#$&*
PHY 202
Your 'measuring atmospheric pressure' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Measuring Atmospheric Pressure_labelMessages **
Your setup for the preceding experiment Raising Water in a Vertical Tube included a vertical tube, with terminating caps on the other two tubes. You will use the vertical tube again in this experiment.
Your kit included two bottlecaps connected by a long tube. The long tube is to be used as a vertical tube, as in the previous experiment.
Each bottlecap has three tubes. One is a short tube; its intended use is to release pressure in the system when and if this becomes necessary. The third is fairly long. This tube is to be used as a 'pressure tube'.
First fill the 'pressure tube' with water. You can do this in any way you wish. One way:
The easiest way to do this is to temporarily disconnect the vertical tube and replace it with the new tube, so that when you squeeze the container you can fill the new tube. Add water to the container until it is nearly full, then screw on the bottlecap.
Hold the open end of the pressure-indicating tube a little higher than the top of the container, near the point where you just connected it, and squeeze the bottle so that water fills the tube. Since the water level in the container is higher than in the preceding experiment, and since the end of the new tube isn't much higher than the water level, this shouldn't require a very hard squeeze.
When the tube is full, maintain the squeeze so the water doesn't return to the container and disconnect it. You will have a tube full of water.
Now empty about half the water from the 'pressure tube'. Cap it and connect it to the system, and replace the vertical tube. You can do this in any way you wish, but one way is described below:
Just raise one end of the full 'pressure tube' and/or lower the other, and some water will flow out.
Once the tube is about half full, place a terminating cap on one end of this tube. This will hold the water in the 'pressure tube'.
You should at this point have:
The vertical tube, extended down into the water and out of the top of the container
The extended pressure-measuring tube, open on one end (through its connection to the newly opened tube in the stopper) to the air inside the container, half full of water, and capped at the other end.
A third tube short through the bottlecap, still closed off at the 'top end'.
In the picture below you see:
the short capped tube (the 'third' tube) sticking out of the top of the stopper,
the 'vertical' tube not yet in a vertical position but extending forward and to the right into a graduated cylinder, and
the pressure-indicating tube half full of caramel-colored liquid (the liquid is cola) and draped over a second graduated cylinder toward the back left. The pressure-indicating tube is capped at its end (hanging down near the tabletop), and the last 25 cm segment of the tube contains no liquid.
The picture below shows how the liquid in the tube comes to a point just below the 'peak' of the tube. This leaves an air column about 25 cm long in the capped end of the tube.
In the new picture the pressure-indicating tube is simply lying on the tabletop so the air column at the capped end is clearly visible.
The figure below shows a sketch of a tube which rises out of the bottle at left, then bends to form a U, then to the right of the U again levels off. The tube continues a ways to the right and is sealed at its right end. Liquid occupies the U up to almost the point of leveling, so that an increase in the pressure of the container will cause the liquid to move into the level region. As is the case in our experiment, the tube is assumed thin enough that the plane of the meniscus remains parallel to the cross-section of the tube (i.e., the meniscus doesn't 'level off' when it moves into a horizontal section of tube).
You should manipulate the pressure tube until its configuration resembles the one shown. The length and depth of the U can vary from that depicted, but the air column at the end of the tube should be at least 15 (actual) cm long. The liquid levels at the left and right ends of U should be very nearly equal.
The basic idea is that as you squeeze the system to raise water in the vertical tube, as in your previous experiment, the pressure in the system increases and compresses the 'air column' in the pressure tube. By measuring the lengths of this 'air column' you can determine relative pressures, and by measuring the heights of the water column in the 'vertical tube' you can determine the actual pressure differences required to support those columns.
Support the end of the vertical tube so that it is more or less vertical, as it was in the previous experiment.
The bottle should be pretty full, but not so full that it covers the open end of the tube to which the pressure tube is connected; the left end of the pressure tube should have an 'open path' to the gas inside the bottle, so that the pressure on the left-hand side of the water column in that tube is essentially equal to the pressure in the bottle.
If you squeeze the container a little, water will rise a little way in the vertical tube and the water in the pressure tube will also move is such a way as to slightly shorten the air column. The harder you squeeze the higher water will rise in the vertical tube and the shorter the air column will become.
Go ahead and observe this phenomenon. There is no need to measure anything yet, just get the 'feel' of the system.
Indicate below how the system behaves (what changes when you do what, how the system's reactions to your actions appear to be related to one another) and how it 'feels'.
----->>>>> behavior
Your answer (start in the next line):
When I squeeze, the water goes up in the vertical tube and moves the air column in the pressure tube.
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Using a measuring device you will measure the relative positions of the meniscus as you vary your squeeze:
One of the ruler copies used in the previous experiment on the distortion of paper rulers should be used here; a reduced copy should be used for greater precision. You may choose the level of reduction at which you think you will achieve the greatest level of precision. Only relative measurements will be important here; it will not be necessary to convert your units to actual millimeters or centimeters.
Indicate below the level of reduction you have chosen, and your reasons for this choice.
----->>>>> level of reduction and reasons
Your answer (start in the next line):
No level of reduction was used, a sewing measuring tape was used for my experiment.
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In the units of the measuring device you have chosen, write down in your lab notebook the readings you used to indicate length of the air column, from the meniscus to the barrier at the capped end. No conversion of the units of your device to standard units (e.g., millimeters or centimeters) is required. Your information should include the marking at one end of the measuring device, and the marking at the other. If necessary two or more copies of paper rulers may be carefully taped together.
Indicate in the first line below the length of the air column in the units of your measuring device.
In the second line explain how you obtained your result, including the readings at the two ends and how you used those readings to indicate the length.
----->>>>> air column length, how obtained incl readings and how used
Your answer (start in the next line):
The air column length is approximately 36 cm. The tape measure was taped to the pressure tube.
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Now place the same measuring device along the tube, positioned so you can observe as accurately as possible the relative positions of the meniscus in the pressure tube.
It is recommended that the initial position of the meniscus be in the vicinity of the center of the measuring device, so that position changes in both directions can be observed.
It is not necessary for the measuring device to extend the entire length of the air column, as long as you know the reading on the measuring device that corresponds to the initial position of the meniscus. From this information and from subsequent readings it will be easy to determine the varying lengths of the air column.
Take whatever precautions are necessary to make sure neither the measuring device nor the pressure tube can move until you have completed the necessary trials.
Mark positions along the vertical tube at 10-cm intervals (actual 10-cm intervals as indicated by a full-sized ruler) above the surface of the water in the bottle.
