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Phy 232
Your 'bottle thermometer' 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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bottle thermometer
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Phy 232
Your 'bottle thermometer' 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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5 hours
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Data program is at http://www.vhcc.edu/dsmith/genInfo/labrynth_created_fall_05/levl1_15\levl2_51/dataProgram.
exe
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You can use the bottle, stopper and tubes as a very sensitive thermometer. This thermometer will have excellent precision, clearly registering temperature changes on the order of .01 degree. The system will also demonstrate a very basic thermal engine and its thermodynamic properties.
Set up your system with a vertical tube and a pressure-indicating tube, as in the experiment on measuring atmospheric pressure. There should be half a liter or so of water in the bottom of the container.
• Refer back to the experiment 'Measuring Atmospheric Pressure' for a detailed description of how the pressure-indicating tube is constructed for the 'stopper' version of the experiment.
For the bottle-cap version, the pressure-indicating tube is the second-longest tube. The end inside the bottle should be open to the gas inside the bottle (a few cm of tube inside the bottle is sufficient) and the other end should be capped.
The figure below shows the basic shape of the tube; the left end extends down into the bottle and the capped end will be somewhere off to the right. The essential property of the tube is that when the pressure in the bottle increases, more force is exerted on the left-hand side of the 'plug' of liquid, which moves to the right until the compression of air in the 'plugged' end balances it. As long as the liquid 'plug' cannot 'leak' its liquid to the left or to the right, and as long as the air column in the plugged end is of significant length so it can be measured accurately, the tube is set up correctly.
If you pressurize the gas inside the tube, water will rise accordingly in the vertical tube. If the temperature changes but the system is not otherwise tampered with, the pressure and hence the level of water in the tube will change accordingly.
When the tube is sealed, pressure is atmospheric and the system is unable to sustain a water column in the vertical tube. So the pressure must be increased. Various means exist for increasing the pressure in the system.
• You could squeeze the bottle and maintain enough pressure to support, for example, a 50 cm column. However the strength of your squeeze would vary over time and the height of the water column would end up varying in response to many factors not directly related to small temperature changes.
• You could compress the bottle using mechanical means, such as a clamp. This could work well for a flexible bottle such as the one you are using, but would not generalize to a typical rigid container.
• You could use a source of compressed air to pressurize the bottle. For the purposes of this experiment, a low pressure, on the order of a few thousand Pascals (a few hudredths of an atmosphere) would suffice.
The means we will choose is the low-pressure source, which is readily available to every living land animal. We all need to regularly, several times a minute, increase and decrease the pressure in our lungs in order to breathe. We're going to take advantage of this capacity and simply blow a little air into the bottle.
• Caution: The pressure you will need to exert and the amount of air you will need to blow into the system will both be less than that required to blow up a typical toy balloon. However, if you have a physical condition that makes it inadvisable for you to do this, let the instructor know. There is an alternative way to pressurize the system.
You recall that it takes a pretty good squeeze to raise air 50 cm in the bottle. You will be surprised at how much easier it is to use your diaphragm to accomplish the same thing. If you open the 'pressure valve', which in this case consists of removing the terminating cap from the third tube, you can then use the vertical tube as a 'drinking straw' to draw water up into it. Most people can easily manage a 50 cm; however don't take this as a challenge. This isn't a test of how far you can raise the water.
Instructions follow:
• Before you put your mouth on the tube, make sure it's clean and make sure there's nothing in the bottle you wouldn't want to drink. The bottle and the end of the tube can be cleaned, and you can run a cleaner through the tube (rubbing alcohol works well to sterilize the tube). If you're careful you aren't likely to ingest anything, but of course you want the end of the tube to be clean.
• Once the system is clean, just do this. Pull water up into the tube. While maintaining the water at a certain height, replace the cap on the pressure-valve tube and think for a minute about what's going to happen when you remove the tube from your mouth. Also think about what, if anything, is going to happen to the length of the air column at the end of the pressure-indicating tube. Then go ahead and remove the tube from your mouth and watch what happens.
Describe below what happens and what you expected to happen. Also indicate why you think this happens.
