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course Phy 202
2/2 10 amPreliminary questions for Class 110131
`q001. Squeezes judged by one experimenter as 1, 4 and 9 on a 1-10 scale resulted in water column heights of 12, 50 and 100 cm.
Squeezes judged by the same experimenter as 2, 5 and 8 on the same 1-10 scale resulted in air column lengths of 29, 27 and 26 cm, where the air column is 30 cm long at atmospheric pressure.
Calculate the additional pressure needed to support the water column, for each of the observed heights.
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`rho=1 g/cm^3 P=1 atm or 100000 Pa
P=`rho*g*h
Squeeze=1= P=1000kg/m^3*9.8 m/s^2*.12 m=1176 Pa
Squeeze 2=4=1000kg/m^3*9.8 m/s^2*.50 m=4900 Pa
Squeeze 3=9=1000 kg/m^3*9.8 m/s^2*1.0 m=9800 Pa
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Calculate the air column pressure, in atmospheres, for each observed length.
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29 cm= a 4% change in length, so if original pressure is 1 atm, then resulting P is 1.04 atm
27 cm= a 10% change, the resulting P is 1.1 atm
26 cm= a 14% change, the resulting P is 1.14 atm
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Sketch a graph of water column pressure vs. estimated squeeze and sketch the straight line you think best fits this graph.
Sketch a graph of air column pressure vs. estimated squeeze and sketch the straight line you think best fits this graph.
Give the slope of each of your straight lines.
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Water column P vs. estimated squeeze slope =~1000 Pa/1 change in squeeze
Air column P vs estimated squeeze=~ .02atm/ 1 change in squeeze
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Based on your two graphs, what would you conclude is the value of atmospheric pressure?
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According to my graphs, atmospheric pressure is approximately 6000 Pa based on the squeeze of about 5.
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@& A squeeze of 5 results in a pressure of 6000 Pa, but a squeeze of 5 doesn't result in an additional atmosphere of pressure (between no squeeze and squeeze 5 the pressure changes by only about .1 atmosphere).*@
@& If you divide the number of pascals per squeeze by the number of atmospheres per squeeze you should get the number of pascals per atmosphere.
1000 Pa / squeeze / (.02 atm / squeeze) = 50 000 Pa / atm.*@
Based on your own similar observations, what would you conclude is the value of the atmospheric pressure?
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`q002. Water exits a cylindrical container, whose diameter is 6 cm, through a hole whose diameter is 0.3 cm. The speed of the exiting water is 1.3 meters / second.
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How long would it take the water to fill a tube of length 50 cm?
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`dt=50 cm/1.3 m/s=50 cm/130 cm/s=.4 sec
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What volume of water exits during this time?
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V=`pi(.15 cm)^2*50 cm=3.5cm^3
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By how much does the water level in the cylinder therefore change?
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Volume=`pi (3cm^2)*`dy=3.5 cm^3=
`dy=3.5cm^3/`pi(3cm^2)=.12 cm
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What is the ratio of the exit speed of the water to the speed of descent of the water surface in the container?
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Speed of exit=130 cm/s speed of decent =.3 cm/s
The ratio= 130 cm/s/.3 cm/s
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`q003. Let point A be the water surface in a cylindrical container of radius 10 cm. Let point B be just outside a hole in the side of the container, 20 cm below point A. The hole has diameter 0.6 cm.
... conceptual ...
One of the three quantities P, v and y in Bernoulli's equation is the same at both points. The other two quantities are each different at A than at B, one being greater at A and the other greater at B.
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Which is constant?
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P
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Which is greater at A?
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y
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Which is therefore greater at B?
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v
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Write down Bernoulli's Equation for this selection of points, and use the equation to determine the 'ideal' velocity of the water as it exits the hole.
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`rho*g*ya+1/2`rho*va^2+Pa=`rho*g*yb+1/2`rho*vb^2+Pb
Pa=Pb
`rhog(ya-yb)=1/2 `rho(vb^2+va^2)
vb^2=2 g(ya-yb)=1/2`rho(va^2) va=0
vb=sqrt(2g(ya-yb)
vb=sqrt(2*9.8 m/s^2*.2m)
vb=1.99 m/s
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If water exits the hole at this rate, at what rate is the surface of the water in the cylinder descending?
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Volume=`pi*(.3 cm^2)*200 cm=57 cm^3= 57 cm^3/s
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@& Good calculation, but that's the rate at which volume is decreasing, not the rate at which the surface of the water is decending.*@
University Physics students: What is the differential equation that relates water depth to exit velocity?
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`q004. Starting at 300 K and atmospheric pressure you heat the gas in a bottle until the added pressure is sufficient to support a column of water 2 meters high, in a tube of negligible volume. You then manage to heat the gas another 100 K.
By what percent does the volume of the gas change, from beginning to end?
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25%
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If the gas has initial volume 5 m^3, then how many m^3 of water will be displaced by the expansion?
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V3=V2*T3/T2
V3=5 m^3*460K/360 K=6.5m^3
`dV=6.5 m^3-5m^3=1.5 m^3
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If this water is collected in a reservoir 2 meters above its initial height, by how much does the gravitational PE of the system increase?
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$$$$$This is where I get confused when the other class comes in$$$
`dPE=wt*`dy
`dPE=mass*g*`dy
`dPE=density*V*g*`dy
`dPE=1000 kg/m^3 *1.5m^3 *9.8 m/s^2* 2 m
=29400 J
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@& Looks good so far.*@
By how much will the total translational KE of the molecules in the gas change during the process?
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KEtrans=3/2nRT=3/2 (65moles)(8.31 J/mol*K)(300K)=243067 J
KEtrans=3/2(65 moles)(8.31 J/mol*K)(460 K)=372703 J
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@& Good, but subtract the results to get the increase.*@
How much work will the gas do against pressure as it expands?
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W=2/3 `dKEtrans=2/3*129636 J=86424 J
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If the gas is diatomic, how much will the total rotational KE of the molecules change during the process?
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`dKErot=2/3`dKEtrans=86424 J (I’m not sure on this one as it doesn’t match the numbers you gave us)
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@& The way my numbers were coming out, you shouldn't worry too much about that.*@
What is the ratio of PE change to the energy added to the gas?
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`dPE=243067 J+86424 J+86424 J=406915 J
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@& This is the thermal energy that goes into the system.
You calculated `dPE previous, about 30 000 J.*@
`q005. A hot rock is dropped into a liter of water, increasing the water's temperature from 10 Celsius to 40 Celsius. How much thermal energy did the water gain from the rock?
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I spent some time on this last night and my question is, shouldn’t we have a mass for the rock in order to solve?
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@& You have the mass and temperature change of the water (mass is 1 kg, which is the mass of a liter of water)
I should have also reminded you that it takes about 4 Joules to raise the temperature of 1 gram of water by 1 Celsius.
1000 grams * 4 J / (g Celsius) * 30 Celsius = 120 000 J.*@
If during the process the water also lost 5000 Joules of energy to the surroundings, how much thermal energy did the rock lose in the process?
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self-critique #$&*
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@& Very good. See my notes.*@