If the bottle is pretty full, as described before, it might be possible to make the first mark on the vertical tube at 10 or 15 cm above the water surface.
Marks may be made using an actual marker, or pieces of tape, or anything else that happens to be convenient.
Write your information in your lab notebook:
Write down the position of the first mark on the vertical tube with respect to the water surface (e.g., 10 cm or 15 cm).
Write down the position of the meniscus in the pressure tube. This position will simply be the reading on your measuring device. For example if the meniscus is at marking 17.35, that is what you write down.
As in all labs, you directly record what you read. Never do any arithmetic between making an observation and recording it.
You will now conduct 5 trials, raising water to the first mark on your vertical tube and reading the position of the meniscus before the squeeze and while water is at the given level.
Squeeze the bottle until water reaches the first mark in the vertical tube, and carefully read the position of the meniscus in the pressure tube. Release the bottle and immediately write down that position.
Repeat, being sure to again write down the position of the meniscus before squeezing the bottle (this position might or might not be the same as before) and the position of the meniscus when the water is at the first mark in the vertical tube.
Repeat three more times, so that you have a total of five trials in which the water was raised to the first mark in the vertical tube. With each repetition you will write down two more numbers.
Record your information below:
Indicate on the first line the vertical position of the first mark on the vertical tube, relative to the water surface, giving a single number in the first line.
On the second line give the length of the air column, as measured in units of the device you used to measure it.
On the third line, give the position of the meniscus before the first squeeze then the position of the meniscus when the water in the vertical tube was at the first mark. Give this information as two numbers, delimited by commas.
On lines four through seven, give the same information for the second through the fifth trials.
Starting in the eighth line give a brief synopsis of the meaning of the information you have given and how you obtained it.
----->>>>> vert pos mark vert tube, air column lgth, meniscus pos 1st trial, same 2d, same 3d, same 4th, same 5th trial, meaning
Your answer (start in the next line):
10 cm
36 cm
36 cm, 35 cm
35 cm, 33.8 cm
35 cm, 33.7 cm
35 cm, 33.7 cm
35 cm, 33.7 cm
Measurements are from the tape measure and measurements were taken as approximates based on the squeezing of the bottle.
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Now repeat the 5-trial process, this time raising water to the second mark. Write down everything as before.
In the space below report your results, using the same format as before:
----->>>>> same for 2d vert pos
Your answer (start in the next line):
20 cm
35 cm
35 cm, 33.5 cm
35 cm, 33.4 cm
35 cm, 33.4 cm
35 cm, 33.3 cm
35 cm, 33.4 cm
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Repeat again, raising water to the highest mark you can manage with normal effort. Remember that this isn't supposed to be a test of strength.
In the space below report your results, using the same format as before:
----->>>>> same for 3d pos
Your answer (start in the next line):
80 cm
35 cm
35 cm, 31 cm
35 cm, 31 cm
35 cm, 31.1 cm
35 cm, 31.1 cm
35 cm, 31.2 cm
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If the highest mark you can easily manage is the third mark, then you may stop. If you have raised the water to a mark higher than the third, then do one more series of 5 trials, this time choosing a mark about halfway between the second and the highest mark.
In the space below report your results, using the same format as before. If you were not able to raise the water higher than your third mark, simply leave these lines empty.
Then report the approximate percent change in the length of the water column for each of the three vertical heights. Report in a single line separated by commas, and in the last line indicate how you got these results, including a sample calculation for the second set of trials.
----->>>>> same for 4th pos if possible
Your answer (start in the next line):
Skip, because I jumped to the 80 cm mark on my third test because that was the highest mark I could reach and reasonably do the lab (i.e., take readings of the pressure gauge while squeezing the bottle.)
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Make your estimate of atmospheric pressure:
What was the maximum height to which the water column was raised?
How much pressure was required to support the column?
Based on the behavior of the air column in the pressure tube, what percent do you think this is of atmospheric pressure?
What therefore do you conclude is atmospheric pressure?
----->>>>> max ht, pressure to support column, percent of atm pressure, conclusion atm pressure
Your answer (start in the next line):
The max I raised it to take a reading was 80 cm, outside of taking measurements I attained somewhere in the ball park of 100 cm.
Not really sure how much pressure is required to support the column. The atmospheric pressure is 101 kPa. Not that that relates to this project in anyway.
Okay, maybe it does relate to this project. I think about 90% of it contributes to the pressure since I didnt add any additional pressure.
Atmospheric pressure in its simplest form is the pressure created by the atmosphere pressing down on things.
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@&
Note that atmospheric pressure doesn't just press down. It presses in a direction perpendicular to any surface.
The pressure results from gravity pulling down on the atmosphere above, but since air is slippery it can be 'squished' in any direction by a downward force, so it 'squishes' directly against any surface with which it is in contact.
*@
*#&!*#&!
@&
Check my note above.
You have good data and should get a good result on this experiment.
*@
#$&*
PHY 202
Your 'measuring atmospheric pressure' 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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Equipment:
If you don't have one you will need to obtain a 2-liter soft drink bottle.
The experiment was originally written for a rubber cork-shaped object called a stopper, with three this tubes protruding from both sides, and a 3-liter container.
There are also several miscellaneous pieces of tubing in your kit, and none should be discarded.
The stopper looked like this, and most of the pictures in these instructions show the stopper in a 3-liter tea container.
The bottlecap looks like this, and includes a short, a medium-length and a long tube.
The lab materials package, as of Spring 2010, actually consists of two such bottlecaps, one bottle cap at each end of the longest tube. Each cap has a short tube and a medium-length tube, inserted in the same manner as in the above picture. NOTE: It is possible that each cap has a single tube 'looped' through two of its holes. If this is the case, you can cut the tube about a centimeter above the top of the cap to form a short tube, leaving the rest to form a longer tube. After the cut you would have a short tube a few inches long extending through the cap (this will be the so-called 'pressure release' tube), a longer tube a couple of feet long (this will be the so-called 'pressure measuring tube'), and of course the tube connecting the two caps. Wait to make the cut until you need to do so; once you've seen the setup below you should understand.
The picture below shows the stopper in a tea container, with the tubes protruding. The short tube is 'capped' with a piece of 1/4-inch tubing, which is closed at one end by a plug of glue. The smaller tubing fits tightly into this 'cap', which seals off the end.
The picture below shows the system set up with a 3-liter bottle.
Initial Experiment
Notes about the tubing:
If necessary you can pull the tubing in one direction or the other, through the holes in the bottlecap. However the tubing should be at a usable length.
The apparatus has been pressure-tested against leakage.
