of density r1 and r2 as shown: Compare the densities of the liquids: r2 r1

  20. (A)r1 < r2 (B)r1 = r2 (C)r1 > r2 Fluids • What happens with two fluids?? • Consider a U tube containing liquids of density r1 and r2 as shown: • At the red arrow the pressure must be the same on either side. r1 x = r2 y • Compare the densities of the liquids: r2 y r1 x

More Related