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Phy 202
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.
** #$&* Your initial message (if any): **
** #$&* Is flow rate increasing, decreasing, etc.? **
The picture below shows a graduated cylinder containing water, with dark coloring (actually a soft drink). Water is flowing out of the cylinder through a short thin tube in the side of the cylinder. The dark stream is not obvious but it can be seen against the brick background.
You will use a similar graduated cylinder, which is included in your lab kit, in this experiment. If you do not yet have the kit, then you may substitute a soft-drink bottle. Click here for instructions for using the soft-drink bottle.
In this experiment we will observe how the depth of water changes with clock time.
In the three pictures below the stream is shown at approximately equal time intervals. The stream is most easily found by looking for a series of droplets, with the sidewalk as background.
Based on your knowledge of physics, answer the following, and do your best to justify your answers with physical reasoning and insight:
As water flows from the cylinder, would you expect the rate of flow to increase, decrease or remain the same as water flows from the cylinder?
Your answer (start in the next line):
As water flows from the cylinder, I expect the rate of flow to decrease.
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As water flows out of the cylinder, an imaginary buoy floating on the water surface in the cylinder would descend.
Would you expect the velocity of the water surface and hence of the buoy to increase, decrease or remain the same?
Your answer (start in the next line):
I would expect the velocity of the water surface and buoy to decrease, since the rate of flow will decrease.
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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? More specifically how could you determine the velocity of the water surface from the values of the other quantities?
Your answer (start in the next line):
The four items listed above are interrelated. The diameter of the cylinder and the hole will determine the velocity of the exiting water, and therefore the velocity of the water surface. The diameter of the cylinder tells us how much water it can hold, while the diameter of the hole lets us know how fast it can exit; determining the velocities.
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The water exiting the hole has been accelerated, since its exit velocity is clearly different than the velocity it had in the cylinder.
Explain how we know that a change in velocity implies the action of a force?
Your answer (start in the next line):
Newtons Second Law of Motion states that the velocity of an object changes when subjected to an external force.
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What do you think is the nature of the force that accelerates the water from inside the cylinder to the outside of the outflow hole?
Your answer (start in the next line):
I believe the nature of the force accelerating the water is the weight of the buoy.
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From the pictures, answer the following and justify your answers, or explain in detail how you might answer the questions if the pictures were clearer:
Does the depth seem to be changing at a regular rate, at a faster and faster rate, or at a slower and slower rate?
Your answer (start in the next line):
The depth seems to be changing at a slower and slower rate. It would be clearer to tell if we knew the time intervals at which each photo was taken.
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What do you think a graph of depth vs. time would look like?
Your answer (start in the next line):
A graph of depth vs. time would look like a downward curve, that is decreasing at a decreasing rate.
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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?
Your answer (start in the next line):
The horizontal distance traveled by the stream, decreases as time goes on.
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Does this distance change at an increasing, decreasing or steady rate?
Your answer (start in the next line):
The distance change seems to be at a decreasing rate, but time intervals are necessary to be sure.
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What do you think a graph of this horizontal distance vs. time would look like? Describe in the language of the Describing Graphs exercise.
Your answer (start in the next line):
A graph of horizontal distance vs. time would also look like a downward curve, decreasing at a decreasing rate.
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You can easily perform this experiment in a few minutes using the graduated cylinder that came with your kit. If you don't yet have the lab materials, see the end of this document for instructions an alternative setup using a soft-drink bottle instead of the graduated cylinder.
Setup of the experiment is easy. You will need to set it up near your computer, so you can use a timing program that runs on the computer. The cylinder will be set on the edge of a desk or tabletop, and you will need a container (e.g., a bucket or trash can) to catch the water that flows out of the cylinder. You might also want to use a couple of towels to prevent damage to furniture, because the cylinder will leak a little bit around the holes into which the tubes are inserted.
Your kit included pieces of 1/4-inch and 1/8-inch tubing. The 1/8-inch tubing fits inside the 1/4-inch tubing, which in turn fits inside the two holes drilled into the sides of the graduated cylinder.