It isn't recommended that you pull the tubing all the way out of the bottlecaps; it can be difficult to reinsert. Should it happen that the tubing does get pulled out (the tubing fits fairly tightly so it's unlikely to be pulled out by accident), it will probably require a little ingenuity and a pair of pliers to pull the end back through.
Short pieces of thin tubing, filled with glue, are included in your materials. These tubes do a good job of sealing the ends of the thicker tubes inserted through the cap. They are usually packed in the cylinder you used for the flow experiment. There are also pieces of thicker tubing, sealed at one end with glue, which can if the situation arises be used to seal the ends of the thinner tubing. Short unsealed tubing pieces of one diameter can be used to connect tubing of the other diameter (i.e., thin tubing can be used to connect two pieces of thicker tubing, or thicker tubing to connect two pieces of thinner tubing).
A 20-penny nail will also for a tight seal in a tube. Some packages may include some cut-off 20-penny nails. However nails tend to rust--no problem for the experiments, but who wants rusty nails around--so their use has probably been discontinued.
For an initial experiment, we are going see how to set up the system and apply force to raise water in a vertical tube.
The longest piece of tubing will be supported vertically above the bottle to make a 'vertical tube':
This 'vertical tube' should either be a few centimeters above, or should just reach, the bottom of the bottle when the cap is screwed on. The other tubes should extend just an inch or two into the container.
You will place about a liter of water in the bottom of the container. Then screw on the cap so that the low end of the vertical tube is submerged in the water.
The 'vertical tube' should be supported so that it is more or less vertical. A little sag or tilt away from vertical won't hurt, but get the tube as nearly vertical as possible without taking a lot of time to do so. The second cap will be near the top of this tube, and will not get in the way of anything (in fact the second cap should make it easier to find a way to support the tube in the vertical position).
The picture below shows the bottle with the vertical tube extending out the top. The tube is actually not all that vertical--you should try to do a little better, but if you can't it should be OK.
Mark heights on the tube and give the bottle a squeeze
Mark the tube or attach small pieces of tape to indicate the points that lie 30 above the level of water in the container, then 40 cm, then 50 cm, 60 cm, etc. to a height of at least 100 cm.
Be sure the caps are still attached to the ends of the other two tubes coming out of the stopper, so that air cannot enter or leave the container through these tubes.
You are going to squeeze the container and make water rise in the vertical tube. If you have reason to believe your hands aren't up to a hard squeeze, you should use a different means of compressing the container. Sitting on the floor and squeezing it between your feet and the wall is one possible alternative.
If you squeeze the container a bit you should see water rising in the tube.
Squeeze the container hard enough so that the water rises to the 30 cm level.
Now squeeze a little harder so that water rises to the 40 cm level.
This requires a significant amount of force. Most people can manage 40 cm, depending on hand size and strength and the shape of the container being used. Most people can't manage much over a meter, maybe two, though of course some people are much stronger than average and can do quite a bit more.
This isn't a test of strength, so stop before your face gets red, and well before you risk a hand or arm injury.
However, continue squeezing to achieve additional 10 cm increments of height in the tube, until water either reaches the top of the tube or you reach the reasonable limits of your ability to raise the water.
In the space below:
Indicate how you perceived the force necessary to raise the water to change with the height of the water column.
Do you think it takes twice the force to raise water twice as high?
Do you think it takes more, less, or the same additional force to raise water from 40 to 50 cm compared to the additional force required to raise it from 30 to 40 cm.
You are answering based on your perception rather than on measurements, and the perceptions of our senses are not generally linear.
Would it be possible to somehow measure the forces required?
If so, how might we do this?
Your answer (start in the next line):
Yes, it takes twice the force to raise the water twice as high. It seems pretty linear, meaning a 1 to 1 increase in water per force of squeeze on bottle.
I think it does take more additional force to raise the water from 40 to 70 cm compared to the additional force required to raise it from 30 to 40 cm.
Force equals mass times acceleration. I would almost think that if you measure the acceleration of the vertical movement of water in the tube and multiply it by the mass of the water then you could calculate the force applied to it?
#$&*
@&
That calculation would give you the additional force applied to the water in the tube, at the lower end of the tube.
However the water isn't accelerating much. A gradually increasing squeeze could, after a short initial period of acceleration, raise the water at constant velocity (until the end, when the squeeze would stop increasing so the system could reach a steady state).
The force you are exerting is, of course, not confined to the lower end of the tube. You exert a force spread out over a significant area of the bottle.
*@
Answer the following in the space below:
At the highest level you achieved, how much do you think the water in the tube weighed?
Do you think the weight of the water in the tube is greater, less than or equal to than the force you had to exert?
Do you think the comparison you make here is obvious? If so what makes you think so?
How might we measure the weight of the water and the force you exerted to make the comparison?
Your answer (start in the next line):
At the highest level achieved, I think the water in the tube weighed more than in the container.
@&
The water in the tube came from the container, and the level of water in the container didn't change much.
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The weight of the water in the tube was less than the force that I had exerted.
Yes, it was obvious in the fact that the force that applied was able to move the water column in the pressure tube.
Again, the force applied divided by the acceleration of the water in vertical tube would equal the mass of the water. I dont think the mass of the air in the bottle would affect the measurement because the vertical tube is submerged in the water.
@&
Per my preceding note, the water in the tube isn't accelerated very much or for very long. In addition, the negative acceleration at the end of the phase cancels the positive acceleration at the beginning. So acceleration does not explain very much of what you feel and observe.
*@
#$&*
If you think the force you exerted is different from the weight of the water, how could this be so? If you think they are the same, then why do you think it is so?
Your answer (start in the next line):
Because the water moved up when pressure was applied. The water did not force its way back down during my squeeze, it only went down when I decreased the force applied.
@&
You had to exert a lot more force than would be required to simply hold up the water column. How could this be?
*@
#$&*
Now set the system so that the tube comes out of the top of the stopper, makes a quick but smooth bend, runs horizontally a foot or so before making a quick but smooth bend to vertical. The tube in the picture below pretty much does this, but the horizontal run is a little curvy, not perfectly horizontal. The tube is hooked around the edge of the monitor and then runs more or less vertically, running out of the picture near the upper right.
You can set up the system, improvising with whatever resources you have handy. Supporting the horizontal run with a board or on a book or a coffee table shelf is one possibility.
The horizontal run doesn't have to be as long as the one shown here. 10 cm or so would be sufficient. The subsequent vertical run can also be as short as about 10 cm.
