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course Phy 242
1-31 2
Report the vertical positions of the three lines, relative to the hole, and the observed clock times.****
!!!! fill pt was 17.5cm from hole, all other measurements in cm are wrt hole.!!!!
1st line 12.5cm, 12.95703sec
2nd line 7cm, 45.89063sec
3rd line 3.5cm, 28.6875sec
dripping pt 0cm, 32.20703sec
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@& The clock keeps running, so clock times would be increasing.
I suspect you've reported time intervals, which is OK since you can add them up, one at a time, to get the clock times.
Starting with clock time 0 for the first line I suspect that clock times would be about 46, 75 and 107 seconds.*@
Repeat, this time observing how far the water stream travels in the horizontal direction as it falls to the sink, the ground, the tub, etc., when the water level is at each of the three marks. Try to make observations accurate to within a centimeter or two.
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1st line approx. 13cm
2nd line approx. 6cm
3rd line approx. 4cm
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Submit in the usual manner. Once we're sure you have reasonable data, we'll get into the analysis.
Hang onto the bottle for a couple of followup experiments (each equally short).
Preliminary 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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Using Bernoulli’s Equ we can calculate pressure needed, knowing that height is the only changing variable and velocity is 0.
(1/2) (v_a)^2 + g(y_a) + P_a = (1/2) (v_b)^2 + g(y_b) + P_b
### velocity squared term cancels out.####
g(y_a) + P_a = g(y_b) + P_b
P_a - P_b = g(y_b) - g(y_a)
P_a - P_b = g(y_b - y_a), this equation states that change in pressure is equal to change in ht* g( density*Force_grav)
Plugging our values into out symbolic solution we get
additional pressure needed = 1,000(kg/m^2)*9.8m/s^2*.12m
= 1,176N/m^2 or 1,176Pa, for ht change of 12cm.
additional pressure needed = 1,000(kg/m^2)*9.8m/s^2*.5m
= 4,900N/m^2 or 4,900Pa, for ht change of 50cm.
additional pressure needed = 1,000(kg/m^2)*9.8m/s^2*1m
= 9,800N/m^2 or 9,800Pa, for ht change of 100cm.
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Calculate the air column pressure, in atmospheres, for each observed length.
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Using formula Pv = nRT, and assuming there is no change in nRT we can solve following…
P_1*v_1 = P_2*v_2, where P_1 is approx. 100,000N/m^2 and v_1 = c.s. area * 30cm = `pi*(r cm) ^2 * 30cm = 30cm^3. (Note c.s. area will be the same for same tube so we can let c.s. area = 1)
100,000N/m^2*30cm^3 = P_2*29cm^3
P_2 = 100,000N/m^2(30cm^3/29cm^3) = 1,034 ????I’m not doing this correctly but not sure where my mistake is?????
Maybe if left in atmospheres then to find P_2
P_2 = 1 atm*(30cm/29cm) = approx. 1.0345 atm, for `ds of 1cm ????
P_2 = 1 atm*(30cm/27cm) = approx. 1.1111 atm, for `ds of 3cm ????
P_2 = 1 atm*(30cm/26cm) = approx. 1.1538 atm, for `ds of 4cm ????
???I guess I’m confused, if we calculate for air column. Never mind maybe I just got it??????
@& you did*@
If we Bernoulli’s Equ again under same circumstances where velocity is equal to zero we have…
g(y_1) + P_1= g(y_2) + P_2, where P_1 = 1atm and y_1=0 so
P_2 = 1atm - g(1cm) = 1atm - ((1,000kg/m^2)*(9.8m/s^2)*(.01m) = 1atm - 98N/m^2 = = (100,000N/m^2)/
@& You wouldn't use 100 000 N/m^2 anywhere in this analysis, since this is the quantity we're trying to estimate.*@
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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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Based on your two graphs, what would you conclude is the value of atmospheric pressure?
@& Your graph would indicate your calculated pressures, which run up to about 10 000 N/m^2 for the 1 meter column, and about 1.15 atmospheres as based on the air column behavior.
With the trend established by these figures, what would be the pressure in N/m^2 when the atmospheric pressure reaches 2 atmospheres?*@
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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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100,000N/m^2
@& This is atmospheric pressure, but it doesn't follow from your graph. *@
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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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Understood
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How long would it take the water to fill a tube of length 50 cm?
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1.3meters/sec = 130cm/sec,
Traveling at 130cm/sec it would take 50cm/(130cm/sec) = approx. .3846sec
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What volume of water exits during this time?
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volume of 50cm of tube…
c.s. area*ht
`pi*(15cm^2)*50cm = approx. 3.534cm^3
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By how much does the water level in the cylinder therefore change?
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mass of water exiting cylinder is …
mass = 1gr/cm^3* 3.534cm^3 = 3.534gr
Volume of 1cm of cylinder..
`pi*3cm^2*1cm = approx. 28.2743cm^3
To find level level that water drops
3.534cm^3/28.2743cm^3 = approx. .125, so when 3.534cm^3 of water is in the tube that water represents 1/8 the volume contained in a section of the cylinder that has a width of 1cm.
So the water level drops approx. 1/8 of an cm when 50cm of the tube is full of water .
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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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It takes .3846sec for the water level to drop 1/8 or .125 of an cm, giving water level velocity of approx. .325cm/sec. Exit velocity is 130cm/sec so…
(130cm/sec)/(.325cm/sec) = 400, so exit velocity is approx. 400 times greater than water level velocity.
@& Good. This should be the ratio of cylinder area to the cross-sectional area of the hole. It is so?
No need to submit an answer if you verify that it's so. If not, you should consider submitting a 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.