Fit a short piece of 1/8-inch tubing inside a short piece of 1/4-inch tubing, and insert this combination into the lower of the two holes in the cylinder. If the only pieces of 1/4-inch tubing you have available are sealed, you can cut off a short section of the unsealed part and use it; however don't cut off more than about half of the unsealed part--be sure the sealed piece that remains has enough unsealed length left to insert and securely 'cap off' a piece of 1/4-inch tubing.
Your kit also includes two pieces of 1/8-inch tubing inside pieces of 1/4-inch tubing, with one end of the 1/8-inch tubing sealed. Place one of these pieces inside the upper hole in the side of the cylinder, to seal it.
While holding a finger against the lower tube to prevent water from flowing out, fill the cylinder to the top mark (this will be the 250 milliliter mark).
Remove your thumb from the tube at the same instant you click the mouse to trigger the TIMER program.
The cylinder is marked at small intervals of 2 milliliters, and also at larger intervals of 20 milliliters. Each time the water surface in the cylinder passes one of the 'large-interval' marks, click the TIMER.
When the water surface reaches the level of the outflow hole, water will start dripping rather than flowing continuously through the tube. The first time the water drips, click the TIMER. This will be your final clock time.
We will use 'clock time' to refer to the time since the first click, when you released your thumb from the tube and allowed the water to begin flowing.
The clock time at which you removed your thumb will therefore be t = 0.
Run the experiment, and copy and paste the contents of the TIMER program below:
Your answer (start in the next line):
1 278.5938 278.5938
2 281.1641 2.570313
3 282.3594 1.195313
4 283.4922 1.132813
5 284.6406 1.148438
6 286.1094 1.46875
7 287.5234 1.414063
8 288.7969 1.273438
9 290.0625 1.265625
10 291.6172 1.554688
11 293.0391 1.421875
12 294.6797 1.640625
13 296.2891 1.609375
14 297.8906 1.601563
15 299.6328 1.742188
16 301.4297 1.796875
17 303.6094 2.179688
18 305.4688 1.859375
19 307.8594 2.390625
20 310.3672 2.507813
21 312.8203 2.453125
22 316.0859 3.265625
23 320.1484 4.0625
24 325.8984 5.75
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Measure the large marks on the side of the cylinder, relative to the height of the outflow tube. Put the vertical distance from the center of the outflow tube to each large mark in the box below, from smallest to largest distance. Put one distance on each line.
Your answer (start in the next line):
0 cm
.9 cm
1.9 cm
2.9 cm
3.9 cm
4.85 cm
5.8 cm
6.75 cm
7.65 cm
8.55 cm
9.45 cm
10.35 cm
11.25 cm
12.15 cm
13.05 cm
13.95 cm
14.85 cm
15.7 cm
16.55 cm
17.40 cm
18.25 cm
19.1 cm
19.9 cm
20.75 cm
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Now make a table of the position of the water surface vs. clock time. The water surface positions will be the positions of the large marks on the cylinder relative to the outflow position (i.e., the distances you measured in the preceding question) and the clock times will as specified above (the clock time at the first position will be 0). Enter 1 line for each event, and put clock time first, position second, with a comma between.
For example, if the first mark is 25.4 cm above the outflow position and the second is 22.1 cm above that position, and water reached the second mark 2.45 seconds after release, then the first two lines of your data table will be
0, 25.4
2.45, 22.1
If it took another 3.05 seconds to reach the third mark at 19.0 cm then the third line of your data table would be
5.50, 19.0
Note that it would NOT be 3.05, 19.0. 3.05 seconds is a time interval, not a clock time. Again, be sure that you understand that clock times represent the times that would show on a running clock.
The second column of your TIMER output gives clock times (though that clock probably doesn't read zero on your first click), the third column gives time intervals. The clock times requested here are those for a clock which starts at 0 at the instant the water begins to flow; this requires an easy and obvious modification of your TIMER's clock times.