Once the system is set up, squeeze the container so water rises to the horizontal bend. Let the water stop before reaching the bend, then try to notice how much additional force seems to be required to move the water through the horizontal section, just up to the point where the tube again begins rising toward vertical.
Then continue squeezing as water once more begins rising vertically.
Compare the additional force required to run the water through the horizontal section with the additional force required when water starts rising vertically.
Once it's up to the horizontal section, is significant additional force required in order to move the water through the horizontal section?
Does it require significant additional force to then move the water through the subsequent vertical section?
Enter your answers in the space below:
Your answer (start in the next line):
No, it appeared to move horizontally much easier.
No, not a significant amount, but a little more than through the horizontal.
#$&*
Repeat. Adjust your squeeze so that water moves at a constant speed in the tube, moving through a few centimeters every second.
If you were to graph your force against time, which of the graphs below do you think would most accurately depict your actions?
Enter your answers in the space below. Include a description of the graph, the reasons you chose the graph you did and the reasons you rejected each of the others.
Your answer (start in the next line):
The first graph would most accurately depict my actions. More pressure gradually applied, not linearly, to push the water up in the vertical tube to create a pretty constant pressure through the horizontal, and then an increase through the vertical section.
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What do you think happens to the pressure of the gas in the bottle as you gradually raise the water?
Your answer (start in the next line):
The pressure would increase.
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Do you think it would take more pressure, less pressure, or about the same pressure in the bottle to raise water to a height of 50 cm in the first setup, where the tube was pretty much vertical, or in the second, where the tube had a horizontal 'run' before returning to vertical?
Your answer (start in the next line):
More pressure for the vertical.
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How much difference do you think there would be in the pressure required to raise water to the 50 cm level in a perfectly vertical tube, and the pressure required in a tube which runs upward but, say, a 10 or 20 degree angle with vertical? What difference would it make if the tube ran at 45 degrees from the vertical?
Your answer (start in the next line):
More pressure/force is required when at a 20 deg angle with vertical.
The 45 deg angle with vertical seemed to take less pressure than the 20 deg angle.
#$&*
What other means might you use to raise the pressure in the bottle?
Your answer (start in the next line):
By putting more pressure into the bottle by blowing in additional air.
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@&
That would work.
Heating the air in the bottle would also work.
*@
Now we're going to use the system to raise some water from its level in the container, to a higher position, thereby increasing the potential energy of the system. As you have already experienced, you're going to have to do some work to accomplish this.
Set up the system so that after a graceful bend the (formerly) vertical tube runs horizontally to the end. Place a container underneath the open end of the tube so that water runs into the container. The picture below shows the tube running out of the stopper, then running in a very nearly horizontal direction to the top of a graduated cylinder. It is not necessary to use the graduated cylinder to catch the water--you may use any container.
Increase the force of your squeeze (and hence the pressure of the gas) gradually and notice that once the water reaches the horizontal segment of the tube, very little extra force is required to move the water through the horizontal segment and maintain the flow.
Continue squeezing steadily until one or two cupfuls of water have been transferred to the container. Try to keep the flow of water slow and steady. Try to remember what the system feels like during the process, so you can answer the following questions:
Once the water begins flowing out, you will notice that you have the squeeze the bottle further and further to displace more and more water. The question is, do you have to exert more and more force to do this, or will a steady force accomplish the purpose?
You might have to repeat the process a couple of times before you are confident in your answer to this question. When you do, insert your answer in the space below:
Your answer (start in the next line):
I had to apply more force to get a stream.
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Now repeat, but this time try to make the water flow faster and faster into the container. Does it take more and more force to increase the speed of the flow?
Your answer (start in the next line):
Yes, it does.
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To what average height was the water raised, relative to the level in the bottle? This will be the difference between the average vertical position height of the water surface in the bottle and the vertical position of the end of the tube. It doesn't matter that the water fell back down after exiting the tube; if we had placed a container at the level of the tube, we could have caught it at that level.
Your answer (start in the next line):
Average vertical position height of water is 12 cm and the end of tube is 100 cm. The average height is 88 cm.
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How much potential energy gain was therefore accomplished, per cm^3 of water raised?
Your answer (start in the next line):
PE = mgh. The mass density would be 998 kg/m^3 = 0.998x10^-3 kg/cm^3.
About half a 2 liter bottle was filled with water, therefore, the volume of the water would be approximately 1000 cm^3. Mass = (0.998x10^-3 kg/cm^3)(1000 cm^3) = 0.998 kg.
PE = 0.998 kg * 981 cm/s^2 * 88 cm = 86 kJ. This does not sound correct at all.
@&
You are correct that this isn't right. It's easy to correct, though, since you did almost everything right.
kg * cm/s^2 * cm = kg cm^2 / s^2.
This is not equivalent to a Joules, which is a kg m^2 / s^2.
Also you based this on raising 1000 cm^3, whereas the question asked for the energy required to raise just one cm^3.
*@
#$&*
A cupful of water has a volume of about 250 cm^3 (very roughly). By how much would the potential energy of the system therefore be increased if you raised a cupful of water to the height of the outflow position?
Your answer (start in the next line):
Mass = (0.998x10^-3 kg/cm^3)(250 cm^3) = .250 kg
PE = .250 kg * 981 cm/s^2 * 88 cm = 22 kJ
#$&*
How much force would you have to exert to raise the water to this position, using a slow steady flow? Simply estimate the force in pounds, then convert to Newtons.
Through how much distance would your hands have to move, from the instant they touch the sides of the bottle?
Estimate the work they would therefore do, and compare to the potential energy increase. Which do you think would be greater, based on your experience with this system, and why?
Your answer (start in the next line):
I am guessing about 50 lbs force which would be equal to approximately 89 N.
@&
50 lbs would be closer to 220 N, but even though they aren't equivalent either 50 lbs or 89 N would be within the fairly broad range of reasonable estimates.
*@
Approximately 10 cm of distance that my hands would have to move from the instant they touch the side of the bottle.
The total work would be 500 J. Not as high as calculated in the question above. W = f * d.
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@&
The bottle is only about 10 cm in diameter, so I don't think you would have compressed it that much.
In any case 100 N through 10 cm would require only 10 Joules of work. Be careful of your units.
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*#&!
@&
You're doing well here, though units and a couple of misconceptions did lead you astray on certain parts of your analysis.
I am going to ask you for a revision on this experiment. I don't think you'll have much difficulty and I don't expect it to take you an inordinate amount of time. if it does, submit with question rather than getting bogged down.
