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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.
** Flow Experiment_labelMessages **
I hope that I completed this correctly. There were so many steps to this one. If I messed up early on,this would have messed up all of my other data.
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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 the level of water decreases in the cylinder. I would anticipate a decrease in the flow rate to accompany a decrease in the water level. The positional energy will decrease with a decrease in water level.
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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):
Based on Archimedes principle, I would assume that the velocity at the waters surface to decrease. When reading the text, I discovered that the force decreases as the depth decreases. I would assume that the depth of the water would not affect the buoy at all. A buoy will float no matter what the depth of water is, unless of course there is no water. Another factor to consider would be the weight of the buoy. If the buoy displaces a large amount of water because of its weight, then it will sink further into the water as the water level decreases.
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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):
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? All of these factors determine the other. For example the velocity would continuously be changing due to the decreasing level of water from the cylinder. As the water drops, the velocity formula would have to be performed for each level dropped in order to correctly calculate the velocity. The diameter of the cylinder would determine the depth of the water. For example, the level of water in a bowl for example would be less than the level in a tall glass with the same amount of water placed in both containers. The diameter of the hole will determine how quickly the water streams out of the hole.
More specifically how could you determine the velocity of the water surface from the values of the other quantities?
To determine the velocity one would have to calculate the area of the water volume within the cylinder. The water level would need to be measured and the radius would need to be found in order to calculate the volume.
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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):
A change in velocity is directly linked to an action of a force. The water will not increase in velocity or decrease in velocity unless a force is applied to it. Meaning, the water being forced from the cylinder will increase in velocity, as the water inside the cylinder will decrease in velocity.
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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 is positional energy combined with the diameter of the hole in the cylinder. If there was no hole, the water would not be able to escape. So both factors are necessary to force the water from 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):
If the pictures had a time stamp we would know at what rate the water is decreasing inside the cylinder. As I can tell from looking at the pictures, I can tell that the depth is changing at a slower and slower rate. I can tell that by looking at the pictures that the water level is decreasing each time. I cant however, tell at what rate it is decreasing.
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What do you think a graph of depth vs. time would look like?
Your answer (start in the next line):
I think that a graph would be in the positive x and y quadrant and that the line would steadily ascend.
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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 of the stream decreases as time passes. As evidenced by the photos above, one can tell that the stream flowing from the cylinder gets less and less forceful, thus decreasing.
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Does this distance change at an increasing, decreasing or steady rate?
Your answer (start in the next line):
This distance changes at an increasing rate. Initially the water is forceful coming out of the cylinder, but quickly decreases as the water level inside the cylinder decreases.
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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 will decrease 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. If you will be using that alternative, read all the instructions, then at the end you will see instructions for modifying the procedure to use a soft drink bottle.
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 17939.64 17939.64
2 17946.58 6.945313
3 17954.29 7.707031
4 17961.64 7.355469
5 17971.5 9.855469
6 17979.55 8.046875
7 17989.57 10.02344
8 18000.5 10.92969
9 18013.05 12.55078
10 18024.83 11.78125
11 18041.88 17.04688
12 18062.59 20.71094
13 18092.98 30.39063
14 18111.78 18.80078
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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):
I had to create a cylinder using the instructions for a bottle at the bottom. I did however have a salt bottle that I used(it was shaped just like a cylinder). I had to color the water black and use a light at the top of my cylinder to see the water inside the cylinder but it worked. I used ½ inch intervals on the side of my cylinder. The distance from the center of my outflow tube to the first marker is 1/2inch. The measurements are as follows(from the center of the outflow tube to the highest marker. I decided to convert inches to cm for the sake of this experiment.
1.2700
2.2225
3.4925
4.7625
5.7150
7.3025
8.2550
9.8425
10.975
12.382
13.335
14.922
15.875
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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):
Clock time in seconds measured from first reading Depth of water (in cm) measured from hole
0 15.875
6.94 14.922
7.71 13.335
7.35 12.382
9.86 10.975
8.05 9.8425
10.02 8.2550
10.93 7.3025
12.55 5.7150
11.78 4.7625
17.05 3.4925
20.71 2.2225
30.39 1.2700
18.8 0
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Your first column contains time intervals, not clock times. Clock times steadiliy increase; time intervals can sometimes increase and sometimes decrease, as they do here.
