Your work on flow experiment 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):
Is flow rate increasing, decreasing, etc.?
The rate of flow should decrease as the water flows out of the cylinder.
Is the velocity of the water surface increasing, decreasing, etc.?
The rate of descension of the water level and buoy would also decrease as the water descended further and further.
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 velocity of the water surface would be slower then that of the exiting water. This is because the velocity must be higher through the smaller hole to ensure an equilibrium because of the reduction of the diameter.
Explain how we know that a change in velocity implies the action of a force:
Force is any mass accelerated. Which if you take a section of water that is contained in the tube then it has a definite mass and that same mass is going faster after it exit the tube. So there was a change of velocity within a change in time giving the acceleration. So that force is in essence expressed as F = m * ('dV/'dt)
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 nature of the force pushing on the water near the outflow hole is the weight of the water above the outflow hole. This is Forceweight = mass above the outflow hole time the acceleration of gravity. The depth seems to be changing at a slower and slower rate because the force is becoming less and less with each time interval.
What do you think a graph of depth vs. time would look like?
The depth vs. time graph for this flow model would be depth decreasing at a decreasing rate as it approacches the depth of the outflow hole.
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 distance decrease s as time goes on because the force accelerating the mass is becoming less and less.
Does this distance change at an increasing, decreasing or steady rate?
The distance changes at a decreasing rate since the acceleration is slowly decreasing at a decreasing rate.
What do you think a graph of this horizontal distance vs. time would look like?
Looking at the graph of horizontal distance versus time we can see that as time moves along the distance travelled horizontal from the outflow hole is less and less. This also shows us that the distance in approaching the 0 distance at which point the flow will stop. Also this approach to 0 is slowing with each time interval so the distance is is decreasing at a decreasing rate.
The contents of TIMER program as you submitted them:
1 855.8789 855.8789
2 858.6211 2.742188
3 860.207 1.585938
4 862.2773 2.070313
5 864.4336 2.15625
6 866.7148 2.28125
7 869.7383 3.023438
8 872.6133 2.875
9 875.7773 3.164063
10 879.7695 3.992188
11 884.0586 4.289063
12 890.9492 6.890625
The vertical positions of the large marks as you reported them, relative to the center of the outflow hole
1.9
3.8
5.7
7.6
9.5
11.4
13.3
15.2
17.1
19.0
20.9
Your table for depth (in cm) vs clock time (in seconds)
0, 20.9
2.742, 19.0
4.327, 17.1
6.397, 15.2
8.553, 13.3
10.834, 11.4
13.857, 9.5
16.732, 7.6
19.896, 5.7
23.888, 3.8
28.177, 1.9
35.067, 0
Is the depth changing at a regular rate, at a faster and faster rate, or at a slower and slower rate?
The depth is changing at a slower and slower rate.
Your description of your depth vs. t graph:
The graph conveys and illustrates the idea that the water depth is changing at a slower and slower rate. This is evident by the fact that the graph is slowly approaching the minimal depth but it is doing so as the slope of the graph decreases a small bit over each time interval.
Your explanation and list of average average velocities:
The average velocity is found using the form (Vo+Vf)/2. However, Vf is Vo + 'dV and the 'dV is found by 'dS/ 'dt which is the change in position divided the change in time. So initially Vo is 0 so for the first interval so if Vf = Vo + 'dV we find the following time intervals' velocity using the reasoning listed above.
Vo 'dV Vf Ave Velocity
1 -2 0cm/s .693cm/s .693cm/s .34565cm/s
2-3 .693 1.02cm/s 1.713cm/s 1.203cm/s
3-4 1.713 .917 2.63 2.17
4-5 2.63 .917 3.547 3.088
5-6 3.547 .884 4.431 3.989
6-7 4.431 .833 5.264 4.484
7-8 5.264 .629 5.893 5.579
8-9 5.893 .662 6.555 6.224
9-10 6.555 .601 7.156 6.855
10-11 7.156 .476 7.632 7.394
11-12 7.632 .444 8.076 7.854
Velocity is decreasing in magnitude, so on any interval v0 and `dv will have opposite signs.
You don't explain how you got your results for `dv.
However the most direct way to calculate average velocity over and interval, based on position vs. clock time data, is to calculate `ds / `dt.
What do you get if you calculate `ds / `dt?
The midpoints of your time intervals and how you obtained them:
1.857
.79
1.035
1.075
1.14
1.51
1.43
1.58
1.99
2.14
3.44
These midpoints were obtained by halving the total interval which gave me the halfway point.
The clock time runs from 0 to about 35 seconds, and is broken into intervals by your timing events.
Your next-to-last interval was from 23.888 s to 28.177 s. The duration of this interval was about 4.3 seconds; its midpoint occurs at approximately t = 25 seconds.
Your table of average velocity of water surface vs. clock time:
1.875, .34565
2.665, 1.203
3.7, 2.17
4.775, 3.088
5.915, 3.989
7.425, 4.484
8.855, 5.579
10.435, 6.224
12.425, 6.855
14.565, 7.394
18.005, 7.854
Your description of your graph of average velocity vs clock time:
The graph shows that the velocity is increasing at a decreasing rate with respect to time
Your explanation of how acceleration values were obtained:
A = 'dV/'dt so using this method we find the following accelerations using the average velocities and the time intervals of the midpoint times:
1 .963cm/s^2
2 .93
3 .857
4 .790
5 .327
6 .765
7 .408
8 .317
9 .252
10 .134
Your acceleration vs clock time table:
There are a few outliers, however, for the most part the acceleration of the water depth descention is decreasing.
According to the evidence here, is acceleration increasing, decreasing, staying the same or is in not possible to tell?
The data coincides with the acceleration of the water surface decreasing as time goes on. I feel that the water surface acceleration is actually decreasing as is presented by the data as well as observed during the experiment.
You appear to have complicated the calculation of average velocities over the intervals, and your results don't make sense (the water surface isn't speeding up, but your calculations indicate that it is).
See my notes and make the necessary modifications. Resubmit the report, starting with that question, and refer to my 7/17 document so I can put the beginning of the report together with the revised end.