`gr99
*@
#$&*
PHY 202
Your 'measuring atmospheric pressure' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Measuring Atmospheric Pressure_labelMessages **
Your setup for the preceding experiment Raising Water in a Vertical Tube included a vertical tube, with terminating caps on the other two tubes. You will use the vertical tube again in this experiment.
Your kit included two bottlecaps connected by a long tube. The long tube is to be used as a vertical tube, as in the previous experiment.
Each bottlecap has three tubes. One is a short tube; its intended use is to release pressure in the system when and if this becomes necessary. The third is fairly long. This tube is to be used as a 'pressure tube'.
First fill the 'pressure tube' with water. You can do this in any way you wish. One way:
The easiest way to do this is to temporarily disconnect the vertical tube and replace it with the new tube, so that when you squeeze the container you can fill the new tube. Add water to the container until it is nearly full, then screw on the bottlecap.
Hold the open end of the pressure-indicating tube a little higher than the top of the container, near the point where you just connected it, and squeeze the bottle so that water fills the tube. Since the water level in the container is higher than in the preceding experiment, and since the end of the new tube isn't much higher than the water level, this shouldn't require a very hard squeeze.
When the tube is full, maintain the squeeze so the water doesn't return to the container and disconnect it. You will have a tube full of water.
Now empty about half the water from the 'pressure tube'. Cap it and connect it to the system, and replace the vertical tube. You can do this in any way you wish, but one way is described below:
Just raise one end of the full 'pressure tube' and/or lower the other, and some water will flow out.
Once the tube is about half full, place a terminating cap on one end of this tube. This will hold the water in the 'pressure tube'.
You should at this point have:
The vertical tube, extended down into the water and out of the top of the container
The extended pressure-measuring tube, open on one end (through its connection to the newly opened tube in the stopper) to the air inside the container, half full of water, and capped at the other end.
A third tube short through the bottlecap, still closed off at the 'top end'.
In the picture below you see:
the short capped tube (the 'third' tube) sticking out of the top of the stopper,
the 'vertical' tube not yet in a vertical position but extending forward and to the right into a graduated cylinder, and
the pressure-indicating tube half full of caramel-colored liquid (the liquid is cola) and draped over a second graduated cylinder toward the back left. The pressure-indicating tube is capped at its end (hanging down near the tabletop), and the last 25 cm segment of the tube contains no liquid.
The picture below shows how the liquid in the tube comes to a point just below the 'peak' of the tube. This leaves an air column about 25 cm long in the capped end of the tube.
In the new picture the pressure-indicating tube is simply lying on the tabletop so the air column at the capped end is clearly visible.
The figure below shows a sketch of a tube which rises out of the bottle at left, then bends to form a U, then to the right of the U again levels off. The tube continues a ways to the right and is sealed at its right end. Liquid occupies the U up to almost the point of leveling, so that an increase in the pressure of the container will cause the liquid to move into the level region. As is the case in our experiment, the tube is assumed thin enough that the plane of the meniscus remains parallel to the cross-section of the tube (i.e., the meniscus doesn't 'level off' when it moves into a horizontal section of tube).
You should manipulate the pressure tube until its configuration resembles the one shown. The length and depth of the U can vary from that depicted, but the air column at the end of the tube should be at least 15 (actual) cm long. The liquid levels at the left and right ends of U should be very nearly equal.
The basic idea is that as you squeeze the system to raise water in the vertical tube, as in your previous experiment, the pressure in the system increases and compresses the 'air column' in the pressure tube. By measuring the lengths of this 'air column' you can determine relative pressures, and by measuring the heights of the water column in the 'vertical tube' you can determine the actual pressure differences required to support those columns.
Support the end of the vertical tube so that it is more or less vertical, as it was in the previous experiment.
The bottle should be pretty full, but not so full that it covers the open end of the tube to which the pressure tube is connected; the left end of the pressure tube should have an 'open path' to the gas inside the bottle, so that the pressure on the left-hand side of the water column in that tube is essentially equal to the pressure in the bottle.
If you squeeze the container a little, water will rise a little way in the vertical tube and the water in the pressure tube will also move is such a way as to slightly shorten the air column. The harder you squeeze the higher water will rise in the vertical tube and the shorter the air column will become.
Go ahead and observe this phenomenon. There is no need to measure anything yet, just get the 'feel' of the system.
Indicate below how the system behaves (what changes when you do what, how the system's reactions to your actions appear to be related to one another) and how it 'feels'.
----->>>>> behavior
Your answer (start in the next line):
When I squeeze, the water goes up in the vertical tube and moves the air column in the pressure tube.
#$&*
Using a measuring device you will measure the relative positions of the meniscus as you vary your squeeze:
One of the ruler copies used in the previous experiment on the distortion of paper rulers should be used here; a reduced copy should be used for greater precision. You may choose the level of reduction at which you think you will achieve the greatest level of precision. Only relative measurements will be important here; it will not be necessary to convert your units to actual millimeters or centimeters.
Indicate below the level of reduction you have chosen, and your reasons for this choice.
----->>>>> level of reduction and reasons
Your answer (start in the next line):
No level of reduction was used, a sewing measuring tape was used for my experiment.
#$&*
In the units of the measuring device you have chosen, write down in your lab notebook the readings you used to indicate length of the air column, from the meniscus to the barrier at the capped end. No conversion of the units of your device to standard units (e.g., millimeters or centimeters) is required. Your information should include the marking at one end of the measuring device, and the marking at the other. If necessary two or more copies of paper rulers may be carefully taped together.
Indicate in the first line below the length of the air column in the units of your measuring device.
In the second line explain how you obtained your result, including the readings at the two ends and how you used those readings to indicate the length.
----->>>>> air column length, how obtained incl readings and how used
Your answer (start in the next line):
The air column length is approximately 36 cm. The tape measure was taped to the pressure tube.
#$&*
Now place the same measuring device along the tube, positioned so you can observe as accurately as possible the relative positions of the meniscus in the pressure tube.
It is recommended that the initial position of the meniscus be in the vicinity of the center of the measuring device, so that position changes in both directions can be observed.
It is not necessary for the measuring device to extend the entire length of the air column, as long as you know the reading on the measuring device that corresponds to the initial position of the meniscus. From this information and from subsequent readings it will be easy to determine the varying lengths of the air column.
Take whatever precautions are necessary to make sure neither the measuring device nor the pressure tube can move until you have completed the necessary trials.
Mark positions along the vertical tube at 10-cm intervals (actual 10-cm intervals as indicated by a full-sized ruler) above the surface of the water in the bottle.
If the bottle is pretty full, as described before, it might be possible to make the first mark on the vertical tube at 10 or 15 cm above the water surface.
