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United States24670 Posts
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+ Show Spoiler +The can will be moving in the same direction as the side in which the hole was punctured. So it will be moving right (because it was punctured on the right side). This is because the vacuum must be filled, and air rushes in to fill it. However, the can also moves because it is not fixed, and moves in the direction that will aid in the air filling the vacuum. Wow that explanation was terrible.
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United States24670 Posts
+ Show Spoiler +On September 27 2008 10:24 ieatkids5 wrote:
The can will be moving in the same direction as the side in which the hole was punctured. So it will be moving right. This is because the vacuum must be filled, and air rushes in to fill it. However, the can also moves because it is not fixed, and moves in the direction that will aid in the air filling the vacuum. Wow that explanation was terrible.
Please spoiler it so people reading don't notice your choice.
edit: ty
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Belgium6771 Posts
+ Show Spoiler +C
it doesnt actually matter where the hole is punctured
a vacuum will just fill the void with the air it gets and perform force in every direction of the inside of the can, not just one side
...in my sleepy mind
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+ Show Spoiler +a) be moving to the left ya ya? It's like the can grew a mouth, and the air like a thin sausage. As the mouth eat the sausage it will move toward it. Or hmm let me see. Can we consider the air and the can as a system? The center of mass of that system does not change, while the mass of air is dispaced to the right into the can, the mass of can must dispace to the left.
edit: added picture
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On September 27 2008 10:27 Xeofreestyler wrote:+ Show Spoiler +C
it doesnt actually matter where the hole is punctured
a vacuum will just fill the void with the air it gets and perform force in every direction of the inside of the can, not just one side
...in my sleepy mind
I just want to say that I agree with this
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Calgary25980 Posts
+ Show Spoiler + The only part that bothers me is "After the vacuum is filled the can will". The can moves to the left, once the system is balanced, no motion will be taking place. We can assume the rate of air entry tapers off, so can's velocity must slow down as the can fills up. I assume at sometime before steady-state is achieved, friction is stronger than the air force, meaning the can is no longer moving.
C
Edit: The answers state "be moving", not "have moved", which is why I believe it's C.
I also disagree with your answer to question 1 due to the fact we can assume the dog is not running while picking up the stick, meaning we would want to throw it forward to maximize the amount of times the dog has to return. I would imagine you would give full credit to a student who explained their answer this way?
Edit: Whoops you wanted to maximize running! Throw them shits backwards son!
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United States24670 Posts
On September 27 2008 10:42 Chill wrote:
I also disagree with your answer to question 1 due to the fact we can assume the dog is not running while picking up the stick, meaning we would want to throw it forward to maximize the amount of times the dog has to return. I would imagine you would give full credit to a student who explained their answer this way?
Edit: Whoops you wanted to maximize running! Throw them shits backwards son!
Why are you assuming that the dog is not running while picking up the stick? There are 1001 ways you can impose your own minor restrictions that change the answer, but you are not using the most simple/obvious interpretation of the question. Also, I would not give a question like this to my class tbh. If I did, it wouldn't be graded for correctness so much as effort, and I suppose you would earn credit in that case.
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+ Show Spoiler +B? It seems to make sense as the air moves in the can moves in the opposite direction. Once it is filled up, it will keep moving in that direction by inertia.
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Closed system: air and can. Center of mass must remain in the same place and air moves "in the left" you say. If this means into the left side of the can, then the can must move to the left as the air is moving to the right. If this means towards the left (i.e. into the right side of the can) then the can moves to the right so that the center of mass is conserved.
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Calgary25980 Posts
On September 27 2008 11:06 micronesia wrote:Show nested quote +On September 27 2008 10:42 Chill wrote:
I also disagree with your answer to question 1 due to the fact we can assume the dog is not running while picking up the stick, meaning we would want to throw it forward to maximize the amount of times the dog has to return. I would imagine you would give full credit to a student who explained their answer this way?
Edit: Whoops you wanted to maximize running! Throw them shits backwards son! Why are you assuming that the dog is not running while picking up the stick? There are 1001 ways you can impose your own minor restrictions that change the answer, but you are not using the most simple/obvious interpretation of the question. Also, I would not give a question like this to my class tbh. If I did, it wouldn't be graded for correctness so much as effort, and I suppose you would earn credit in that case.
