PHY232
Your 'flow experiment' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Your initial message (if any): **
I was almost done with this assinment and the form cleared on me? Starting over.
note the advice to compose your responses in a text editor or word processor so you don't lose them if your computer, your ISP, a bad Internet router between you and VHCC, a power outtage or even a malfunction in the VHCC computerbreaks your connection.
** Is flow rate increasing, decreasing, etc.? **
The rate of flow will decrease due to the potential energy in the water column decreasing. Pressue is equal to rho(density of the fluid) * gravity (9.81m/s^2) * the height of the water in the column or tube.
It's very good that you're thinking in terms of PE and pressure. The cause and effect is not quite as you say, though:
The PE decrease doesn't cause the flow rate to decrease. The two are both the result of the decreasing water level.
Decreasing level implies decreasing pressure which results in the decreasing flow rate.
Decreasing level implies decrease in PE.
But decreasing PE does not cause the decreasing pressure that causes the decreasing flow rate.
** Is the velocity of the water surface increasing, decreasing, etc.? **
The velocity will decrease as the potential energy decreases.
** How would the velocity of the water surface, the velocity of the exiting water, the diameter of the cylinder and the diameter of the hole be interrelated? **
The water surface, velocity of the exiting wate, and the diameter of the hoel are interrelated by the continuity equaiton.
** Explain how we know that a change in velocity implies the action of a force: **
Force is equal to mass times accleration. The downward force due to the potential energy of the water column is changing, casuing a change in the veolcity.
The PE of the water column doesn't cause the pressure. If the cylinder had 1000 times the cross-sectional area it would have 1000 times the PE at a given level, but the pressure at the hole would be the same.
It's the pressure, not the PE, that exerts the force that causes the change in velocity.
The PE change of the entire system for any time interval is (in the ideal case) equal to the KE change during that interval, so there is a relationship between the PE change and the exit velocity; however this relationship also involves the mass of the exiting water.
** Does the depth seem to be changing at a regular rate, at a faster and faster rate, or at a slower and slower rate **
The depth is changing at slower rate.
** What do you think a graph of depth vs. time would look like? **
A parabolic curve that starts with a steep slope and decreases with time.
** Does the horizontal distance (the distance to the right, ignoring the up and down distance) traveled by the stream increase or decrease as time goes on? **
The horizontal distance will decrease over time due to the velocity slowing down which in turn is due to the potential energy.
** Does this distance change at an increasing, decreasing or steady rate? **
The deistanc is changing at a decreasng rate.
** What do you think a graph of this horizontal distance vs. time would look like? **
Same as the vertical distance vs. time. Parabolic with a steep slope initialy, slope decreasing with time.
** The contents of TIMER program as you submitted them: **
1 73.70313 73.70313
2 77.1875 3.484375
3 82.20313 5.015625
4 82.8125 .609375
5 83.46875 .65625
6 84.0625 .59375
7 84.85938 .796875
8 85.92188 1.0625
9 86.65625 .734375
10 90.78125 4.125
** The vertical positions of the large marks as you reported them, relative to the center of the outflow hole **
30,0
20,3.5
10,8.5
9,9.1
8,9.8
7,10.4
6,11
5,12
4,13
0,17
** Your table for depth (in cm) vs clock time (in seconds) **
30,0
20,3.5
10,8.5
9,9.1
8,9.8
7,10.4
6,11
5,12
4,13
0,17
The columns in your table appear to be reversed. y vs. x has x in the first column. Your table would therefore represent clock time vs. depth.
** Is the depth changing at a regular rate, at a faster and faster rate, or at a slower and slower rate? **
The depth is chaning at a slower rate with an increase in time.
** Your description of your depth vs. t graph: **
The graph is a parabolic curve with an intial steep slope decreasing at the water level approaches zero.
** Your explanation and list of average average velocities: **
30,0
20,2.86
10,1.18
9,0.11
8,0.10
7,0.10
6,0.09
5,0.08
4,0.08
0,0.24
Calculated the distance between marks and divide by the time interval.
** The midpoints of your time intervals and how you obtained them: **
It took 17 seconds for all of the water to exit the tube, so 17/2 = 8.5 At 8.5 seconds the velocity is equal to 1.1 cm/sec
&#You have not indicated midpoint clock times. You might be reporting half the time interval, but time intervals and clock times are two different things.
For example if clock times are 2, 5, 15 and 17 seconds, then the time intervals are respectively 3, 10 and 2 seconds. The latter are time intervals, not clock times. The clock starts at the beginning of the observation and continues running through the end, and clock times are the times showing on the clock. The clock cannot run backward, so clock times cannot decrease.
The midpoint clock times are 3.5 seconds (halfway between clock times 2 and 5 seconds), 10 seconds (halfway between 5 and 15 seconds) and 16 seconds (halfway between 15 and 17 seconds).
Half the time intervals would give you 1.5, 5 and 1 second.
It should be clear that midpoint clock times cannot decrease, whereas time intervals can either increase or decrease. &#
** Your table of average velocity of water surface vs. clock time: **
A parabolic curve downward.
** Your description of your graph of average velocity vs clock time: **
The velocity starts off at a fast rate and decreases with time
** Your explanation of how acceleration values were obtained: **
20,3.5
10,8.5
9,9.1
8,9.8
7,10.4
6,11
5,12
4,13
0,17
The acceleration would be equal to the derivitive of the average velocity.
If you had a velocity function that would work; however you don't have the velocity function to take the derivative of.
** Your acceleration vs clock time table: **
20,3.5
10,8.5
9,9.1
8,9.8
7,10.4
6,11
5,12
4,13
0,17
I think the table is backwards. You don't specify units, but the accelerations would be much smaller than anything you report here.
** According to the evidence here, is acceleration increasing, decreasing, staying the same or is in not possible to tell? **
The acceleraton of the water surface is constant.
** **
1 Hr. The form cleared some how, so I had to repeat.
Most of your work is good, but be sure to see my notes. You don't give reasonable values for accelerations and don't explain how you got them from the data.
&#Please see my notes and 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). &#