Marks may be made using an actual marker, or pieces of tape, or anything else that happens to be convenient.
Write your information in your lab notebook:
Write down the position of the first mark on the vertical tube with respect to the water surface (e.g., 10 cm or 15 cm).
Write down the position of the meniscus in the pressure tube. This position will simply be the reading on your measuring device. For example if the meniscus is at marking 17.35, that is what you write down.
As in all labs, you directly record what you read. Never do any arithmetic between making an observation and recording it.
You will now conduct 5 trials, raising water to the first mark on your vertical tube and reading the position of the meniscus before the squeeze and while water is at the given level.
Squeeze the bottle until water reaches the first mark in the vertical tube, and carefully read the position of the meniscus in the pressure tube. Release the bottle and immediately write down that position.
Repeat, being sure to again write down the position of the meniscus before squeezing the bottle (this position might or might not be the same as before) and the position of the meniscus when the water is at the first mark in the vertical tube.
Repeat three more times, so that you have a total of five trials in which the water was raised to the first mark in the vertical tube. With each repetition you will write down two more numbers.
Record your information below:
Indicate on the first line the vertical position of the first mark on the vertical tube, relative to the water surface, giving a single number in the first line.
On the second line give the length of the air column, as measured in units of the device you used to measure it.
On the third line, give the position of the meniscus before the first squeeze then the position of the meniscus when the water in the vertical tube was at the first mark. Give this information as two numbers, delimited by commas.
On lines four through seven, give the same information for the second through the fifth trials.
Starting in the eighth line give a brief synopsis of the meaning of the information you have given and how you obtained it.
----->>>>> vert pos mark vert tube, air column lgth, meniscus pos 1st trial, same 2d, same 3d, same 4th, same 5th trial, meaning
Your answer (start in the next line):
10 cm
36 cm
36 cm, 35 cm
35 cm, 33.8 cm
35 cm, 33.7 cm
35 cm, 33.7 cm
35 cm, 33.7 cm
Measurements are from the tape measure and measurements were taken as approximates based on the squeezing of the bottle.
#$&*
Now repeat the 5-trial process, this time raising water to the second mark. Write down everything as before.
In the space below report your results, using the same format as before:
----->>>>> same for 2d vert pos
Your answer (start in the next line):
20 cm
35 cm
35 cm, 33.5 cm
35 cm, 33.4 cm
35 cm, 33.4 cm
35 cm, 33.3 cm
35 cm, 33.4 cm
#$&*
Repeat again, raising water to the highest mark you can manage with normal effort. Remember that this isn't supposed to be a test of strength.
In the space below report your results, using the same format as before:
----->>>>> same for 3d pos
Your answer (start in the next line):
80 cm
35 cm
35 cm, 31 cm
35 cm, 31 cm
35 cm, 31.1 cm
35 cm, 31.1 cm
35 cm, 31.2 cm
#$&*
If the highest mark you can easily manage is the third mark, then you may stop. If you have raised the water to a mark higher than the third, then do one more series of 5 trials, this time choosing a mark about halfway between the second and the highest mark.
In the space below report your results, using the same format as before. If you were not able to raise the water higher than your third mark, simply leave these lines empty.
Then report the approximate percent change in the length of the water column for each of the three vertical heights. Report in a single line separated by commas, and in the last line indicate how you got these results, including a sample calculation for the second set of trials.
----->>>>> same for 4th pos if possible
Your answer (start in the next line):
Skip, because I jumped to the 80 cm mark on my third test because that was the highest mark I could reach and reasonably do the lab (i.e., take readings of the pressure gauge while squeezing the bottle.)
#$&*
Make your estimate of atmospheric pressure:
What was the maximum height to which the water column was raised?
How much pressure was required to support the column?
Based on the behavior of the air column in the pressure tube, what percent do you think this is of atmospheric pressure?
What therefore do you conclude is atmospheric pressure?
----->>>>> max ht, pressure to support column, percent of atm pressure, conclusion atm pressure
Your answer (start in the next line):
The max I raised it to take a reading was 80 cm, outside of taking measurements I attained somewhere in the ball park of 100 cm.
Not really sure how much pressure is required to support the column. The atmospheric pressure is 101 kPa. Not that that relates to this project in anyway.
Okay, maybe it does relate to this project. I think about 90% of it contributes to the pressure since I didnt add any additional pressure.
Atmospheric pressure in its simplest form is the pressure created by the atmosphere pressing down on things.
#$&*
@&
Note that atmospheric pressure doesn't just press down. It presses in a direction perpendicular to any surface.
The pressure results from gravity pulling down on the atmosphere above, but since air is slippery it can be 'squished' in any direction by a downward force, so it 'squishes' directly against any surface with which it is in contact.
*@
*#&!*#&!
@&
Check my note above.
You have good data and should get a good result on this experiment.
*@
#$&*
PHY 202
Your 'measuring atmospheric pressure' 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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Equipment:
If you don't have one you will need to obtain a 2-liter soft drink bottle.
The experiment was originally written for a rubber cork-shaped object called a stopper, with three this tubes protruding from both sides, and a 3-liter container.
There are also several miscellaneous pieces of tubing in your kit, and none should be discarded.
The stopper looked like this, and most of the pictures in these instructions show the stopper in a 3-liter tea container.
The bottlecap looks like this, and includes a short, a medium-length and a long tube.
The lab materials package, as of Spring 2010, actually consists of two such bottlecaps, one bottle cap at each end of the longest tube. Each cap has a short tube and a medium-length tube, inserted in the same manner as in the above picture. NOTE: It is possible that each cap has a single tube 'looped' through two of its holes. If this is the case, you can cut the tube about a centimeter above the top of the cap to form a short tube, leaving the rest to form a longer tube. After the cut you would have a short tube a few inches long extending through the cap (this will be the so-called 'pressure release' tube), a longer tube a couple of feet long (this will be the so-called 'pressure measuring tube'), and of course the tube connecting the two caps. Wait to make the cut until you need to do so; once you've seen the setup below you should understand.
The picture below shows the stopper in a tea container, with the tubes protruding. The short tube is 'capped' with a piece of 1/4-inch tubing, which is closed at one end by a plug of glue. The smaller tubing fits tightly into this 'cap', which seals off the end.
The picture below shows the system set up with a 3-liter bottle.
Initial Experiment
Notes about the tubing:
If necessary you can pull the tubing in one direction or the other, through the holes in the bottlecap. However the tubing should be at a usable length.
The apparatus has been pressure-tested against leakage.