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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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Using equation
(1/2) (v_a)^2 + g(y_a) + P_a = (1/2) (v_b)^2 + g(y_b) + P_b
where P_a=P_b and v_a is approx. 0 plugging into equation….
(1/2) (v_b)^2 + g(y_b) = g(y_a), solving for v_b we get
(1/2) (v_b)^2 = g(y_a) - g(y_b) = g(y_a - y_b)
(1/2)(v_b)^2 = g(y_a - y_b)
v_b^2 = 2 g(y_a - y_b)
v_b = `sqrt(2 g(y_a - y_b)), where y_a – y_b = 20cm and g = 9.8m/s^2
So..
v_b = `sqrt(2 *9.8m/s^2*.2m) = `sqrt(3.92m/s) = approx. 1.9799m/s or approx. 198cm/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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Using equation
(1/2) (v_a)^2 + g(y_a) + P_a = (1/2) (v_b)^2 + g(y_b) + P_b
where P_a=P_b and plugging known values into the equation….
(1/2) (v_b)^2 + g(y_b) = (1/2) (v_a)^2 + g(y_a), solving for v_a we get
(1/2) (v_a)^2 =(1/2) (v_b)^2 + g(y_b) - g(y_a)
(1/2)(v_a)^2 = (1/2)(v_b)^2 + g(y_b - y_a)
v_a^2 = (v_b)^2 + 2 g(y_b - y_a)
v_a = `sqrt((v_b)^2 + 2 g(y_b - y_a)), where (y_a – y_b) = 20cm so (y_b - y_a) must equal -20cm. Again g = 9.8m/s^2 and of course v_b = 198cm/s. Now plugging in values we can solve.
v_a = `sqrt((v_b)^2 + 2 g(y_b - y_a))
= `sqrt(198cm/s^2 + (2*980cm/s^2*-20cm))
=`sqrt(39,204cm^2/s^2 - 39,200cm^2/s^2) = `sqrt(4cm^2/s^2)
v_a = 2cm/s
So water level is descending at a rate of 2cm/s.
@& Good use of the equation, but your result 2 cm/s is more likely the result of roundoff errors in these calculations, which involve much larger numbers, than an accurate result from Bernoulli's equation. Among other things, your original calculation of v_b assumed that v_a was zero.
What do you get if you consider the cross-sectional areas of the hole and the cylinder?*@
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University Physics students: What is the differential equation that relates water depth to exit velocity?
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The volume of water exiting thru hole is
c.s. area( of hole)* (v(t)*`dt), which is a column of water with a width = c.s. area and length = vel.*`dt
So water level would be
w.l. = -k*vel_hole + c, where k = 1/100 and c is init water level.
@& Let y be the water level relative to the hole. Then y ' = - k * v_exit.
v_exit is itself a function of y; by Bernoulli's equation v_exit = sqrt( 2 g y ).
k is the ratio of hole area to cylinder area.
Thus
y ' = - k sqrt( 2 g y). *@
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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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Pressure needed to support water 2meters high in tube we find weight….
2meters of tube, = 200cm^3*980cm/s^2 = 200gr*980cm/s^2 = 196000g*cm/s^2 or 1,960N
Pressure required to support water column 2meters or 200cm is approx. 20,000N/m^2
nR/v is constant so we have P/T, which we set up as
P_1/T_1 = P_2/T_2, which we can solve for T_2
(100,000N/m^2)/300k = (120,000N/m^2)/T_2
T_2 = [(120,000N/m^2)/(100,000N/m^2)]*300k
=1.2*300k = 360k
So temp has total change of 460k which is a difference of
460k/300k = approx. 1.5333
So volume will change will be approx. 153%
@& Volume doesn't begin to change significantly until the temperature is 360 K. Until that point the water is just filling the tube, which has negligible 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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Init volume mult by change
5m^3*1.5333 = 7.6667m^3, when init 5m^3 is deducted gas gains 2.6667 m^3 volume
So displaced water is 2.6667 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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It’s mass is
2.6667m^3 or 266.67cm^3, when 1g/cm^3*266.67cm^3 = 266.67g
@& 1 m^3 = (100 cm)^3 = 1 000 000 cm^3, not 100 cm^3*@
weight is
266.67g*980cm/s^2 = .26667kg*9.8m/s^2 = approx. 2.6134N
`dPE = 2.6134N*2m = 5.2267J
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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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n= Pv/RT = [1atm*5m^3]/[8.31(J/mol*k)*300k] = (506,625 J)/(2493J/mol) = approx. 203.219mol
For monatomic gas
KE_trans = (3/2)nRT
KE_trans = [3/2*(203.219mol)*(8.31(J/mol*k)*300k] – [3/2*(203.219mol)*(8.31(J/mol*k)*4600k]
= approx. 1,165,237.424J - 759,937.4505J
= 405,299.9736J
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How much work will the gas do against pressure as it expands?
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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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For diatomic gas
KE_trans = (5/2)nRT
So we can..
405,299.9736J/(3/2) = 270199.9824J
270199.9824J*(5/2) = 675,499.956J
KE_trans(diatomic) = 675,499.956J
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What is the ratio of PE change to the energy added to the gas?
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405,299.9736J/675,499.956J = .6
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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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It takes one caleroie of heat energy to raise one gr of a substance 1k. When we convert Celsius to kelvin we get
10 Celsius =283 K, 40 Celsius 313 K, which is difference of 30k
1 liter = 1000g
So 30*1000 = 30000, so it will take 30,000calories to raise temp to 40c
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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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????
@& Good. I didn't give you all the necessary information for this one.
1 calorie is about 4 Joules.
With this, you would have gotten it.*@
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self-critique #$&*
#$&* self-critique
self-critique rating
&#This looks good. See my notes. Let me know if you have any questions. &#