When I sucked the water into the vertical tube, and put the cap on the pressure valve and took my mouth away, the water level in the vertical tube dropped down into the bottle, but retained some of the water within the tube. I expected to see the water level in the vertical tube to remain at the level that I had brought it up to initially because of the pressure created in the bottle, but this was not the case. I didn’t notice any appreciable difference in the length of the air column, but that doesn’t mean that it didn’t move.
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Now think about what will happen if you remove the cap from the pressure-valve tube. Will air escape from the system? Why would you or would you not expect it to do so?
Go ahead and remove the cap, and report your expectations and your observations below.
I observed that when the pressure valve was released, the small amount of water in the vertical tube dropped back into the bottle. I expected this because the pressure built up in the system from my sucking in the tube was allowed to release into the surrounding which it did with a slight hiss. I didn’t notice any appreciable difference in the air column.
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Now replace the cap on the pressure-valve tube and, while keeping an eye on the air column in the pressure-indicating tube, blow just a little air through the vertical tube, making some bubbles in the water inside the tube. Blow enough that the air column in the pressure-indicating tube moves a little, but not more than half a centimeter or so. Then remove the tube from your mouth, keeping an eye on the pressure-indicating tube and also on the vertical tube.
• What happens?
When I blew into the vertical tube, the length of the air column decreased and water was pulled up into the vertical tube where it remained in the same place.
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• Why did the length of the air column in the pressure-indicating tube change length when you blew air into the system? Did the air column move back to its original position when you removed the tube from your mouth? Did it move at all when you did so?
The air column most likely changed length because I was putting pressure into the system which must have pushed the water in front of the air column. The air column did not move back to its original position when I removed the tube from my mouth, but it did move backwards a little bit then remained constant a little further from the original length.
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• What happened in the vertical tube?
As stated before, the water moved up into the vertical tube and stayed at a constant level.
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• Why did all these things happen? Which would would you have anticipated, and which would you not have anticipated?
The water moved into the vertical tube because there was a pressure in the tube caused by my blowing, and when I removed my mouth, that pressure was not as strong, so the water moved up into the tube to fill it. I expected this to happen. In addition I expected the air column to move because pressure was applied to it. I believe that the air column moved backward because the pressure was a little lessened when some of it moved back into the vertical tube, but it is clear that there is still pressure in the system because it has not escape route.
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• What happened to the quantities P, V, n and T during various phases of this process?
Pressure increased because I blew into the tube, volume decreased because of increased pressure, the amount of moles of gas remained constant, and due to the pv=nrt relationship, temperature must have increased with an increase in pressure.
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When you blow into the tube you increase the amount of gas in the bottle. This would increase the pressure even if temperature and volume remained unchanged.
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Place the thermometer that came with your kit near the bottle, with the bulb not touching any surface so that it is sure to measure the air temperature in the vicinity of the bottle and leave it alone until you need to read it.
Now you will blow enough air into the bottle to raise water in the vertical tube to a position a little ways above the top of the bottle.
• Use the pressure-valve tube to equalize the pressure once more with atmospheric (i.e., take the cap off). Measure the length of the air column in the pressure-indicating tube, and as you did before place a measuring device in the vicinity of the meniscus in this tube.Replace the cap on the pressure-valve tube and again blow a little bit of air into the bottle through the vertical tube. Remove the tube from your mouth and see how far the water column rises. Blow in a little more air and remove the tube from your mouth. Repeat until water has reached a level about 10 cm above the top of the bottle.
• Place the bottle in a pan, a bowl or a basin to catch the water you will soon pour over it.
• Secure the vertical tube in a vertical or nearly-vertical position.
The water column is now supported by excess pressure in the bottle. This excess pressure is between a few hundredths and a tenth of an atmosphere.
The pressure in the bottle is probably in the range from 103 kPa to 110 kPa, depending on your altitude above sea level and on how high you chose to make the water column. You are going to make a few estimates, using 100 kPa as the approximate round-number pressure in the bottle, and 300 K as the approximate round-number air temperature. Using these ball-park figures:
• If gas pressure in the bottle changed by 1%, by how many N/m^2 would it change?
1000 Pa or n/m^2
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• What would be the corresponding change in the height of the supported air column?