It isn't recommended that you pull the tubing all the way out of the bottlecaps; it can be difficult to reinsert. Should it happen that the tubing does get pulled out (the tubing fits fairly tightly so it's unlikely to be pulled out by accident), it will probably require a little ingenuity and a pair of pliers to pull the end back through.
Short pieces of thin tubing, filled with glue, are included in your materials. These tubes do a good job of sealing the ends of the thicker tubes inserted through the cap. They are usually packed in the cylinder you used for the flow experiment. There are also pieces of thicker tubing, sealed at one end with glue, which can if the situation arises be used to seal the ends of the thinner tubing. Short unsealed tubing pieces of one diameter can be used to connect tubing of the other diameter (i.e., thin tubing can be used to connect two pieces of thicker tubing, or thicker tubing to connect two pieces of thinner tubing).
A 20-penny nail will also for a tight seal in a tube. Some packages may include some cut-off 20-penny nails. However nails tend to rust--no problem for the experiments, but who wants rusty nails around--so their use has probably been discontinued.
For an initial experiment, we are going see how to set up the system and apply force to raise water in a vertical tube.
The longest piece of tubing will be supported vertically above the bottle to make a 'vertical tube':
This 'vertical tube' should either be a few centimeters above, or should just reach, the bottom of the bottle when the cap is screwed on. The other tubes should extend just an inch or two into the container.
You will place about a liter of water in the bottom of the container. Then screw on the cap so that the low end of the vertical tube is submerged in the water.
The 'vertical tube' should be supported so that it is more or less vertical. A little sag or tilt away from vertical won't hurt, but get the tube as nearly vertical as possible without taking a lot of time to do so. The second cap will be near the top of this tube, and will not get in the way of anything (in fact the second cap should make it easier to find a way to support the tube in the vertical position).
The picture below shows the bottle with the vertical tube extending out the top. The tube is actually not all that vertical--you should try to do a little better, but if you can't it should be OK.
Mark heights on the tube and give the bottle a squeeze
Mark the tube or attach small pieces of tape to indicate the points that lie 30 above the level of water in the container, then 40 cm, then 50 cm, 60 cm, etc. to a height of at least 100 cm.
Be sure the caps are still attached to the ends of the other two tubes coming out of the stopper, so that air cannot enter or leave the container through these tubes.
You are going to squeeze the container and make water rise in the vertical tube. If you have reason to believe your hands aren't up to a hard squeeze, you should use a different means of compressing the container. Sitting on the floor and squeezing it between your feet and the wall is one possible alternative.
If you squeeze the container a bit you should see water rising in the tube.
Squeeze the container hard enough so that the water rises to the 30 cm level.
Now squeeze a little harder so that water rises to the 40 cm level.
This requires a significant amount of force. Most people can manage 40 cm, depending on hand size and strength and the shape of the container being used. Most people can't manage much over a meter, maybe two, though of course some people are much stronger than average and can do quite a bit more.
This isn't a test of strength, so stop before your face gets red, and well before you risk a hand or arm injury.
However, continue squeezing to achieve additional 10 cm increments of height in the tube, until water either reaches the top of the tube or you reach the reasonable limits of your ability to raise the water.
In the space below:
Indicate how you perceived the force necessary to raise the water to change with the height of the water column.
Do you think it takes twice the force to raise water twice as high?
Do you think it takes more, less, or the same additional force to raise water from 40 to 50 cm compared to the additional force required to raise it from 30 to 40 cm.
You are answering based on your perception rather than on measurements, and the perceptions of our senses are not generally linear.
Would it be possible to somehow measure the forces required?
If so, how might we do this?
Your answer (start in the next line):
Yes, it takes twice the force to raise the water twice as high. It seems pretty linear, meaning a 1 to 1 increase in water per force of squeeze on bottle.
I think it does take more additional force to raise the water from 40 to 70 cm compared to the additional force required to raise it from 30 to 40 cm.
Force equals mass times acceleration. I would almost think that if you measure the acceleration of the vertical movement of water in the tube and multiply it by the mass of the water then you could calculate the force applied to it?
#$&*
@&
That calculation would give you the additional force applied to the water in the tube, at the lower end of the tube.
However the water isn't accelerating much. A gradually increasing squeeze could, after a short initial period of acceleration, raise the water at constant velocity (until the end, when the squeeze would stop increasing so the system could reach a steady state).
The force you are exerting is, of course, not confined to the lower end of the tube. You exert a force spread out over a significant area of the bottle.
*@
Answer the following in the space below:
At the highest level you achieved, how much do you think the water in the tube weighed?
Do you think the weight of the water in the tube is greater, less than or equal to than the force you had to exert?
Do you think the comparison you make here is obvious? If so what makes you think so?
How might we measure the weight of the water and the force you exerted to make the comparison?
Your answer (start in the next line):
At the highest level achieved, I think the water in the tube weighed more than in the container.
@&
The water in the tube came from the container, and the level of water in the container didn't change much.
*@
The weight of the water in the tube was less than the force that I had exerted.
Yes, it was obvious in the fact that the force that applied was able to move the water column in the pressure tube.
Again, the force applied divided by the acceleration of the water in vertical tube would equal the mass of the water. I dont think the mass of the air in the bottle would affect the measurement because the vertical tube is submerged in the water.
@&
Per my preceding note, the water in the tube isn't accelerated very much or for very long. In addition, the negative acceleration at the end of the phase cancels the positive acceleration at the beginning. So acceleration does not explain very much of what you feel and observe.
*@
#$&*
If you think the force you exerted is different from the weight of the water, how could this be so? If you think they are the same, then why do you think it is so?
Your answer (start in the next line):
Because the water moved up when pressure was applied. The water did not force its way back down during my squeeze, it only went down when I decreased the force applied.
@&
You had to exert a lot more force than would be required to simply hold up the water column. How could this be?
*@
#$&*
Now set the system so that the tube comes out of the top of the stopper, makes a quick but smooth bend, runs horizontally a foot or so before making a quick but smooth bend to vertical. The tube in the picture below pretty much does this, but the horizontal run is a little curvy, not perfectly horizontal. The tube is hooked around the edge of the monitor and then runs more or less vertically, running out of the picture near the upper right.
You can set up the system, improvising with whatever resources you have handy. Supporting the horizontal run with a board or on a book or a coffee table shelf is one possibility.
The horizontal run doesn't have to be as long as the one shown here. 10 cm or so would be sufficient. The subsequent vertical run can also be as short as about 10 cm.
