dan0105 Posted October 18, 2009 Report Share Posted October 18, 2009 you know it's time to improve your french WHEN... danny speaks better than you thanks for that. and great! now people figured out dan is short for danny! everyone knows who I am now!!!!! Link to comment Share on other sites More sharing options...
The Law Posted October 18, 2009 Author Report Share Posted October 18, 2009 lol dan... because it wasn't really obvious to people before. 11 on PS is impressive Law. The first time I wrote it I got 11 on PS too. And I consider myself kinda good at physics so good on ya! Thanks bud. I slaved away to get that score last time around! Link to comment Share on other sites More sharing options...
avenir001 Posted October 19, 2009 Report Share Posted October 19, 2009 thanks for that. and great! now people figured out dan is short for danny! everyone knows who I am now!!!!! i kid, i kid like how they found out where i live thanks to u. now we're even Link to comment Share on other sites More sharing options...
dan0105 Posted October 19, 2009 Report Share Posted October 19, 2009 i kid, i kid like how they found out where i live thanks to u. now we're even I don't recall ever spilling your home address, you lie! Link to comment Share on other sites More sharing options...
avenir001 Posted October 20, 2009 Report Share Posted October 20, 2009 well u said burnaby and i've been living in fear ever since. i mean, it's not that hard to spot the prettiest girl in burnaby Link to comment Share on other sites More sharing options...
future_doc Posted October 20, 2009 Report Share Posted October 20, 2009 well u said burnaby and i've been living in fear ever since. i mean, it's not that hard to spot the prettiest girl in burnaby see you in Metrotown! Link to comment Share on other sites More sharing options...
avenir001 Posted October 20, 2009 Report Share Posted October 20, 2009 see, all these premeds are closing in on me Link to comment Share on other sites More sharing options...
future_doc Posted October 20, 2009 Report Share Posted October 20, 2009 You will recognize me b/c I will be carrying a sign saying 'rayven' and then you can introduce yourself if you like Link to comment Share on other sites More sharing options...
The Law Posted October 20, 2009 Author Report Share Posted October 20, 2009 Does anybody want to try to explain the relationships in Bernoulli's equation a bit more clearly to me? I find moving fluids confuzzling. Also, here's a question: Note: Pretend B and C are in the same depth and that A is on the surface. Which of the following has the most pressure? A) A B C) C D) Pressure equal So, I picked B because obviously it has more than A.. but between B and C I had a bit of problem. I figured, since C is moving fluid, well close to the point where fluid is rushing out... due to bernoulli's equation, we'd expect lower pressure there. Anyway, the answer said that points A and C are equal to atmospheric pressure. I understand A, but why C? Also... if the container was closed, and B and C are in the same plane... the pressure then would be the same, correct? Also, I got this q right but I'm a bit confused why it seems so complicated. "AN ideal fluid with pressure P flows through a horizontal pipe with radius r. If the radius of the pipe increased by a factor of 2, which of the following most likely gives the new pressure? A) P 4P C) 16P D) Not enough info given" Answer: D I picked D because I was thinking, pressure is the force per unit area. If you increase r by 2, then P should accordingly go down by 4... and 1/4 P is not there, so that's why I went with D. However, the book is offering an explanation saying... Based on Q = Av, the velocity increases by 4 when the radius is increased by 2 (makes sense.). But then it says based on Bernoulli, we'd think K = P + 1/2dv^2 (I don't know why they took out the 3rd term in bernoulli's eqn). and "since we know velocity increases by 4, but we don't know the amount, we don't know how much P decreases by. I'm confused, why we took out the third term... and why we can assume the increase in second term is directly proportional to the decrease in P. Link to comment Share on other sites More sharing options...