Since I am using a 2 liter bottle and there is approximately 1 liter of water in the bottle,
P1v1=p2v2
100Kpa(1000cm^3)=101kPa(x) x=990cm^3 The change in volume would be about 10cm^3 which is about 2.1 cm.
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If the tube is thin enough the water in it will occupy very little volume, but will still require additional pressure to support it. In this case the volume of the system would be very nearly constant. The excess pressure would be due to additional gas being forced into the bottle.
The initial air pressure is 100 kPA. A 1% increase in air pressure would amount to an extra 1 kPa or 1000 Pa.
An extra 1000 Pa would support an additional 10 cm column of water, so in a very thin tube (in which the displaced water would occupy very little volume, so that the volume of gas in the system would essentially be unchanged) the water column would rise 10 cm.
In the present case it turns out that a 10 cm length of the tube holds about 0.7 cm^3 of water, so as this water is displaced into the tube the volume of the system increases by 0.7 cm.
To what height would water therefore be expected to rise in the tube?
&&&& I am assuming that the statement “the volume of the system increases by 0.7cm” is supposed to say 0.7cm^3 because cm is not a measure of volume. If that is the case, then a proportion could be set up P1V1=P2V2
100kPa(1000cm^3)=x(1000.7cm^3)
X=99.93kPa which would be the pressure change because of changing the volume. Since 1000Pa was added to the system, the final pressure would be 99.93kPa+1000Pa=100.93kPa
A new proportion could then be set up
101kPa/10cm=100.93kPa/x
X=9.99cm
I could solve this based on the knowledge that 1000Pa moves the water column 10cm, but I don’t know how I am supposed to assume this. Essentially I don’t know where this value came from.
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• By what percent would air temperature have to change to result in this change in pressure, assuming that the container volume remains constant?
100Kpa/300k=101kpa/x
X=303K =1%increase
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Continuing the above assumptions:
• How many degrees of temperature change would correspond to a 1% change in temperature?
3 degrees
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• How much pressure change would correspond to a 1 degree change in temperature?
100Kpa/300K=x/301k
X=100.3Kpa so the change would be 0.3 Kpa
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• By how much would the vertical position of the water column change with a 1 degree change in temperature?
300K/1000cm^3=301k/x x=1003.3cm^3 The change would be about 3 cm^3.
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Vertical position in the tube is measured in cm, not in cm^3.
I would agree that a 1 degree change in gas temperature is associated with about a 3 cm change in the vertical position of the water column, but since a 3 cm column will not contain 3 cm^3 of water, this is not the result of expansion of the gas.
&&&& 100kPa/300K=x/301K
X=100.3kPa
Since an addition of 1000Pa raises the water column 10cm
1000Pa/10cm=300Pa/x
X=3cm
Again, this is easy to determine because of the assumption that 1000Pa raises the water height 10cm, but I don’t know where that value came from
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• How much temperature change would correspond to a 1 cm difference in the height of the column?
300K/10cm=x/11cm
X=330K so the temperature difference would be 30 degrees. This doesn’t really seem right to me, but I believe that is the right way to calculate it.
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Your calculation is unrelated to the gas laws. That 11 cm figure is not associated with any quantity proportional to any of the associated variables.
So to your credit, your intuition that this doesn't seem right is correct.
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1000Pa/10cm=x/11cm
X=1100Pa
100kPa/300K=101kPa/x
X=303K
101kPa/303K=101.1kPa/x
X=303.3K
The difference in temperature would be 0.3 degrees.
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• How much temperature change would correspond to a 1 mm difference in the height of the column?
300K/10cm=x/10.1cm
X=303K=3 degrees temperature change
I didn’t get the right answer for the temperature change values, and I don’t know what I did wrong.
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A change in temperature of 1 Kelvin or Celsius degree in the gas inside the container should correspond to a little more than a 3 cm change in the height of the water column. A change of 1 Fahrenheit degree should correspond to a little less than a 2 cm change in the height of the water column. Your results should be consistent with these figures; if not, keep the correct figures in mind as you make your observations.
The temperature in your room is not likely to be completely steady. You will first see whether this system reveals any temperature fluctuations:
• Make a mark, or fasten a small piece of clear tape, at the position of the water column.