Because what kind of dog picks something up while running? Shouldn't common sense be the obvious restriction?
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+ Show Spoiler +b. the puncture of the can will reduce the air pressure on the left side. there will momentarily be a higher pressure on the right side of the can forcing the can left. As the air fills the can it will push on the inside right wall, but it will still be of a lower pressure than the outside, so the can will continue to accelerate left until the pressure equalises inside the can. at this moment, the can will still be traveling to the left.
I also think it is reasonable to assume that taking and throwing the stick does not take 0 time, and in that time, the dog will be waiting for the stick to be thrown. physics gets annoying when people rely on nitpicking - here is the above with some common-sense assumptions written out in full, so that they cannot be misinterpreted, see how it is a bit unnecessary.
+ Show Spoiler +b. the puncture of the can will reduce the air pressure on the left side. there will momentarily be a higher pressure on the right side of the can forcing the can left. (assuming outside the can is not of near vacuum, such that the amount of air involved is one the number of molcules scale) As the air fills the can, it will push on the inside right wall, (assuming the walls of the can do not react with the air in a chemical reaction, so as to consume the air) but it will still be of a lower pressure than the outside, so the can will continue to accelerate left until the pressure equalises inside the can. at this moment (assuming low enough air resistance so as not to slow the can appreciably, but of high enough resistance and viscosity such that the higher pressure of air gains purchase on the can as is assumed in the rocket can example), the can will still be travelling to the left. all this is assuming that the pressure gradient caused by the intake of air is greater than that which is created in the opposite direction by the can moving to the left (like a bow-wave). this is assuming the can is in an open space of isotropic pressure which maitains a global pressure value but not a local one to begin with and by "air fills the can" you mean that it is at the same pressure as this global air pressure. the moving can will eventually stop due to air resistance, and by the act that as it moves, the pressure in the can will get higher due to its movement forward with a hole at the front, and that the body of air in the can will be moving in the opposite direction. eventually, the can will stop due to this.
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United States24670 Posts
On September 27 2008 11:55 Chill wrote:Show nested quote +On September 27 2008 11:06 micronesia wrote:On September 27 2008 10:42 Chill wrote:
I also disagree with your answer to question 1 due to the fact we can assume the dog is not running while picking up the stick, meaning we would want to throw it forward to maximize the amount of times the dog has to return. I would imagine you would give full credit to a student who explained their answer this way?
Edit: Whoops you wanted to maximize running! Throw them shits backwards son! Why are you assuming that the dog is not running while picking up the stick? There are 1001 ways you can impose your own minor restrictions that change the answer, but you are not using the most simple/obvious interpretation of the question. Also, I would not give a question like this to my class tbh. If I did, it wouldn't be graded for correctness so much as effort, and I suppose you would earn credit in that case. Because what kind of dog picks something up while running? Shouldn't common sense be the obvious restriction?
At this level, it becomes impossible for there to be a 'correct' answer to any question like this :-/
The more physics problems you do, the more obvious it becomes which types of unstated restrictions are assumed by the question writer, and which are not even being considered. The example you are coming up with is a restriction where the question writer would specifically state that time was spent picking up the stick each time.
How do you prove who is right when arguing about what should or shouldn't be considered, giving the wording of a physics problem? Unfortunately, the best, or perhaps least bad way to do it is to interview a bunch of physicists and look for a consensus. Also I didn't make up the answer... taken straight from a published physicist.
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I just want to say that even though I am not answering the questions and providing reasons, I am following your blog and I appreciate you doing this. Keep up the questions micronesia, you are great! 
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The best things in life are free, and being punctured isn't the best thing that can happen to you, so the can owes someone money.
+ Show Spoiler +
I want to consider that most cans can't take that kind of pressure and would just implode.
Also, the problem's wording is poor. Air rushes in the left. Into the can from the left side? Into the can, moving in the direction "left"? Typically, such wording is correlated to the wording from the example given, but in this case the wording is "blows to the right" compared to "blows in the right." "To" and "in" could both be relative to the can or the observer, so that doesn't help much.