avenir001 Posted October 20, 2009 Report Share Posted October 20, 2009 aww look at poor law trying to get his beloved thread back on topic.. intuitively u'd expect a little bit less pressure at the opening than around it cuz the molecules moving through the opening experience less collisions..but the container's open to the atm at both points so Patm is the main contributor to the pressure. and looking at the bernoulli equation P + pgh + 1/2pv^2 = some constant K, u can figure out which terms are contributing at which points. at point A, v=0 so only the first 2 terms count. at point C, h=0 so the middle term drops out. so basically u have: P1+ pgh1 = P2 + 1/2pv^2 (1= point A, 2= point C) now P1 = P2 = Patm cuz they're open to the atm. so what this means is that the potential energy of height at point A is converted to the kinetic energy of motion at point C. for the 2nd question, either the book is wrong...or you copied wrong again? remember the flow rate is the same. increasing r by a factor of 2, increases the area by a factor of 4...and so decreases the velocity by a factor of 4 (cuz flow rate = Av = constant which means A and v are inversely proportional and if u know the factor by which one has increased u'll know the factor by which the other one has decreased and vice versa). ok so we know v has decreased by a factor of 4 so P must increase..but we don't know by what factor..because the bernoulli eqn is not a proportionality eqn (it's got terms..ie the plus signs) so without knowing the actual numbers, u don't know exactly how much the other term has changed (u know the direction of change, but not the magnitude so to speak). make sense? and u take out the middle term cuz the pipe is horizontal so height doesn't play into it and u can set h = 0 as your reference point. Link to comment Share on other sites More sharing options...
avenir001 Posted October 20, 2009 Report Share Posted October 20, 2009 You will recognize me b/c I will be carrying a sign saying 'rayven' and then you can introduce yourself if you like i'm still in the closet and c'mon, show rayven some love already. Link to comment Share on other sites More sharing options...
The Law Posted October 20, 2009 Author Report Share Posted October 20, 2009 aww look at poor law trying to get his beloved thread back on topic.. intuitively u'd expect a little bit less pressure at the opening than around it cuz the molecules moving through the opening experience less collisions..but the container's open to the atm at both points so Patm is the main contributor to the pressure. and looking at the bernoulli equation P + pgh + 1/2pv^2 = some constant K, u can figure out which terms are contributing at which points. at point A, v=0 so only the first 2 terms count. at point C, h=0 so the middle term drops out. so basically u have: P1+ pgh1 = P2 + 1/2pv^2 (1= point A, 2= point C) now P1 = P2 = Patm cuz they're open to the atm. so what this means is that the potential energy of height at point A is converted to the kinetic energy of motion at point C. Thanks EM, that makes sense now. Would I be right in saying at point B, both, velocity and height are zero.. thus maximizing pressure? for the 2nd question, either the book is wrong...or you copied wrong again? remember the flow rate is the same. increasing r by a factor of 2, increases the area by a factor of 4...and so decreases the velocity by a factor of 4 (cuz flow rate = Av = constant which means A and v are inversely proportional and if u know the factor by which one has increased u'll know the factor by which the other one has decreased and vice versa). ok so we know v has decreased by a factor of 4 so P must increase..but we don't know by what factor..because the bernoulli eqn is not a proportionality eqn (it's got terms..ie the plus signs) so without knowing the actual numbers, u don't know exactly how much the other term has changed (u know the direction of change, but not the magnitude so to speak). make sense? Okay eshgh... so, yeah, ideal moving fluid = Q is constant. Since, radius doubled, Area is up 4x. This means velocity down by 4. This means, 1/2dv^2 is up by 16... but it's not proportional to pressure, so we can't tell by how much. I get it now! I just gotta get a bit more confident using bernoulli's and then I think I'll be okay. Let's say this was not a moving fluid though. Would pressure then correspond to the 1/change in area? Link to comment Share on other sites More sharing options...
The Law Posted October 20, 2009 Author Report Share Posted October 20, 2009 I'm confused about this question. Can someone explain the principles that I am not understanding here? "At atmospheric pressure, air is 7x heavier than helium. A sealed helium balloon rises into the atmosphere. If the helium remains in thermal equilibrium with its surroundings, the balloon will rise: A) Until the pressure inside the balloon = pressure of the surrounding atmosphere until the pressure in the balloon is 7x greater than the pressure of the surrounding atmosphere C) until the pressure inside the balloon is seven times less than the pressure of the surrounding atmosphere D) the balloon will not rise if the temperature of the helium = the temp of its surroundings Answer: B" and it's sister question: As a hot air balloon rises into the atmosphere its volume does not change. It is opened at the bottom to allow air exchange with atmosphere. Ignoring the weight of the balloon, which of the following will stop a hot air balloon from rising? A) The pressure inside bal. = pressure atm. volume in balloon = vol atm C) temp inside balloon = temp atm D) number of moles of air inside baloon is equal number of moles of air in atm Answer: B Link to comment Share on other sites More sharing options...
