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PHY 122
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.
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9/7 7pm
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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):
I would expect the rate of flow to decrease as water flows from the cylinder. As the water flows from the container the amount of water in the container decreases therefore the weight of the water decreases and causes less force to be applied at the location of the exiting water.
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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 or buoy to decrease as the water flows out of the cylinder due to amount of water leaving the cylinder decreasing.
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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):
By the velocity of the exiting water,and the diamater of the hole, we could determine how much water was exiting the hole in any time frame. If the diameter of the cylinder was know we could determine the rate at which the water surface is droping by knowing who much water a particular length of the cylinder would hold, and therefore be able to determine at what rate the water level was droping.
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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):
In order for the water to accelerate the velocity must change. In order for the velocity of the water to change it must be acted upon by some 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):
The gravitational pull on the water pulling it towards the ground along with the change in diamater size of the hole from the cylinder to the exiting hole being smaller causes the force that excellerates the water exiting the hole.
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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. This can be observed by the exiting water seen flowing against the background of the sidewalk. As the water level decreases the stream leaving the cylinder falls of to a slower and slower rate, as observed by the exiting water falling closer and closer to the cylinder.
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What do you think a graph of depth vs. time would look like?
Your answer (start in the next line):
The graph would start on the left at its highest point, then decrease at a decreasing rate as the graph progressed to right until the depth reached zero.
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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 changes at a decreasing rate.
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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):
The graph would start at its max on the left, then decrease at a decreasing rate as it progresses to the right until the y value reaches zero.
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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 54.375 54.375
2 56.45313 2.078125
3 58.48438 2.03125
4 60.67188 2.1875
5 63.01563 2.34375
6 65.70313 2.6875
7 68.45313 2.75
8 71.60938 3.15625
9 75.07813 3.46875
10 79.35938 4.28125
11 84.71875 5.359375
12 94.39063 9.671875
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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):
0.8
2.75
4.65
6.55
8.45
10.3
12.1
13.85
15.55
17.25
18.95
20.55
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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):
0,20.55
2.07813,18.95
4.10938,17.25
6.29688,15.55
8.64063,13.85
11.32813,12.1
14.07813,10.3
17.23439,8.45
20.703813,6.55
24.98438,4.65
30.34375,2.75
40.01563,0.8
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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 depth is changing at a slower and slower rate, it is taking longer and longer for the water to reach the evenly spaced marks as time progresses.
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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 starts at its max point at (0,20.55) and decreases at a dereasing rate through points (4.11,18.95) and so on as the graph continues to decrease at a decreasing rate as the graph moves to righ until the graph reaches it final point at (40.02,0.80).
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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):
1.039,-.7699
3.094,-.8369
5.203,-.7771
7.469,-.7253
9.984,-.6512
12.70,-.6545
15.66,-.5861
18.97,-.5476
22.84,-.4439
27.66,-.3545
35.18,-.2016
I found these velocities by using the Difference Quotients vs. Midpoint Values program. The average velocities are shown in the second column with there associated midpiont time interval in the first comlumn.
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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):
1.039 Midpoint time interval for time 0 through 2.08
3.094 Midpoint time interval for time 2.08 through 4.11
5.203 Midpoint time interval for time 4.11 through 6.30
7.469 Midpoint time interval for time 6.30 through 8.64
9.984 Midpoint time interval for time 8.64 through 11.33
12.70 Midpoint time interval for time 11.33 thruogh 14.08
15.66 Midpoint time interval for time 14.08 through 17.23
18.97 Midpoint time interval for time 17.23 through 20.70
22.84 Midpoint time interval for time 20.70 through 24.78
27.66 Midpoint time interval for time 24.98 through 30.34
35.18 Midpoint time interval for time 30.34 through 40.02
These clock time midpoints were found using the Difference Quotients vs. Midpoint Values program.The midpoint clock times are in the first column and the associated clock times (that are rounded) are shown in the second column.
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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):
1.039,-.7699
3.094,-.8369
5.203,-.7771
7.469,-.7253
9.984,-.6512
12.70,-.6545
15.66,-.5861
18.97,-.5476
22.84,-.4439
27.66,-.3545
35.18,-.2016
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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 begins on the left at its third lowest point (1.039,-.7699) and increases through point (3.094,-8.369) then begins decreasing at a steadaly decreasing rate through points (5.203,-.7771),(7.469,-.7253) to point (9.984,-.6512) where it begins to increase at a steady rate to point (12.70,-.6545) where it begins to decrease at a decreasing rate through points (15.66,-.5861),(18.97,-.5476),(22.81,-.4439),(27.66,-.3545), and finaly though point (35.18,-.2016) at the experiments end.
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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):
2.067, .03260
4.149,-0.02835
6.336,-0.02285
8.727,-0.02946
11.34, 0.001215
14.18,-0.02311
17.32,-.01163
20.91,-0.02679
25.25,-0.01855
31.42,-0.02033
The average acceleration was determined by inserting the data from the midpoint time and average velocity found using the Difference Quotients vs. Midpoint Values program and then runing that data through the Difference Quotients vs. Midpoint Values program agian with the clock time shown in the first column and the agverage acceleration shown in the second column.
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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):
2.067, .03260
4.149,-0.02835
6.336,-0.02285
8.727,-0.02946
11.34, 0.001215
14.18,-0.02311
17.32,-.01163
20.91,-0.02679
25.25,-0.01855
31.42,-0.02033
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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 given above for acceleration is somewhat inconculsive, however overall the data seems to predict that the accleration is decreasing due to fact that although at a few points in time the data shows an increase, the data for acceleration is mostly decreasing.
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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):
The slope of the best fit line is 0.0179, this represents the rate of change between the velocity and clock time or acceleration. The best fit line is within 5 percent for all the data points except the first which is roughly 8 percent off and the last which is roughly 10 percent off. The best fit line or model does a fairly well job of predicting the data except at the very beging and very end of the graph.The model or best fit line shows decreasing acceleration due to fact that the velocity is decreasings as time passes.
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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):
3 hours
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You may add any further comments, questions, etc. below:
Your answer (start in the next line):
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Copy your document into the box below, be sure you have filled in your name and other identifying information at the top of this form, and submit:
Soft-drink bottle alternative
For students who use the soft-drink bottle as an alternative to the graduated cylinder:
Use a bottle which has a uniform cylindrical section. Most bottles are tapered at the top and the bottom, with a uniform cylindrical section in between.
Remove the label so you can easily observe the water level in the cylindrical section.
If you have a drill and a small drill bit (between about 1/16 inch and 1/8 inch), drill a hole near the bottom of the cylindrical section. If not, use something sharp and try to create a hole about 1/8 inch on a side; it's generally not difficult to make a triangular hole using a sharp blade. You can practice on a spare bottle, or if you prefer near the very top of the cylindrical section.
Instead of the milliliter markings, use a ruler or a ruler copy (the latter is a printable copy of a number of rulers; you may cut out a section and tape it to your bottle, which gives you a convenient scale to use for the experiment).
If you use this alternative, you should include the shape and dimensions of the hole, as best you can determine them.
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Copyright © 1999 [OrganizationName]. All rights reserved.
Revised: 07/02/10
&#Very good work on this lab exercise. Let me know if you have questions. &#