Physics 207 – 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.
Physics 207: Lecture 30, Pg 1
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.
Physics 207: Lecture 30, Pg 2
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Physics 207 – Lecture 30
History Graph:
2
-2
2
3
4
6 5 time (sec)
7 Physics 207: Lecture 30, Pg 3
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?
Physics 207: Lecture 30, Pg 4
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Physics 207 – Lecture 30
I= R
Psource 4 πR 2
Psource=35 W R=50 m The power hitting the microphone is:
Pmicrophone= I Amicrophone Physics 207: Lecture 30, Pg 5
Intensity of sounds If we were asked to calculate the intensity level in
decibels:
I β = 10log10 I0 I0: threshold of human hearing I0=10-12 W/m2
Physics 207: Lecture 30, Pg 6
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Physics 207 – Lecture 30
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?
I1 I2 β1 = 10log10 β 2 = 10log10 I0 I0 I1 β1 − β 2 = 10log10 = 3 I2 I1 = 10 0.3 = 2 I2 Physics 207: Lecture 30, Pg 7
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 Physics 207: Lecture 30, Pg 8
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Physics 207 – Lecture 30 First, use the ideal gas law to find temperatures
4Pi
8Ti
4Ti
Q=90J
Pi
2Ti
Ti
Q=25J
2Vi
Vi
From the right branch, we have:
nCV T=90 J n(3R/2)6Ti=90J
nRTi=10J Physics 207: Lecture 30, Pg 9
Work output is the area enclosed by the curve:
4Pi
8Ti
4Ti
Q=90J
Pi
2Ti
Ti
Q=25J
2Vi
Vi
Wout=area=3PiVi=3nRTi=30J
Physics 207: Lecture 30, Pg 10
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Physics 207 – Lecture 30
From energy conservation:
Wout=QH-QC Wout=30J QC=115J
QH=145J
The thermal efficiency is:
=0.2 Physics 207: Lecture 30, Pg 11
The Carnot Engine
Carnot showed that the thermal efficiency of a Carnot engine is:
ηCarnot cycle = 1−
Tcold Thot
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. Physics 207: Lecture 30, Pg 12
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Physics 207 – Lecture 30
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
Physics 207: Lecture 30, Pg 13
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
Physics 207: Lecture 30, Pg 14
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Physics 207 – Lecture 30
The average translational kinetic energy is: avg=(1/2)
mvrms2=(3/2) kBT
Mean free path is the average distance particle moves between collisions:
λ=
1 4 2π (N /V )r 2
N/V: particles per unit volume
The specific heat for a monatomic gas is: CV=3R/2 (monatomic gas) Physics 207: Lecture 30, Pg 15
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 Physics 207: Lecture 30, Pg 16
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Physics 207 – Lecture 30 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 m
k -2
0( Xeq)
2 Physics 207: Lecture 30, Pg 17
The speed when the mass reaches the equilibrium position:
½ k A 2=½ m vmax2 vmax=A
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)v new2=½ k A new2 Anew=A/ 3 Physics 207: Lecture 30, Pg 18
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Physics 207 – Lecture 30
Fluids What happens with two fluids?? Consider a U tube containing liquids of
ρ2
density ρ1 and ρ2 as shown:
ρ1
Compare the densities of the liquids: (A) ρ1 < ρ2
(B) ρ1 = ρ2
(C) ρ1 > ρ2
Physics 207: Lecture 30, Pg 19
Fluids
What happens with two fluids??
Consider a U tube containing liquids of density ρ1 and ρ2 as shown:
ρ2
x ρ1
At the red arrow the pressure must be the same on either side. ρ1 x = ρ2 y Compare the densities of the liquids: (A) ρ1 < ρ2
(B) ρ1 = ρ2
y
(C) ρ1 > ρ2
Physics 207: Lecture 30, Pg 20
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