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course Phy 242
February 12 around 2pm. I DID revise this assignment, even though it was optional. Thanks for the help.
Questions 110131-revised#$&*
course Phy 242
February 10 around 9pm.
110131q001. 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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12cm=1200 N/m^2
26cm=2600 N/m^2
27cm=2700 N/m^2
29cm=2900 N/m^2
30cm=3000 N/m^2
50cm=5000 N/m^2
100cm=10 000 N/m^2
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Calculate the air column pressure, in atmospheres, for each observed length.
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[1 atm = 100,000 N/m^2] and [1 Pa = 1 N/m^2]
12cm=101 200 Pa
26cm=102 600 Pa
27cm=102 700 Pa
29cm=102 900 Pa
30cm=103 000 Pa
50cm=105 000 Pa
100cm=1 010 000 Pa
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@& Your calculations are in Pa, not in atmospheres.*@
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0.9988 atm
1.0126 atm
1.0136 atm
1.0155 atm
1.0165 atm
1.0363 atm
9.9679 atm
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@& Your calculations should be based on the assumption that the length of the original air column corresponds to 1 atmosphere.
Those calculations would come out pretty close to the ones you have here; the trend wouldn't vary a lot from the one you get here. However that last calculation is way off. The air column would have had to shorten to 3 or 3 cm to get that kind of pressure.*@
@& Your last pressure is still in error. You don't get 9.9679 atm out of this data.
The others look OK.*@
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 pressure vs. squeeze: slope=1100 [(8800 N/m^2)/(8 squeeze)]
Air column pressure vs. squeeze: slope=50 [300 Pa/6 squeeze]
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@& Air column slope should be in atmospheres per squeeze.*@
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(0.0003 atm/6 squeeze)
=0.00005 atm/squeeze
@& It's not clear where a difference of .0003 atm comes from or how that corresponds to 6 on the 'squeeze scale'.*@
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Based on your two graphs, what would you conclude is the value of atmospheric pressure?
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What exactly is this question and the next asking? How do I look at the graph and answer these?
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@& First get your air column pressures in atmospheres.
Then think about the units of the two slopes and what they are telling you.
Then see if you can figure out this answer. Don't get too hung up on it; if you need to ask questions, by all means do so.*@
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Well if you do (1100 N/m^2 / squeeze) / (0.00005 atm/squeeze)
=2.2 * 10^7
@& That would be a good calculation, though your .00005 atm/squeeze can't be right.
What would be the units?
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But if you do (0.00005 atm/squeeze) / (1100 N/m^2 / squeeze)
=4.5 * 10^(-8)
[1 N/m^2] = [9.869 * 10^(-6) atm]
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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.
How long would it take the water to fill a tube of length 50 cm?
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dt= (50cm)/(1.3m/s)
=(50cm)/(130cm/s)
=0.385 seconds
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What volume of water exits during this time?
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Volume= (c.s area * length)
= pi * (0.15cm)^2 * (50cm)
=3.53 cm^3
AND
dy= (3.53cm^3)/(pi*(3cm)^2)
=0.125 cm
Volume= pi * (3cm)^2 * 0.125cm
=3.53 cm^3
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By how much does the water level in the cylinder therefore change?
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pi * (0.3cm)^2 * (130cm)
=36.75cm^3
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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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Flow rate= 36.75cm^3/sec
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@& The speed at which water descends is in cm/s, not in cm^3 / s.
36.75 cm^3 / s is the rate at which the volume of water in the cylinder decreases, and the rate at which tube volume is filled.*@
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I think I would do (50cm)/(0.385 sec) to get the speed at which water descends.
=129.9 cm/s
@& If the water surface moved 130 cm, the change in volume inside the cylinder would be very much more than 36.75 cm^3.
What is the cross-sectional area of the container and what does it have to do with the present question?*@
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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.
Which is constant?
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P is constant
P_A = P_B =1 atm
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Which is greater at A?
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V and y differ from A to B.
Y DECREASES.
Y_B < Y_A
Then “y” decreases from A to B.
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Which is therefore greater at B?
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See above.
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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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P + [(1/2)*rho*v^2] + [rho*g*h] = Bernoulli’s Equation
rho*g(y_A-y_B) = (1/2)*rho*(v_B^2 + v_A^2)
v_B^2 = 2g(y_A-y_B) + [(1/2)*rho*v_A^2]
v_B= sqrt (2g* (y_A-y_B)) and v_A^2=0
v_B=2 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*(0.3cm)^2*200cm
=60cm^3
Flow rate: 60cm^3/sec
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University Physics students:
`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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P0=1 atm, P1=1 atm+20kPa
By heating: T2/T1=1.2
T2=300K * (120/100)
=360K
=(5m^3)*(460/360)
=6.5m^3
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@& Right. What is the increase as a percent of the original volume?*@
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Going from (5m^3) to (6.5m^3):
5/6.5=0.7692
About 77%
This doesn’t seem right. I don’t know.
OR
The increase is (1.5m^3) which is 30% of the original (5m^3)?!
@& 30% is correct. You find the increase in volume as a percent of the original volume.*@
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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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dV=6.5-5
=1.5m^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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Gravitational PE: (weight * dy)
=mass * g * dy
=density * volume * g * dy
Approx 30 000 Joules
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By how much will the total translational KE of the molecules in the gas change during the process?
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KE_trans gas= (3/2)nRT [n=PV/RT]
At 300K= 243067.5 J
At 460K= 372703.5 J
372703.5 - 243067.5= 129636 J changed during the process
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How much work will the gas do against pressure as it expands?
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dW= (2/3)*dKE_trans
=(2/3)*(129 636 J)
=86424?!
OR would it be:
=(2/2)*n*R*dT
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@& Both would give you the same result.*@
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OK!
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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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=(2/3)*(dKE_trans)
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@& RIght. But you had a number of `dKE_trans, so you can get a number here.*@
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(2/3) * (129 636 J)
=86424 J (same as the work gas will do against pressure during expansion)
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What is the ratio of PE change to the energy added to the gas?
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@& You have, or can easily get, the numbers for PE change and energy added to the gas, so this shouldn't be a difficult question to answer.*@
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(30 000 J) / (86424 J)
=0.35
I think this is right...
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`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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=(5/2)*n*R*dT
=(5/2)*(45 mol)*(8.31 J/mol K)*(100 K)
=93 500 J
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@& 5/2 n R `dT applies to a gas. There's no gas involved here.*@
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Increased 30 C:
30C = 303.15 K
@& A change in temperature of 30 C is not the same as a change in temperature of 303 K.
For example the temperature could have changed from 20 C to 50 C. That's a change of 30 C.
Measured in Kelvin, the change would be from 293 K to 323 K, a change of 30 K.
Since a Kelvin degree is the same size as a Celsius degree, the change in Celsius temperature is the same as the change in Kelvin temperature.*@
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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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@& It takes 4 Joules to raise a the temperature of a gram of water by 1 degree Celsius.
This is all you need to answer the given question.*@
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@& You've got a lot of this, but you're on the wrong track with that last question.
See my notes and revise as indicated. I think that will put you in good shape.
&#Please see my notes and, unless my notes indicate that revision is optional, submit a copy of this document with revisions and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes.
If my notes indicate that revision is optional, use your own judgement as to whether a revision will benefit you.
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*@"
Self-critique (if necessary):
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Self-critique rating:
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02-12-2011
02-17-2011
@& See my notes, and please submit one more revision. This time indicate insertions by ####.*@