• Observe, at 30-second intervals, the temperature on your alcohol thermometer and the height of the water column relative to the mark or tape (above the tape is positive, below the tape is negative).
• Try to estimate the temperatures on the alcohol thermometer to the nearest .1 degree, though you won't be completely accurate at this level of precision.
• Make these observations for 10 minutes.
Report in units of Celsius vs. cm your 20 water column position vs. temperature observations, in the form of a comma-delimited table below.
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-0.1, 21.5
-0.1,21.5
-0.1, 21.5
-0.1,21.5
-0.1, 21.5
-0.1,21.5
-0.2,21.4
-0.2,21.4
-0.2,21.4
-0.2,21.4
-0.2,21.4
-0.2, 21.4
-0.2, 21.4
-0.2, 21.4
-0.2, 21.4
-0.2, 21.4
-0.3,21.3
-0.3,21.3
-0.3,21.3
-0.3,21.3
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Describe the trend of temperature fluctuations. Also include an estimate (or if you prefer two estimates) based on both the alcohol thermometer and the 'bottle thermometer' the maximum deviation in temperature over the 10-minute period. Explain the basis for your estimate(s):
The temperature seems to fluctuate by about 0.3 degrees Celsius over a 10 minute period, so it seems to me that the maximum deviation would be 0.3 degrees Celsius. I don’t know, however, whether or not this was simply because the pressure was leaking and dropping ever so slightly.
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Now you will change the temperature of the gas in the system by a few degrees and observe the response of the vertical water column:
• Read the alcohol thermometer once more and note the reading.
• Pour a single cup of warm tap water over the sides of the bottle and note the water-column altitude relative to your tape, noting altitudes at 15-second intervals.
• Continue until you are reasonably sure that the temperature of the system has returned to room temperature and any fluctuations in the column height are again just the result of fluctuations in room temperature. However don't take data on this part for more than 10 minutes.
Report your results below:
21.5 C
10cm
8cm
6cm
5cm
4cm
3.5cm
3.0cm
2.7cm
2.4cm
2.1cm
1.9cm
1.7cm
1.5cm
1.2cm
1.0cm
0.9cm
0.8cm
0.7cm
0.6cm
0.5cm
0.4cm
0.3cm
0.25cm
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If your hands are cold, warm them for a minute in warm water. Then hold the palms of your hands very close to the walls of the container, being careful not to touch the walls. Keep your hands there for about a minute, and keep an eye on the air column.
Did your hands warm the air in the bottle measurably? If so, by how much? Give the basis for your answer:
My hands did measurably affect the temperature. The column of air moved about 1 cm which would lead me to believe that the air temperature increased by about 0.3 degrees based on the fact that a 3cm movement of the air column corresponds to a 1 degree temperature increase.
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Now reorient the vertical tube so that after rising out of the bottle the tube becomes horizontal. It's OK if some of the water in the tube leaks out during this process. What you want to achieve is an open horizontal tube,, about 30 cm above the level of water in the container, with the last few centimeters of the liquid in the horizontal portion of the tube and at least a foot of air between the meniscus and the end of the tube.
The system might look something like the picture below, but the tube running across the table would be more perfectly horizontal.
Place a piece of tape at the position of the vertical-tube meniscus (actually now the horizontal-tube meniscus). As you did earlier, observe the alcohol thermometer and the position of the meniscus at 30-second intervals, but this time for only 5 minutes. Report your results below in the same table format and using the same units you used previously:
21.5, 0.0
21.4, 0.0
21.5,0.0
21.4, -0.1
21.5,-0.1
21.5,0.0
21.5,0.0
21.5,-0.1
21.5,-0.1
21.5,0.0
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Repeat the experiment with your warm hands near the bottle. Report below what you observe:
When I put my hands near the bottle, the water level moved horizontally, but it didn’t seem to stop at any point, it just kept moving. The water level went much further than the vertical distance.
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When in the first bottle experiment you squeezed water into a horizontal section of the tube, how much additional pressure was required to move water along the horizontal section?
• By how much do you think the pressure in the bottle changed as the water moved along the horizontal tube?
