PHY 211 Final Exam for 12/10/03

The final exam will take place at 8 am on Wed, Dec 10. You will have a full three hours to do the exam. If you have another exam immediately after this one, please let me know ahead of time so that we can make arrangements. The exam will consist of 10 to 12 problems. About half of the questions will be on the material covered on Exams I, II and III, with the remaining half over the new material. The final is worth 200 pts.

Below are a few sample problems from a previous exam, for the material not covered on the first three exams. As usual, DO NOT ASSUME that a subject not covered in the sample exam below will not show up on the exam! You will be provided with an equation sheet.

Sample Physics 211 Final Exam (new material)

  1. 50 gm of ice at a temperature of -10 oC is place in an insulating cup with 100 gm of water initially at 20 oC. Describe the equilibrium state.
    (Cice = .5 cal/goC, LF = 80 cal/gm, Cwater = 1.0 cal/goC)

    DQwarm ice = (50(.5)(10) = 250 cal, DQmelt ice = (50)(80) = 4,000 cal, and DQcool water = (100)(1.0)(20) = 2,000 cal. So there will be 2,000 - 250 = 1750 cal = mice(80) --> mice = 22 gms of ice melted.

  2. The engine cycle below consists of four processes; isobaric, isochoric, isothermal, and adiabatic, respectively. Identify each and fill in the table with "+", "0", or "-".

    1 to 2 is isobaric, and all follow as listed. Note that the adiabat is steeper than the isotherm.


  3. An ideal gas expands from 3 liters to 6 liters at a constant pressure of 1.5 atmospheres. The pressure is then dropped to .5 atmospheres at a constant volume. How much work was done by the gas?

    W = (1.5 x 105 Pa)(.006 - .003 m3) + 0 = 450 J

  4. A steam engine operates between the temperature extremes of 120 oC and 45 oC.
    a) What is the maximum possible effeciency for this engine?
    b) If the engine generates 3,000 Watts of power, at what rate is heat wasted? (Assume the efficiency calculated in part a.)

    Eff = 1 - (273+45)/(273+120) = .19 = Work/Qin --> Qin = 3000 W/.19 = 15, 700 W and Qout = Qin - W = 12, 700 W. Note that we are really using power = Q/t in all cases. You can simply think of it as 3000 J of work every second, etc.

  5. A 1000 kg car is suspended on its axels by springs. When four 70 kg students pile in, the car lowers 3 cm. What is the effective spring constant of the suspension and what is the period of oscillation of the loaded car?

    F = 4(700 N) = k(.03) --> k = 9.3 x 104 N/m. Period = 2pi x Sqrt(m/k) = .74 seconds. (Note that the car's mass must be included for the period!)

  6. An ocean wave moves by as you are floating. You move through a total vertical distance of 2 m from the crest to the adjacent trough of the wave in just 1.2 seconds. You note that two adjacent crests are 7 m apart. What is the amplitude, frequency, wavelength and velocity of the wave?

    Amplitude = 1/2 of total vertical motion = 1 m. The period is 2.4 seconds, so f = .42 Hz. Wavelength is 7 m and so v = lf = 2.9 m/s

  7. A trombone player plays a note that has an intensity level of 50 dB at a distance of 10 m from the player.
    a) What would the decibel level be 100 m away from the player?
    b) How many identical trombones players would be needed to increase the level to 60 dB at 10 m?

    a) Since P ~ 1/r2, the intensity will drop by 102 = 100. That is a drop of 20 dB. (A factor of 10 for every 10 dB.) So the dB level at 100m is 30 dB.
    b) Increase of 10 dB requires a tenfold increase in intensity ... so 10 trombones! You may also do this problem the long way using 10 log(I/Io).

  8. A mast wire is held under tension 1000N. If the 10 m length of wire has mass .8 kg what is the frequency of the fundamental (lowest) mode of vibration in the wire?

    v = sqrt(T/m) = sqrt(1000N/(.8 kg/10m)) = 112 m/s. The lowest mode of vibration is 1/2 wavelength between the ends, so l = 20 m and f = v/l = 5.6 Hz

  9. You are walking at 1 m/s away from a stationary trumpet player playing a 512 Hz note. At the same instant an identical trumpeter is walking at 1 m/s towards you. What frequency do you hear from each and what other effect do you hear hear?

    f' = [(343-1)/343] 512 = 511 Hz for the first trumpet.
    f' = [(343 +1)/(343-1)]512 = 515 Hz for the second trumpet
    You will hear these two tones and the beat between them of 4 Hz.