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On September 27 2008 13:27 BottleAbuser wrote:The best things in life are free, and being punctured isn't the best thing that can happen to you, so the can owes someone money. + Show Spoiler +
I want to consider that most cans can't take that kind of pressure and would just implode.
Also, the problem's wording is poor. Air rushes in the left. Into the can from the left side? Into the can, moving in the direction "left"? Typically, such wording is correlated to the wording from the example given, but in this case the wording is "blows to the right" compared to "blows in the right." "To" and "in" could both be relative to the can or the observer, so that doesn't help much.
read more physics >_> I dare say you're very not familiar with how physics is run conventionally so the confusion is understandible.
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My first interpretation is that the air is moving in the left direction, not entering the can's left side, and my intuition tells me it is what was meant. But if I was grading this, and someone was arguing with me about the wording, I'd fold. Dunno about Micronesia.
I'm just pointing out that there is real ambiguity here. I prefer unambiguous notation, and if you must use a natural language, be sure not to use unambiguous terms at all. In particular, the word "in" when used around containers is somewhat dangerous, if you do not actually refer to the inside of the container.
+ Show Spoiler +Since the wording is "after the can is filled," the can will no longer move. Consider that for it to be moving, some air somewhere must be moving also. The air inside the can would be moving in the same direction at the same speed, so that wouldn't help.
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Yeah I know that you're pointing out there is a real ambiguity. What I'm saying is if you deal with physics on a regular basis there should be very little ambiguity in this question. I find this question perfectly worded and I'm sure my fellow physics friends would agree too.
For comparison, imagine now I tell you PvZ you can go 10/12 gate. Now this is perfectly clear what I mean right? But somebody who's not playing starcraft very often might be confused as to "Oh you build 10 gateways over 12 gateways ? what's going on?". That does not mean my wording is ambiguous, that just mean that person, (in this case, you), is not familiar with the conventions of the subject
But lemme reword it so maybe it'll help, because convention shouldn't held people back from thinking recreatively:
Can is vacuumed out.
You poke a hole on the left of the can
Obviously, since the can is vacuum, it'll start sucking in air from the left side cuz there's a hole on the left side.
Which way does the can move as a result?
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Much clearer.
+ Show Spoiler +However, I'm not confident that it's the correct interpretation. Either way, the answer does not change, as we're interested in velocity of the can when it has been filled, ie no longer moving.
By the way Chill, you don't need friction. The momentum of air is enough.
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So if the hole is in the left side, my answer would be reversed 
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On September 27 2008 12:38 micronesia wrote:Show nested quote +On September 27 2008 11:55 Chill wrote:On September 27 2008 11:06 micronesia wrote:On September 27 2008 10:42 Chill wrote:
I also disagree with your answer to question 1 due to the fact we can assume the dog is not running while picking up the stick, meaning we would want to throw it forward to maximize the amount of times the dog has to return. I would imagine you would give full credit to a student who explained their answer this way?
Edit: Whoops you wanted to maximize running! Throw them shits backwards son! Why are you assuming that the dog is not running while picking up the stick? There are 1001 ways you can impose your own minor restrictions that change the answer, but you are not using the most simple/obvious interpretation of the question. Also, I would not give a question like this to my class tbh. If I did, it wouldn't be graded for correctness so much as effort, and I suppose you would earn credit in that case. Because what kind of dog picks something up while running? Shouldn't common sense be the obvious restriction? At this level, it becomes impossible for there to be a 'correct' answer to any question like this :-/
Nah, Chill's right. Maybe he should be the physics teacher.
Stick to bowling and silly ads! :p
Btw the real answer is that you could just pretend to throw the stick and yubee would just keep running forever.
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don't u tell Micronesia what to do ;(
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+ Show Spoiler +
My answer is C, even if gas is ideal and there is no friction. The only moving parts of this system are can and gas that blows inside the can and eventually they are forced to move in one direction. So the only mean to fulfill momentum conservation law is to assume that they will not move at all. So the can will be just shifted in the process.
In this solution i assumed that gas will blow only inside the can and not outside which based just on common sense because actual description of this process is quite difficult.