avenir001 Posted October 21, 2009 Report Share Posted October 21, 2009 Thanks EM, that makes sense now. Would I be right in saying at point B, both, velocity and height are zero.. thus maximizing pressure? yeah, pressure at B is equal to Patm + pgy where y is the depth from the surface. Okay eshgh... so' date=' yeah, ideal moving fluid = Q is constant. Since, radius doubled, Area is up 4x. This means velocity down by 4. This means, 1/2dv^2 is up by 16... but it's not proportional to pressure, so we can't tell by how much. I get it now! I just gotta get a bit more confident using bernoulli's and then I think I'll be okay. Let's say this was not a moving fluid though. Would pressure then correspond to the 1/change in area?[/quote'] correct, P = F/A so if u have the same amount of water exerting the same amount of force on a smaller area, u got less pressure...i mean, MORE pressure. duhhh Link to comment Share on other sites More sharing options...
avenir001 Posted October 21, 2009 Report Share Posted October 21, 2009 I'm confused about this question. Can someone explain the principles that I am not understanding here? "At atmospheric pressure, air is 7x heavier than helium. A sealed helium balloon rises into the atmosphere. If the helium remains in thermal equilibrium with its surroundings, the balloon will rise: A) Until the pressure inside the balloon = pressure of the surrounding atmosphere until the pressure in the balloon is 7x greater than the pressure of the surrounding atmosphere C) until the pressure inside the balloon is seven times less than the pressure of the surrounding atmosphere D) the balloon will not rise if the temperature of the helium = the temp of its surroundings Answer: B" are u sure the answer is not A? if the pressure is greater than that of the surrounding atm, shouldn't it fall? but yeah, as it rises, it cools and the volume gets smaller while the mass remains constant, so p=m/V increases and so does P until it equals P of the surrounding..at which point it should stop rising. i would think and it's sister question: As a hot air balloon rises into the atmosphere its volume does not change. It is opened at the bottom to allow air exchange with atmosphere. Ignoring the weight of the balloon' date=' which of the following will stop a hot air balloon from rising? A) The pressure inside bal. = pressure atm. volume in balloon = vol atm C) temp inside balloon = temp atm D) number of moles of air inside baloon is equal number of moles of air in atm Answer: B[/quote'] this one's easier cuz if it's open to the atm, air is gonna rush into the balloon as it rises untillll there's as much air inside the balloon as there would be if it (the balloon) wasn't even there. get it? Link to comment Share on other sites More sharing options...
The Law Posted October 22, 2009 Author Report Share Posted October 22, 2009 are u sure the answer is not A? if the pressure is greater than that of the surrounding atm, shouldn't it fall? but yeah, as it rises, it cools and the volume gets smaller while the mass remains constant, so p=m/V increases and so does P until it equals P of the surrounding..at which point it should stop rising. i would think The answer is definitely B. I thought about this question a lot (with help from The_ and now, I think I totally get it (mathematically). My conclusions: 1) Just because pressure inside = pressure outside, that doesn't matter. What matters is density of the objects. If the two are the same density, the balloon will stop "moving" through the atmosphere (thinking of the atmosphere as another fluid). 2) Think about it in terms of PV=nRT... which is the same as PV=(m/M)RT Since m/v = density (d), we can rewrite gas law as PM=dRT. We know that molar mass M, for Helium, is 1/7 that of air. However, we want the density of Helium to be the same as that of the atmosphere. To do this, since we cannot alter M, we would have to thereby increase the pressure 7x. I don't know how write this is, but it kind of makes sense, I think? lol Link to comment Share on other sites More sharing options...
The Law Posted October 22, 2009 Author Report Share Posted October 22, 2009 this one's easier cuz if it's open to the atm, air is gonna rush into the balloon as it rises untillll there's as much air inside the balloon as there would be if it (the balloon) wasn't even there. get it? I still don't really get this one. I mean, I understand eventually the mixing will cause them to basically have the same composition. lol.. but how can the volume of the balloon be the same as the volume of air? Why can't it be A? I find this one kind of harder than the last one lol. Link to comment Share on other sites More sharing options...
