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AQA Physics PHYA5 - Thursday 18th June 2015 [Exam Discussion Thread]

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Original post by Protoxylic
dt is the time between collisions. So this is the overall time it takes to change the momentum of the particle from it's initial position (wall A) by 2mux in the x direction. So that when time dt is elapsed, the particle is at the same position as it was, but now with the opposite momentum. This requires that the time of the collision is very small in comparison to the length of time between the start and end points of which you're describing the change in momentum/force. Otherwise if the collision takes place over a larger time interval, then it HAS to be true that the speed is not constant (which is the case in reality) and this dt cannot be calculated using vx as it isn't constant, it changes with time as there is a force acting on it at the boundary.


Ok, if dt is not the time during which each particle and the container wall are in contact then we cannot substitute this into the momentum formula because the force found is not the force exerted by the each particle on the wall but is the deriving force supplied by the energy of the particle to change its momentum, if I'm not wrong


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Original post by Mehrdad jafari
Ok, if dt is not the time during which each particle and the container wall are in contact then we cannot substitute this into the momentum formula because the force found is not the force exerted by the each particle on the wall but is the deriving force supplied by the energy of the particle to change its momentum, if I'm not wrong


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Basically dt is not the true time duration of each collision, it is the time between collisions which requires the assumption that the time duration of each collision is small.
Original post by CD223
[br]ΔQ=mcΔT[br][br]\Delta Q = mc \Delta T[br]

[br]c=ΔQmΔT[br][br]c = \dfrac{\Delta Q}{m \Delta T}[br]

[br]ΔQ=7.5×1043.0×104[br][br]\Delta Q = 7.5 \times 10^4 - 3.0 \times 10^4[br]

[br]c=7.5×1043.0×1041.2×(32784)[br][br]\therefore c = \dfrac{7.5 \times 10^4 - 3.0 \times 10^4}{1.2 \times (327-84)}[br]

[br]c=155Jkg1K1[br][br]\therefore c = 155 Jkg^{-1}K^{-1}[br]


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Why is the change in temperature 327-84? Kind of confused
Original post by Protoxylic
Basically dt is not the true time duration of each collision, it is the time between collisions which requires the assumption that the time duration of each collision is small.


That makes perfect sense but do you not think assuming that would give us the wrong overall pressure?


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Original post by Mehrdad jafari
That makes perfect sense but do you not think assuming that would give us the wrong overall pressure?


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The pressure is of the order of KPa normally, so this assumption is justified by the fact that the difference between the actual pressure and the calculated pressure isn't that much I presume
Original post by Protoxylic
Offer holder for NatSci at Cambridge :colondollar:. My passion is in Chemistry, however. But (grades permitting) I get to continue with Physics, Chem, Maths and take up Material Sciences at Cambridge :tongue:


Ahh awesome!! Well all the best hopefully you get in


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Do any of you guys know why 1c uses the change in temperature as 327-84? http://filestore.aqa.org.uk/subjects/AQA-PHYA51-W-QP-JUN10.PDF
Just started revision for this. How likely is it to get A* after a day's revision?
Original post by Protoxylic
The pressure is of the order of KPa normally, so this assumption is justified by the fact that the difference between the actual pressure and the calculated pressure isn't that much I presume


Well, that's true, we tend to give the priority to the experimental observations rather than the theoretical understanding of it.


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Original post by SuperMushroom
Ahh awesome!! Well all the best hopefully you get in


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Cheers! And I hope you get the grades you need for wherever you want to go :smile:
Absolutely bricking it! I know the spec inside out I can just tell its gonna be a bad exam


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Someone ask me some turning point questions


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Reply 1892
Original post by JJBinn
Why is the change in temperature 327-84? Kind of confused


Because that's the temperature that the mould decreases to, from its initial temperature of 327 degrees.


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Original post by ubisoft
Just started revision for this. How likely is it to get A* after a day's revision?


Just how likely you think it is :smile:


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Original post by Mehrdad jafari
Just how likely you think it is :smile:


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How likely do I think it is?
Original post by gcsestuff
Someone ask me some turning point questions


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ether, a hypothetical medium once thought to necessary to support the propagation of electromagnetic waves throughout space, cannot explain why a light shone towards a detector on a moving train traveling in the same direction as the incident light takes the same amount of time to reach the detector as if the train was stationary. Explain why :smile:



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Original post by ubisoft
How likely do I think it is?


Yeah, but i didn't ask you a question


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Original post by Mehrdad jafari
Yeah, but i didn't ask you a question


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I did
Original post by ubisoft
I did


Lol, yeah, but you asked yourself not me :smile:


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Original post by Mehrdad jafari
Lol, yeah, but you asked yourself not me :smile:


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I asked you how likely do you think that I think it is?

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