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Q2: The compressed air can moved because there was a lot force trapped into the can in the form of compression. Because a can is a vacuum assuming it doesn't just suck into itself the air shouldn't move it at all since the only force is suction into the can, which shouldn't move the can. If it does, it will move to the left because the air comes in the left the can should occupy that space. But I don't think thats the right answer.
Q1: I disagree with you on this micro. If you throw the ball behind you every time when you walk, at the end of the 15 minutes the dog will have to catch up to you since he would have been running the opposite direction. So his t will be t+15
MN t=14:30min UB
<------- --------->
... t=15min
Home+MN <---------UB
If you throw it in front UB will actually be home faster than MN and be running less, but i guess he still has to return it but he will have to run less distance.
EDIT: throw sideways would work but not as well as shown in the diagram: MN has moved after throwing stick and UB has just reached it.
....................^
................ MN
.................|3m
1UB>=====|
........4m
He will only run 5 m because of Pythagoras' Theorem. If you through it backwards the same diagram will be thus:
MN
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3m
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|4m
UB
Path will be 3+4 = 7m.
and if you through in front: it will be 4-3 = 1m. By the time you get home the dog will have run less because your walking will help him. Not to mention all the amount of extra time you were picking up a stick because the dog got to you much faster.
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+ Show Spoiler +
As ppl have pointed out, the correct answer must be "not moving"
The system starts in rest, nothing is moving. Then air will rush in from the left, moving the can the opposite direction. But AFTER the can is filled, the air inside the can cannot move in any direction other than the direction the can is moving (macroscopically speaking).
Since the system started out in rest, the added velocities of the air and can will be 0. Since they both has to move in the same direction, the answer must be that both velocities are 0.
System starts in rest: v_air + v_can = 0
Restriction after can is filled: v_air = v_can
Put the two together: 2*v_can = 0
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+ Show Spoiler +c
The vaccum is "filled" the instant that any air enters that can, so immediately after the puncture. At that time, the can won't be moving yet. Shortly thereafter, the can will move towards the left.
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+ Show Spoiler +"The air blows in the left as it enters the can."
- The air blows in [through] the left [hole] as it enters the can?"
- The air blows to the left [direction] as it enters the can?
I think the second is the most intuitive, which means that the hole is on the right side.
I guess to the right. Before the hole opens, the atmospheric pressure on the can is constant everywhere; summing up the forces gives you zero. Once the hole opens, though, there's a surface on the right side of the can that no longer has leftward pressure exerted on it, since the interior of the can exerts no pressure on the outside. The net force on the metal of the can is then rightward.
Obviously if its blowing in through the left direction hole reverse the answer.
I think.
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+ Show Spoiler +it wont be moving, it will just be filled with air. Because for it to be able to get moving there has to be acceleration, and for acceleration there has to be a force, but there is no force applied to the can.
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On September 27 2008 17:00 Slayer91 wrote:
Q1: I disagree with you on this micro. If you throw the ball behind you every time when you walk, at the end of the 15 minutes the dog will have to catch up to you since he would have been running the opposite direction. So his t will be t+15
MN t=14:30min UB
<------- --------->
... t=15min
Home+MN <---------UB
If you throw it in front UB will actually be home faster than MN and be running less, but i guess he still has to return it but he will have to run less distance.
EDIT: throw sideways would work but not as well as shown in the diagram: MN has moved after throwing stick and UB has just reached it.
....................^
................ MN
.................|3m
1UB>=====|
........4m
He will only run 5 m because of Pythagoras' Theorem. If you through it backwards the same diagram will be thus:
MN
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3m
|
|
|4m
UB
Path will be 3+4 = 7m.
and if you through in front: it will be 4-3 = 1m. By the time you get home the dog will have run less because your walking will help him. Not to mention all the amount of extra time you were picking up a stick because the dog got to you much faster.
but you are miss interpreting the question. It says that yubee walks for 15 minutes, and you have to assume that the dog also runs for 15 minutes. As I also said in the other blog, you have to do some assumptions, such as the dog runs with constant velocity, and that the time it takes to pick up the stick is negligible. Otherwise you would be able to come up with countless scenarios, and there wouldn't be a "correct" answer.