The Law Posted October 23, 2009 Author Report Share Posted October 23, 2009 Sweet just did EK physics exam for fluids and got a 10! Gonna look over my answers tomorrow. Link to comment Share on other sites More sharing options...
avenir001 Posted October 23, 2009 Report Share Posted October 23, 2009 The answer is definitely B. I thought about this question a lot (with help from The_ and now, I think I totally get it (mathematically). My conclusions: 1) Just because pressure inside = pressure outside, that doesn't matter. What matters is density of the objects. If the two are the same density, the balloon will stop "moving" through the atmosphere (thinking of the atmosphere as another fluid). 2) Think about it in terms of PV=nRT... which is the same as PV=(m/M)RT Since m/v = density (d), we can rewrite gas law as PM=dRT. We know that molar mass M, for Helium, is 1/7 that of air. However, we want the density of Helium to be the same as that of the atmosphere. To do this, since we cannot alter M, we would have to thereby increase the pressure 7x. I don't know how write this is, but it kind of makes sense, I think? lol aah i get it! u're setting the densities as = so it's like (PM/RT) for baloon = (PM/RT) for surroundings. tricky! and are u saying The_B is not only the p101 love doc but he's also good at physics? wow Link to comment Share on other sites More sharing options...
rmorelan Posted October 23, 2009 Report Share Posted October 23, 2009 aah i get it! u're setting the densities as = so it's like (PM/RT) for baloon = (PM/RT) for surroundings. tricky! and are u saying The_B is not only the p101 love doc but he's also good at physics? wow That makes sense - things only float if they weigh less than the weight of the volume of fluid (air or water etc) that they displace by being there. Link to comment Share on other sites More sharing options...
avenir001 Posted October 23, 2009 Report Share Posted October 23, 2009 I still don't really get this one. I mean, I understand eventually the mixing will cause them to basically have the same composition. lol.. but how can the volume of the balloon be the same as the volume of air? Why can't it be A? I find this one kind of harder than the last one lol. well according to what u said for the last question, pressure inside = pressure outside doesn't really matter. it's density that matters. and the reason i chose B is that we're assuming the baloon is submerged in the atm, and remember a submerged object displaces its volume in fluid. the volume of the balloon isn't the same as the volume of the surrounding air...that's funny wording. what it means i think is that the balloon fills with air and rises until it's displaced the same volume of air as it's own volume. Link to comment Share on other sites More sharing options...
The Law Posted October 23, 2009 Author Report Share Posted October 23, 2009 aah i get it! u're setting the densities as = so it's like (PM/RT) for baloon = (PM/RT) for surroundings. tricky! and are u saying The_B is not only the p101 love doc but he's also good at physics? wow Did I just explain something to EM???! The world is going to fall apart. lol Link to comment Share on other sites More sharing options...
The Law Posted October 23, 2009 Author Report Share Posted October 23, 2009 well according to what u said for the last question, pressure inside = pressure outside doesn't really matter. it's density that matters. and the reason i chose B is that we're assuming the baloon is submerged in the atm, and remember a submerged object displaces its volume in fluid. the volume of the balloon isn't the same as the volume of the surrounding air...that's funny wording. what it means i think is that the balloon fills with air and rises until it's displaced the same volume of air as it's own volume. D'oh of course. I get it now. The wording tripped me up. Once the volumes are the same, if the air is mixed, the densities are the same too. Merci. Link to comment Share on other sites More sharing options...
The Law Posted November 1, 2009 Author Report Share Posted November 1, 2009 I hate circuits! Link to comment Share on other sites More sharing options...
The Law Posted November 1, 2009 Author Report Share Posted November 1, 2009 Note: each battery has a voltage of 12V. What is the voltage between points A and B? A) 0 12V C) 6V D) 24 V I thought it was zero because the currents are opposing each other so I though they'd cancel out or something... The answer is 12V. Question... are they opposing currents? Or was I wrong about that? Merci pour the help! Link to comment Share on other sites More sharing options...
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