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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):
Yes, my data supports my opinion. 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 decreases at a decreasing rate.
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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):
I obtained the average velocity by first finding the radius of the cylinder. The diameter of the cylinder is 2.75 inches so the radius will be 1.375. To find the area is will simply use the following equation A=pir^2 so A=(3.14)(1.375)^2=5.94in^2. I will then multiply each marked distance by the area to give me a volume for each level. These will follow:
21.83
20.52
18.34
17.03
15.09
13.53
11.35
10.04
7.86
6.55
4.80
3.06
1.75
0
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This will give you the volume information, and is well done.
However you were asked for the velocity of the water surface, which for any time interval is the change in position divided by the change in clock time.
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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):
3.47
7.325
7.53
8.605
8.955
9.035
10.475
11.74
12.165
14.415
18.88
25.55
24.595
18.8
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If your clock times as reported earlier were correct, then these would be the correct midpoint clock times, so you've got the idea of the midpoint clock time. This is good.
However, as I indicated previously, what you reported as clock times were in fact time intervals.
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This will need to be corrected.
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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):
Average velocity Midpoint clock time
21.83 3.47
20.52 7.325
18.34 7.53
17.03 8.605
15.09 8.955
13.53 9.035
11.35 10.475
10.04 11.74
7.86 12.165
6.55 14.415
4.80 18.88
3.06 25.55
1.75 24.595
0 18.8
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Note that a table of average velocity vs. midpoint clock time would have the midpoint clock time in the first column.
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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 decreases at a constant 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):
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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):
Clock time at midpoint Average acceleration
3.47 6.29
7.325 2.80
7.53 2.44
8.605 1.98
8.955 1.69
9.035 1.50
10.475 1.08
11.74 0.86
12.165 0.65
14.415 0.45
18.88 0.25
25.55 0.12
24.595 0.07
18.8 0
I achieved my average acceleration by dividing my average velocity/ midpoint in time. For example the second row in my table is the first calculated average acceleration. I took the first found average velocity (21.83) and divided it by the first midpoint in clock time (3.47) which gives me 6.29 average acceleration.
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Average acceleration is average rate of change of velocity with respect to clock time, which is change in velocity divided by change in clock time.
You have simply divided velocity by clock time, which does not give you average acceleration.
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You'll need to divide the change between two velocities by the change between the two corresponding midpoint clock times.
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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 graph indicates that the acceleration fluctuated very little, but still fluctuated. The acceleration seems to be decreasing as the time passes.
I think that by the data presented the acceleration of the water surface is 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):
My slope was calculated by dividing the rise/run, which equals -0.41. The slope is decreasing at a decreasing rate.
I believe that my straight line represents the actual behavior of the system fairly well. The graph demonstrates a decrease in the average velocity and times, therefore this information directly correlates with the data I was able to accrue by performing this experiment.
The graph for average velocity vs. midpoint clock time is more consistent with decreasing acceleration. As one can see by looking at both the table and the graph I have drawn, the acceleration decreases as each midpoint in time occurs.
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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):
I broke this experiment up into several sessions. I would say overall it took me somewhere between 2 ½ to 3 hours to complete.
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You've done very good work with most steps in your report.
However you'll need to redo a lot of your calculations because you've incorrectly used time intervals where you need to have used clock times. I believe this issue came up on a previous assignment as well (perhaps the on on formatting).
Before you do any extensive recalculation, send me a copy of your data
1 17939.64 17939.64
2 17946.58 6.945313
3 17954.29 7.707031
4 17961.64 7.355469
5 17971.5 9.855469
6 17979.55 8.046875
7 17989.57 10.02344
8 18000.5 10.92969
9 18013.05 12.55078
10 18024.83 11.78125
11 18041.88 17.04688
12 18062.59 20.71094
13 18092.98 30.39063
14 18111.78 18.80078
and a corrected list of your clock times. I don't want you waste time by going through the entire process with incorrect clock times.
Once you've got the clock times right it should be reasonably easy to make subsequent corrections.
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