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course PHY 232
11:46 am 2/04 I made some changes from my original initial answers concerning the squeeze and it's graph marked by ####
QA 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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In atm
P_1 = 1.0116
P_4= 1.0484
P_9= 1.0968
@& You don't need an additional atmosphere of pressure to support the water column. You've already got an atmosphere of pressure at the base of the column.
How much pressure do you have to add, in Pascals?*@
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In Pa
P_1 = 101160 Pa
P_4 = 104840 Pa
P_9 = 109680 Pa
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Calculate the air column pressure, in atmospheres, for each observed length.
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Using P1V1=P2V2 where V=A*h then V1= 0.3*A & P1 = 1 atm
V2=0.29*A -->0.29*A*P2=0.3*A*1--> P2= 1.03
P_2= 1.03 atm
P_5= 1.11 atm
P_8= 1.15 atm
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P_2= 103000 Pa
P_5= 111000 Pa
P_8= 115000 Pa
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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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H20= 0.01
Air=0.02
@& Your additional water column pressures would be calculated in Pascals, not atmospheres.
It looks like water column pressure changed by something on the order of 1 000 Pa per unit of squeeze, though somewhat less than this figure.*@
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Based on your two graphs, what would you conclude is the value of atmospheric pressure?
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My two lines very nearly intersect my pressure axis at 1. The H2O is a little above and the air below. So it seems that, within error, my graph shows 1.
@& Your conclusion is that 1 atmosphere of water pressure is equal to 1 atmosphere of pressure. In other words, at this point, your reasoning is circular.
However if you use the specified units for water pressure, you get a result in those units.*@
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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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I'm not certain about this question. I've been working in atmospheres and I guess this refers to Pa maybe? I haven't made enough observations to deduce that.
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@& Good. As you are beginning at this point to suspect, you should have calculated water pressures in Pa.*@
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Rescaling my graph by 1x100000 still gives me the same coordinates and slopes and 100,000 Pa for atm P.
@& This is because when calculating water pressures you assumed 100 000 Pa as the atmospheric pressure, rather than using just your observations and the known density of water.
I could have made that clearer at the beginning, but I'm still debating on whether I should have. The point being 'why would we calculated pressure in atmospheres when we can calculate it in Pascals'?
OK, now I know what to do. I'll state at the beginning that at no time should a value for atmospheric pressure be assumed. Or some version of that.
Thanks for helping me figure that out.*@
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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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t=x/v= 0.5/1.3 = 0.4 s
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What volume of water exits during this time?
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V_t= π*r^2*x= 0.25*π*(0.3)^2*50=3.53 cm^3 = 3.53 mL
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By how much does the water level in the cylinder therefore change?
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∆V_c= 0.25*π*36*∆y--> ∆y= 0.12 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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∆y/∆t= 0.12/0.4 = 0.3 cm/s = 0.003 m/s
v=k*v_y --> v= (433.3)*v_y or 1:433
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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
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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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P+rho*g*y+1/2rho*v^2 here this becomes: g*y=1/2v^2
g*y_A + 0 = 1/2v_B^2
v_B=1.98 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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gy=1/2v^2 ---> v= √(2gy)
∆V=π*r_c^2*∆y
∆V= v*CSA*∆t= v*π*r_h^2*∆t= √(2gy)*π*r_h^2*∆t
∴ √(2gy)*π*r_h^2*∆t = π*r_c^2*∆y
Solving for ∆y/∆t
∆y/∆t= √(2gy)(r_h/r_c)^2
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University Physics students: What is the differential equation that relates water depth to exit velocity?
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Easily enough since dy/dt is the limit of the above expression so:
dy/dt= √(2gy)(r_h/r_c)^2
which could be rewritten in the form:
y'= √(2g)*(r_h/r_c)^2*√y
or more generally:
y'= -k√y
k need be negative since dy/dt is slowing down as y gets lower and ∆y would be negative in the original expression
a solution here:
y= (-1/2*k*t+C)^2
C= y_0
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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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P(2m)= rho*g*y= 20000
300/100,000= T/120000
P_0*V_0= P*V ---> ∆V%= +20%
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The tube has negligible volume, and the container doesn't expand significantly. So during this phase the volume of the gas will not change significantly.
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T1= 360 K---> ∆T= 100 k --> T2= 460 K
T2/T1= V2/V1 = 460/360 = 1.28= V2/V1 -->∆V1/2 %= +28%
1.28(1.2V_0)= V3 ---Net ∆V%= 53.6%
@& If the pressure is held constant then the volume must increase, and during the constant-pressure phase the gas does expand by 28%.*@
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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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(5 m^3)0.536 = 2.68 m^3
@& See previous.
The gas expands by only 28%, not 53.6%.
The increase in volume is 28% of 5 m^3, or 1.4 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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I'm assuming that all 2.68 m^3 are now in the reservoir, which means the ∆y= 2m.
Thus ∆PE= -W= mgy = rho*V*g*y= 1000*2.68*9.81*2= 52582 J
@& Right, except that the expansion is only 1.4 m^3 so `dPE is more like 30 000 J.*@
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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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`dKE_tr = 3/2nR`dT= 3/2 `dP`dV= 3/2(20000)(2.68)=80400 J
@& KE_trans = 3/2 n R T so
`dKE_trans = 3/2 n R `dT.
With n = 45 moles and `dT = 160 K, we get
3/2 n R `dT = 3/2 * 45 mol * 8.31 J / (mol K) * 160 K = 90 000 Joules.
n R T = P V
so
d(n R T) = d( P V )
so with n constant
n R dT = P dV + V dP.
If P is constant then dP = 0 and this becomes
n R `dT = P `dV (constant pressure)
This applies to the constant-pressure phase, the second phase of the heating.
If V is constant then `dV = 0, and
n R `dT = V `dP (constant volume).
In the present example this doesn't work out work a darn:
V `dP = 5 m^3 * 20 000 Pa = 100 000 J
during the first (constant-volume) phase and
P `dV = 120 000 Pa * 1.4 m^3 = 168 000 J
during the second phase.
I'll have to write this down to figure out what the problem is, and can't do so right now.
I'll get back to you soon.
Note, however, than when P and V both change together, with neither P nor V constant, we have to resort to integration to evaluate the contributions of P dV and V dP.
More later.
*@
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How much work will the gas do against pressure as it expands?
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`dW= P`dV= 20,000*2.68=53600 J or W= 2/3*KE_trans= 2/3*80400= 53600 J
@& The pressure of the gas is 120 000 Pa.*@
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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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`dKE_rot= 2/3 KE_trans = 2/3(80400) = 53600 J
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What is the ratio of PE change to the energy added to the gas?
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`dPE/`dE= `dPE/(W+`dKE_tr+KE_rot)= 52582/187600= 0.28
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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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`dQ= mc`dT= (1 kg)(4190 J/kg•K)(30 K)= 125700 J
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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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Rock lost 125700 originally but if water lost 5000 then the rock would lose an additional 5000 J to 'equalize' with the new energy level of the water. At least I think that's true.
`dQ_rock= -125700-5000= -130700 J
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@& Check my notes. There are some discrepancies in my explanations, and I'm a little fuzzy this afternoons o I'm going to have to break down and write things out. Should get back to you soon.*@