For example if your TIMER reported clock times of 223, 225.45, 228.50 these would be converted to 0, 2.45 and 5.50 (just subtract the initial 223 from each), and these would be the times on a clock which reads 0 at the instant of the first event.
Do not make the common error of reporting the time intervals (third column of the TIMER output) as clock times. Time intervals are the intervals between clicks; these are not clock times.
Your answer (start in the next line):
0, 20.75
2.57, 19.9
3.765, 19.1
4.898, 18.25
6.046, 17.4
7.515, 16.55
8.929, 15.7
10.202, 14.85
11.468, 13.95
13.022, 13.05
14.444, 12.15
16.085, 11.25
17.694, 10.35
19.296, 9.45
21.038, 8.55
22.835, 7.65
25.015, 6.75
26.874, 5.8
29.265, 4.85
31.772, 3.9
34.225, 2.9
37.491, 1.9
41.553, 0.9
47.303, 0
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You data could be put into the following format:
clock time (in seconds, measured from first reading) Depth of water (in centimeters, measured from the hole)
0 14
10 10
20 7
etc. etc.
Your numbers will of course differ from those on the table.
The following questions were posed above. Do your data support or contradict the answers you gave above?
Is the depth changing at a regular rate, at a faster and faster rate, or at a slower and slower rate?
Your answer (start in the next line):
The data I gathered supports my answers from above. The depth is changing at a slower and slower rate.
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Sketch a graph of depth vs. clock time (remember that the convention is y vs. x; the quantity in front of the 'vs.' goes on the vertical axis, the quantity after the 'vs.' on the horizontal axis). You may if you wish print out and use the grid below.
Describe your graph in the language of the Describing Graphs exercise.
Your answer (start in the next line):
The graph of water depth vs. clock time moves in a downward curve from the high end on the left to the low end on the right. The curve is therefore decreasing at a decreasing rate as time goes on.
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caution: Be sure you didn't make the common mistake of putting time intervals into the first column; you should put in clock times. If you made that error you still have time to correct it. If you aren't sure you are welcome to submit your work to this point in order to verify that you really have clock times and not time intervals
Now analyze the motion of the water surface:
For each time interval, find the average velocity of the water surface.
Explain how you obtained your average velocities, and list them:
Your answer (start in the next line):
The average velocities were found by finding the differences between the depths and clock times and then dividing the depth differences by the time intervals.
.33 cm/sec
.6695 cm/sec
.7502 cm/sec
.7404 cm/sec
.5786 cm/sec
.6011 cm/sec
.6677 cm/sec
.7106 cm/sec
.5792 cm/sec
.6329 cm/sec
.5484 cm/sec
.5594 cm/sec
.5618 cm/sec
.5166 cm/sec
.5008 cm/sec
.4128 cm/sec
.5108 cm/sec
.3973 cm/sec
.3789 cm/sec
.4077 cm/sec
.3062 cm/sec
.2462 cm/sec
.1565 cm/sec
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Assume that this average velocity occurs at the midpoint of the corresponding time interval.
What are the clock times at the midpoints of your time intervals, and how did you obtain them? (Give one midpoint for each time interval; note that it is midpoint clock time that is being requested, not just half of the time interval. The midpoint clock time is what the clock would read halfway through the interval. Again be sure you haven't confused clock times with time intervals. Do not make the common mistake of reporting half of the time interval, i.e., half the number in the third column of the TIMER's output):
Your answer (start in the next line):
1.285 sec
3.1675 sec
4.3315 sec
5.472 sec
6.7805 sec
8.222 sec
9.5655 sec
10.835 sec
12.245 sec
13.733 sec
15.2645 sec
16.8895 sec
18.495 sec
20.167 sec
21.9365 sec
23.925 sec
25.9445 sec
28.0695 sec
30.5185 sec
32.9985 sec
35.858 sec
39.522 sec
44.428 sec
I obtained the midpoint values, by finding the difference between each known time, dividing by 2, then adding the answer to the lower of the times.