There was no need for additional pressure, just constant force. I don’t think the pressure changed at all.
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• If the water moved 10 cm along the horizontal tube, whose inner diameter is about 3 millimeters, by how much would the volume of air inside the system change?
The volume would change by pi (0.15cm)^2(10cm)=0.707cm^3
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Right. Note that this is the same figure given in one of my notes.
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• By what percent would the volume of the air inside the container therefore change?
Since there is 1 liter of air in the bottle and 1 liter is 1000cm^3 then the volume of air changed by .707cm^3/1000cm^3=0.07%
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• Assuming constant pressure, how much change in temperature would be required to achieve this change in volume?
1000cm^3/300K=1000.707/x= 300.21 so the change would be 0.21 degrees.
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At constant pressure, which is the case when the tube is horizontal, this is correct.
If the tube is not horizontal, this change in temperature would be associated with a much smaller movement of the water in the tube, since with that configuration the change in temperature would mostly result in a change in pressure.
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• If the air temperature inside the bottle was 600 K rather than about 300 K, how would your answer to the preceding question change?
If the air temperature was 600K rather than 300K, I don’t think the answer would change because assuming that the volume was still 1000cm^3 to begin with and the pressure was constant, there would still be the same change in temperature because it would still be the same change in volume.
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Note that at 600 C as opposed to 300 C, a 1% change in temperature would be 6 Celsius rather than 3 Celsius. A 1% change in temperature would still result in a 1% change in pressure at constant volume, or a 1% change in volume at constant pressure.
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There were also changes in volume when the water was rising and falling in the vertical tube. Why didn't we worry about the volume change of the air in that case? Would that have made a significant difference in our estimates of temperature change?
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The changes in volume were so miniscule that they could be discounted for causing any significant temperature change.
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As your figures will show you the volume changes aren't completely insignificant, but their effect was an order of magnitude smaller than changes associated with pressure.
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If the tube was not completely horizontal, would that affect our estimate of the temperature difference?
For example consider the tube in the picture below.
Suppose that in the process of moving 10 cm along the tube, the meniscus moves 6 cm in the vertical direction.
• By how much would the pressure of the gas have to change to increase the altitude of the water by 6 cm?
According to calculations earlier it would take 2000 Pa of pressure to increase the altitude by 6 cm.
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10 cm along the tube would correspond to volume change 0.7 cm^3.
A 6 cm rise would correspond to about 600 Pa in pressure change.
What temperature change would be required to achieve both of these effects?
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To find the temperature change associated with this, the gas laws are needed.
100kPa/300K=100.6kPa/x
X would be 301.8K so the temperature change would be 1.8 degrees K
For the volume change,
1000cm^3/300K=1000.7cm^3/x
X=300.21K so the increase would be 0.21 degrees.
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• Assuming a temperature in the neighborhood of 300 K, how much temperature change would be required, at constant volume, to achieve this pressure increase?
100kPa/300K=102kPa/x=306K so the temperature would change by 6 degrees.
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If the pressure change was 2 kPa then this wold be correct.
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100kPa/300K=100.6kPa/x
X would be 301.8K so the temperature change would be 1.8 degrees K
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• The volume of the gas would change by the additional volume occupied by the water in the tube, in this case about .7 cm^3. Assuming that there are 3 liters of gas in the container, how much temperature change would be necessary to increase the gas volume by .7 cm^3?
3000cm^3/300K=3000.7cm^3/X x=300.07k so the temperature would change by 0.07K
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Continue to assume a temperature near 300 K and a volume near 3 liters:
• If the tube was in the completely vertical position, by how much would the position of the meniscus change as a result of a 1 degree temperature increase?
300K/3000cm^3=301K/x x=3010cm^3 so the meniscus would change by 10cm^3.
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• What would be the change if the tube at the position of the meniscus was perfectly horizontal? You may use the fact that the inside volume of a 10 cm length tube is .7 cm^3.
I don’t know how to calculate this. I don’t quite understand how horizontal movement is different from vertical movement.
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• A what slope do you think the change in the position of the meniscus would be half as much as your last result?
I would assume that a 45 degree slope would generate half simply based on the calculations I did with the sloping tube.
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