As someone mentioned in the last blog, it's all about time*velocity = distance, and the question was, did it matter in which direction the stick was thrown? no, because the time is still constant, and so is the velocity with the assumptions we already made.
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+ Show Spoiler +On September 27 2008 10:17 micronesia wrote:
...After the vacuum is filled... C - Since the can has been filled there will be no pressure and the can will be motionless.
On September 27 2008 10:17 micronesia wrote:
Makhno was the second person who seemed to give a nice explanation! :D
This seriously made my day ^^, I love this blog.
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+ Show Spoiler +I think options C that it wont move at all since we can assume the pressure and temperature outside the can will remain constant even after the it has filled the can with air.
Then we can assume quasistatic and adiabatic (this hasnt to be assumed but makes the calulations much simpler) then:
TdS = dW + dQ = -P_o*dV (this is then entropy of the system where the can is included) =>
-P_o*dV = |irreverseble process, not really but there are ways around this, as we are only looking for the states of equilibrium) then TdS = 0 =>-P_o*dV = 0. This means the net force acting on the can is zero, thus no movement.
I might be wrong, if so please correct me 
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+ Show Spoiler +C. I hate your tricky questions. After the can completely fills up, the system will once again have found an equilibrium with respect to its center of mass, so the can will not be moving after it is completely filled.
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+ Show Spoiler +after it's filled the forces are balanced so the air resistance would very quickly stop any motion. C
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yes, some things need to be clarified:
is this an open or closed system? is it ST&P so that the speed of sound slow enough to set up pressure gradients?
what do you mean by 'filled'?
filled=of the same pressure as the outside? (this is what I think you mean)
filled=with any air inside? (not really relevant: true vacuums are not really possible, because the inside of the can will evaporate in a true vacuum to create some independent molecules)
filled=a steady state with no acceleration? - the answer will always be C because the total momentum at the start is 0, and must be the same at the end steady state, without any consideration of vacuums or air pressure. So I'm guessing this is not what you mean.
you should specify whether friction is negligible, or if the air is viscous, because if air resistance exists, the answer will always be C after time. So I'm guessing you intend there to be no air resistance.
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On September 27 2008 10:41 fusionsdf wrote:Show nested quote +On September 27 2008 10:27 Xeofreestyler wrote:+ Show Spoiler +C
it doesnt actually matter where the hole is punctured
a vacuum will just fill the void with the air it gets and perform force in every direction of the inside of the can, not just one side
...in my sleepy mind
I just want to say that I agree with this
This one wins
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United States24670 Posts
On September 27 2008 23:48 betaben wrote:
yes, some things need to be clarified:
is this an open or closed system? is it ST&P so that the speed of sound slow enough to set up pressure gradients?
what do you mean by 'filled'?
filled=of the same pressure as the outside? (this is what I think you mean)
filled=with any air inside? (not really relevant: true vacuums are not really possible, because the inside of the can will evaporate in a true vacuum to create some independent molecules)
filled=a steady state with no acceleration? - the answer will always be C because the total momentum at the start is 0, and must be the same at the end steady state, without any consideration of vacuums or air pressure. So I'm guessing this is not what you mean.
you should specify whether friction is negligible, or if the air is viscous, because if air resistance exists, the answer will always be C after time. So I'm guessing you intend there to be no air resistance.
I took this question from someone who's better at this than me, so I'm trying not to change which clarifications are/aren't made. Usually the most obvious assumptions are the ones that are meant to be made.
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United States24670 Posts
On September 27 2008 15:40 evanthebouncy! wrote:
don't u tell Micronesia what to do ;(
Was this directed at Sonuvbob? If so, don't worry and just reread what he said hehe.
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Do as you would on an exam, state clearly all assumptions made.
If the question is unclear and it turns out it makes assumptions different from yours, you can complain in the next topic and keep talking about how right you were
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Much wrong use of the laws of Newton in this thread.
If someone finds anything wrong with this please poke it, but for now I believe that most in this thread are wrong:
+ Show Spoiler +Actually, I am pretty sure that we would have a net force on the can during the fillup process, and once it is full we would have an equilibrium again and there is no way air resistance would stop it in an instant.