Once the system is set up, squeeze the container so water rises to the horizontal bend. Let the water stop before reaching the bend, then try to notice how much additional force seems to be required to move the water through the horizontal section, just up to the point where the tube again begins rising toward vertical.
Then continue squeezing as water once more begins rising vertically.
Compare the additional force required to run the water through the horizontal section with the additional force required when water starts rising vertically.
Once it's up to the horizontal section, is significant additional force required in order to move the water through the horizontal section?
Does it require significant additional force to then move the water through the subsequent vertical section?
Enter your answers in the space below:
Your answer (start in the next line):
No, it appeared to move horizontally much easier.
No, not a significant amount, but a little more than through the horizontal.
#$&*
Repeat. Adjust your squeeze so that water moves at a constant speed in the tube, moving through a few centimeters every second.
If you were to graph your force against time, which of the graphs below do you think would most accurately depict your actions?
Enter your answers in the space below. Include a description of the graph, the reasons you chose the graph you did and the reasons you rejected each of the others.
Your answer (start in the next line):
The first graph would most accurately depict my actions. More pressure gradually applied, not linearly, to push the water up in the vertical tube to create a pretty constant pressure through the horizontal, and then an increase through the vertical section.
#$&*
What do you think happens to the pressure of the gas in the bottle as you gradually raise the water?
Your answer (start in the next line):
The pressure would increase.
#$&*
Do you think it would take more pressure, less pressure, or about the same pressure in the bottle to raise water to a height of 50 cm in the first setup, where the tube was pretty much vertical, or in the second, where the tube had a horizontal 'run' before returning to vertical?
Your answer (start in the next line):
More pressure for the vertical.
#$&*
How much difference do you think there would be in the pressure required to raise water to the 50 cm level in a perfectly vertical tube, and the pressure required in a tube which runs upward but, say, a 10 or 20 degree angle with vertical? What difference would it make if the tube ran at 45 degrees from the vertical?
Your answer (start in the next line):
More pressure/force is required when at a 20 deg angle with vertical.
The 45 deg angle with vertical seemed to take less pressure than the 20 deg angle.
#$&*
What other means might you use to raise the pressure in the bottle?
Your answer (start in the next line):
By putting more pressure into the bottle by blowing in additional air.
#$&*
@&
That would work.
Heating the air in the bottle would also work.
*@
Now we're going to use the system to raise some water from its level in the container, to a higher position, thereby increasing the potential energy of the system. As you have already experienced, you're going to have to do some work to accomplish this.
Set up the system so that after a graceful bend the (formerly) vertical tube runs horizontally to the end. Place a container underneath the open end of the tube so that water runs into the container. The picture below shows the tube running out of the stopper, then running in a very nearly horizontal direction to the top of a graduated cylinder. It is not necessary to use the graduated cylinder to catch the water--you may use any container.
Increase the force of your squeeze (and hence the pressure of the gas) gradually and notice that once the water reaches the horizontal segment of the tube, very little extra force is required to move the water through the horizontal segment and maintain the flow.
Continue squeezing steadily until one or two cupfuls of water have been transferred to the container. Try to keep the flow of water slow and steady. Try to remember what the system feels like during the process, so you can answer the following questions:
Once the water begins flowing out, you will notice that you have the squeeze the bottle further and further to displace more and more water. The question is, do you have to exert more and more force to do this, or will a steady force accomplish the purpose?
You might have to repeat the process a couple of times before you are confident in your answer to this question. When you do, insert your answer in the space below:
Your answer (start in the next line):
I had to apply more force to get a stream.
#$&*
Now repeat, but this time try to make the water flow faster and faster into the container. Does it take more and more force to increase the speed of the flow?
Your answer (start in the next line):
Yes, it does.
#$&*
To what average height was the water raised, relative to the level in the bottle? This will be the difference between the average vertical position height of the water surface in the bottle and the vertical position of the end of the tube. It doesn't matter that the water fell back down after exiting the tube; if we had placed a container at the level of the tube, we could have caught it at that level.
Your answer (start in the next line):
Average vertical position height of water is 12 cm and the end of tube is 100 cm. The average height is 88 cm.
#$&*
How much potential energy gain was therefore accomplished, per cm^3 of water raised?
Your answer (start in the next line):
PE = mgh. The mass density would be 998 kg/m^3 = 0.998x10^-3 kg/cm^3.
About half a 2 liter bottle was filled with water, therefore, the volume of the water would be approximately 1000 cm^3. Mass = (0.998x10^-3 kg/cm^3)(1000 cm^3) = 0.998 kg.
PE = 0.998 kg * 981 cm/s^2 * 88 cm = 86 kJ. This does not sound correct at all.
@&
You are correct that this isn't right. It's easy to correct, though, since you did almost everything right.
kg * cm/s^2 * cm = kg cm^2 / s^2.
This is not equivalent to a Joules, which is a kg m^2 / s^2.
Also you based this on raising 1000 cm^3, whereas the question asked for the energy required to raise just one cm^3.
*@
#$&*
A cupful of water has a volume of about 250 cm^3 (very roughly). By how much would the potential energy of the system therefore be increased if you raised a cupful of water to the height of the outflow position?
Your answer (start in the next line):
Mass = (0.998x10^-3 kg/cm^3)(250 cm^3) = .250 kg
PE = .250 kg * 981 cm/s^2 * 88 cm = 22 kJ
#$&*
How much force would you have to exert to raise the water to this position, using a slow steady flow? Simply estimate the force in pounds, then convert to Newtons.
Through how much distance would your hands have to move, from the instant they touch the sides of the bottle?
Estimate the work they would therefore do, and compare to the potential energy increase. Which do you think would be greater, based on your experience with this system, and why?
Your answer (start in the next line):
I am guessing about 50 lbs force which would be equal to approximately 89 N.
@&
50 lbs would be closer to 220 N, but even though they aren't equivalent either 50 lbs or 89 N would be within the fairly broad range of reasonable estimates.
*@
Approximately 10 cm of distance that my hands would have to move from the instant they touch the side of the bottle.
The total work would be 500 J. Not as high as calculated in the question above. W = f * d.
#$&*
@&
The bottle is only about 10 cm in diameter, so I don't think you would have compressed it that much.
In any case 100 N through 10 cm would require only 10 Joules of work. Be careful of your units.
*@
*#&!
@&
You're doing well here, though units and a couple of misconceptions did lead you astray on certain parts of your analysis.
I am going to ask you for a revision on this experiment. I don't think you'll have much difficulty and I don't expect it to take you an inordinate amount of time. if it does, submit with question rather than getting bogged down.
&#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. &#
*@