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Make a table of average velocity vs. clock time. The clock time on your table should be the midpoint clock time calculated above.
Give your table below, giving one average velocity and one clock time in each line. You will have a line for each time interval, with clock time first, followed by a comma, then the average velocity.
Your answer (start in the next line):
1.285, .33
3.1675, .6695
4.3315, .752
5.472, .7404
6.7805, .5786
8.222, .6011
9.5655, .6677
10.835, .7106
12.245, .5792
13.733, .6329
15.2645, .5484
16.8895, .5594
18.495, .5618
20.167, .5166
21.9365, .5008
23.925, .4128
25.9445, .5108
28.0695, .3973
30.5185, .3789
32.9985, .4077
35.858, .3062
39.522, .2462
44.428, .1565
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Sketch a graph of average velocity vs. clock time. Describe your graph, using the language of the Describing Graphs exercise.
Your answer (start in the next line):
The graph type is a scatter plot, but a trend line can be estimated. The trend line is in a descending curve. The curve seems to descending at an increasing rate.
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For each time interval of your average velocity vs. clock time table determine the average acceleration of the water surface. Explain how you obtained your acceleration values.
Your answer (start in the next line):
.2568 cm/sec/sec
.2114
.1732
.1353
.0853
.0731
.0698
.0656
.0473
.0461
.0359
.0331
.0304
.0256
.0228
.0173
.0197
.0142
.0124
.0124
.0085
.0062
.0035
The average acceleration of the water surface was determined by dividing the average velocity by the midpoint clock times.
Acceleration is change in velocity divided by change in clock time, not change in velocity divided by clock time.
You will need to correct these results and subsequent results which depend on them.
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Make a table of average acceleration vs. clock time, using the clock time at the midpoint of each time interval with the corresponding acceleration.
Give your table in the box below, giving on each line a midpoint clock time followed by a comma followed by acceleration.
Your answer (start in the next line):
1.285, .2568
3.1675, .2114
4.3315, .1732
5.472, .1353
6.7805, .0853
8.222, .0731
9.5655, .0698
10.835, .0656
12.245, .0473
13.733, .0461
15.2645, .0359
16.8895, .0331
18.495, .0304
20.167, .0256
21.9365, .0228
23.925, .0173
25.9445, .0197
28.0695, .0142
30.5185, .0124
32.9985, .0124
35.858, .0085
39.522, .0062
44.428, .0035
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Answer two questions below:
Do your data indicate that the acceleration of the water surface is constant, increasing or decreasing, or are your results inconclusive on this question?
Do you think the acceleration of the water surface is actually constant, increasing or decreasing?
Your answer (start in the next line):
The data indicates that the acceleration of the water surface is decreasing.
I think the acceleration of the water surface is actually decreasing due to the increased time intervals.
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Go back to your graph of average velocity vs. midpoint clock time. Fit the best straight line you can to your data.
What is the slope of your straight line, and what does this slope represent? Give the slope in the first line, your interpretation of the slope in the second.
How well do you think your straight line represents the actual behavior of the system? Answer this question and explain your answer.
Is your average velocity vs. midpoint clock time graph more consistent with constant, increasing or decreasing acceleration? Answer this question and explain your answer.
Your answer (start in the next line):
-.15 cm/sec / 10 sec
The slope indicates that the velocity decreases as time progresses.
The straight line is fairly accurate to the actual behavior of the system, but a slightly curved line would be more accurate.
The graph is consistent with decreasing acceleration, due to the decreasing velocity.
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Your instructor is trying to gauge the typical time spent by students on these experiments. Please answer the following question as accurately as you can, understanding that your answer will be used only for the stated purpose and has no bearing on your grades:
Approximately how long did it take you to complete this experiment?
Your answer (start in the next line):
2 hours 10 minutes
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You may add any further comments, questions, etc. below:
Your answer (start in the next line):
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Very good data, and good analysis except for the error you made in finding accelerations. That will need to be corrected.
&#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).
Be sure to include the entire document, including my notes. &#