This do not break any of newtons laws since when it is absorbing air it will reflect less molecules and thus while it will be moving towards the hole the closed system would start moving in the other direction due to the air pressure getting slightly lower on one end. (Or the average of the gas will be moving in the direction the can is not moving)
The easiest way to picture this is to imagine a square. Remove one of the walls and imagine that its interior gets filled by gas molecules, until the molecules who entered fills the whole square up the air pressure on the other side of it will create a net force and thus speed it up, and the moment it is full it will have a net force of 0 on both sides if we are neglecting the air resistance(And we can do that since air resistance will not slow it to a dead stop in such a limited time interval). The net force on the square gets exactly evened out by the net loss of reflected air from the removal of one of the walls.
If you want an open system just imagine that we have an artificial air around it, all with a speed directed towards the can and only in a volume to have exactly enough air to hit the can untill it has reached the critical point were we stop observing it. In such a system we would get a gas sphere around the can except were the hole is, there we would also get an air hole in the sphere, and that is were the momentum gained by the can is taken from.
In the end, the answer is B: It is moving to the right.
I can not believe that so many missed this...
Also yes even if the hole is extremely small, the fact that there is no air resistance unless the object is moving, and that until the last of the air have moved in there is always a net force working on the can from the higher pressure from the wall wo a hole it got to be moving at the time when equilibrium is reached.
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I'm with Klockan on this one.
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On September 28 2008 10:25 Klockan3 wrote:Much wrong use of the laws of Newton in this thread. If someone finds anything wrong with this please poke it, but for now I believe that most in this thread are wrong: + Show Spoiler +Actually, I am pretty sure that we would have a net force on the can during the fillup process, and once it is full we would have an equilibrium again and there is no way air resistance would stop it in an instant.
This do not break any of newtons laws since when it is absorbing air it will reflect less molecules and thus while it will be moving towards the hole the closed system would start moving in the other direction due to the air pressure getting slightly lower on one end. (Or the average of the gas will be moving in the direction the can is not moving)
The easiest way to picture this is to imagine a square. Remove one of the walls and imagine that its interior gets filled by gas molecules, until the molecules who entered fills the whole square up the air pressure on the other side of it will create a net force and thus speed it up, and the moment it is full it will have a net force of 0 on both sides if we are neglecting the air resistance(And we can do that since air resistance will not slow it to a dead stop in such a limited time interval). The net force on the square gets exactly evened out by the net loss of reflected air from the removal of one of the walls.
If you want an open system just imagine that we have an artificial air around it, all with a speed directed towards the can and only in a volume to have exactly enough air to hit the can untill it has reached the critical point were we stop observing it. In such a system we would get a gas sphere around the can except were the hole is, there we would also get an air hole in the sphere, and that is were the momentum gained by the can is taken from.
In the end, the answer is B: It is moving to the right.
I can not believe that so many missed this...
Also yes even if the hole is extremely small, the fact that there is no air resistance unless the object is moving, and that until the last of the air have moved in there is always a net force working on the can from the higher pressure from the wall wo a hole it got to be moving at the time when equilibrium is reached.
air resistance is not the only way the can can slow down. the air that enters the bottle will have a net velocity to the back of the can, which on average should slow it down (although the pressure would equalise before this happens, as it must pass through the equilibrium point before it reaches a higher pressure at the back wall.) thus, at equal pressure, the system is still moving, with the air which has entered the bottle possibly still moving with a net velocity in the opposite direction to the bottle wrt the bottle. the bottle moves because of the pressure on the side opposite to the hole, but the air entering the hole has it's own net velocity in the opposite direction, as it is forced in by the pressure outside the hole.
+ Show Spoiler +my previous answer confused the letter with the direction. I think the bottle will move towards the hole, will still be moving as the pressure equalises, but the air inside will be moving opposite to the can to slow it down. momentum has to be conserved eventually.
this is why it needs to be decided what 'filled' means.
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On September 28 2008 12:39 betaben wrote:
air resistance is not the only way the can can slow down. the air that enters the bottle will have a net velocity to the back of the can, which on average should slow it down
The fact that the air moves faster in one direction relative to the bottle compared to the other is air resistance and there are no other slowing forces.
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EPT
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