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Phy 242
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 **
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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 waterflow out of the cylinder to decrease as time goes on.
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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 expect that the velocity at which the water lever descends would be at a decreasing rate.
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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):
If you determined the rate at which water was exiting the bottle, the diameter of the cylinder, and the diameter of the hole you could use these values to calculate the velocity of the lowering water lever. To do this you could use the rate at which the water is flowing out, the diameter of the cylinder, and your knowledge of calculus and related rated to find the rate at which the water level is moving.
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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):
We know that the change in velocity means a change in the forces because a change in velocity is known as the system being accelerated and force equals mass times acceleration.
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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 that the nature of force is actually the loss of the normal force from the side of the cylinder. The water directly near the hole has the force of gravity pulling downwards and the force of the other water particles pushing it outward from the center and when the cylinder wall is not there to achieve equilibrium, the water will accelerate away from these forces.
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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):
It appears as though the depth is changing at a faster and faster rate according to the pictures.
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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 at the initial water level and slowly start to decrease at first, but then decrease faster and faster until it reaches 0 or the height at which no more soda will come out.
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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):
This 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 the original distance the water spurts and slope down once the water level has dropped far enough to change the force and then it will slope down faster until it reaches a near minimum and steadies out until the water is finished spilling.
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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):
8 434.6484 239.9414
9 441.0313 6.382813
10 447.2695 6.238281
11 452.793 5.523438
12 461.3867 8.59375
13 471.6211 10.23438
14 485.8008 15.17969
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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):
2cm
4.3cm
6.6cm
8.5cm
11cm
13.4cm
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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, 13.4
6.382813, 11
12.621094, 8.5
18.144532, 6.6
26.738282, 4.3
36.972662, 2
52.152352, 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 depth is changing at a decreasing rate. That contradicts my the pictures but agrees with my first guess.
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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):
My graph begins at the height 13.4 and slopes down at about 1/3 cm per second until it reaches around the 20s on the x axis and it begins to decrease at a decreasing rate until it is nearly horizontal and at the x value 52.2 it reaches zero.
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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):
-.37926
-.40076
-.34399
-.26764
-.22473
-.13175
I found the average velocities by diving the change in height by the elapsed time of that interval.
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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):
3.19
9.50
15.38
22.44
31.86
44.56
I found the clock time in seconds by dividing the interval by 2 and adding it onto the first number in the interval.
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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):
3.19, -.37926
9.50, -.40076
15.38, -.34399
22.44, -.26764
31.86, -.22473
44.56, -.13175
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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 is decreasing at a decreasing rate. The time and rate are directly rated, as the time goes on the rate increases (becomes more negative). The y values begin at -.37 and increase a tad then decrease and slowly the slope decreases creating a concave down line.
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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):
-.00341
.00966
.01081
.00456
.00732
I divided the difference in velocities by the difference in times (final minus initial rate/ final minus initial time).
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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):
6.345 , -.00341
12.44, .00966
18.91, .01081
27.15, .00456
38.21, .00732
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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):
My answers either suggest that at first it increased (maybe because it was adjusting in the bottle or I started my clock a little early) and then decreased until it reached zero. I dont know that I would consider them inconclusive, but it would be beneficial to repeat the exercise and see if the results were similar.I believe the acceleration of the water surface is increasing in the opposite direction of the velocity, so if we made the axis positive towards the hold then the acceleration would be decreasing.
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Good.
Does the 'Deterioration of Difference Quotients' phenomenon shed any light on the reliability of your results?
Please respond by appending your answer to a copy of this document, as a revision.
&#Please see my notes and submit a copy of this document with revisions, comments 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. &#
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