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AQA PHYA4 ~ 13th June 2013 ~ A2 Physics

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Original post by Jack93o
on a scale of 1 - 10

how sure are you that the answer is 'horizontal, BLV'?

...**** I'm getting paranoid


10/10 for me

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Reply 761
Original post by Jack93o
on a scale of 1 - 10

how sure are you that the answer is 'horizontal, BLV'?

...**** I'm getting paranoid


If i've got the details correct, magnetic field to the right? (or left, doesn't make a difference), the rod was perpendicular to the field from the side view which mean the current was, so i would rate 10 for horizontal, i'm fairly sure it was BLV but not 100% so 8 for that
Reply 762
Original post by Qwertish
Lower than normal, I think. Section A was a bit harder than normal, and loads of people would have got 1c or the peak emf one. QWC question was pretty straightforward though, so that might balance it out.


just to check, the 6 marker was question 4 wasn't it? (like, it wasn't 4a, 4b, etc)
This is ridiculous. Jan13 was the hardest one ive ever done but it had the highest boundaries out of the ones in the past. In order of the latest ones, the full ums has been 63, 63, 62, 62, 62, 60, 62. Im guessing its gonna be 63 this time but what does everyone else think????

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Reply 764
for the Dees question, work out velocity by:
mv^2/r = BQv
v = BQr/m and just used 0.48T and the radius they give with Q and m values for a proton

direction of magnetic field was into plane of the paper.

to work out the Pd between the two plates just did E = V/d with E being the breaking force thing.



with capacitance greater it was RC larger therefore greater time between sparks, but greater brightness as more charge can be stored so more energy stored.

for multiple choice:
- there is a couple on the coil (for the electric motor one)
- 3.6 for ratio of earth and moon radius or something
- 2mu for particle
(edited 10 years ago)
Reply 765
Original post by The H
If i've got the details correct, magnetic field to the right? (or left, doesn't make a difference), the rod was perpendicular to the field from the side view which mean the current was, so i would rate 10 for horizontal, i'm fairly sure it was BLV but not 100% so 8 for that


I definitely remember the magnetic field was shown by an arrow to the right
Reply 766
Original post by cooldudeman
This is ridiculous. Jan13 was the hardest one ive ever done but it had the highest boundaries out of the ones in the past. In order of the latest ones, the full ums has been 63, 63, 62, 62, 62, 60, 62. Im guessing its gonna be 63 this time but what does everyone else think????

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ha....thats what I thought as well when I was doing the past papers

couldn't believe the boundary was higher than the easier papers like the ones in 2010 (and one in 2011 I think)
Original post by foximus
for the Dees question, work out velocity by:
mv^2/r = BQv
v = BQr/m and just used 0.48T and the radius they give with Q and m values for a proton

direction of magnetic field was into plane of the paper.

to work out the Pd between the two plates just did E = V/d with E being the breaking force thing.



with capacitance greater it was RC larger therefore greater time between sparks, but greater brightness as more charge can be stored so more energy stored.

for multiple choice:
- there is a couple on the coil (for the electric motor one)
- 3.6 for ratio of earth and moon radius or something
- 2mu for particle


no there was an equal and opposite force in either side of the coil.
Did anyone else get 3.5*10 to the something for the capacitance question? And also can anyone else remember the other parts of the question, it got brighter as there was more energy?
Reply 769
Original post by Jack93o
I definitely remember the magnetic field was shown by an arrow to the right


BLV is the area of flux cut per second, equivalent of BA, so yeah i'm confident with the answer
I really hope I improved on my January score, I'd be surprised if I didn't as in January I couldn't answer half the questions and ended up with a D. I actually managed to do all the questions in this paper (not saying I got them all right by any means, but it should still be an improvement)

Screwed up the ratio one with the 1st question, did the 1.9/2.0 bit and 20 x 2 but forgot to minus by 2 to account for the first oscillation! Oh well hopefully a method mark for the ratio bit :smile: Just blanked out at the 2 situations when particle is stationery question, now I think about it is so obvious.

Hopefully that B will be coming my way on August 16th! :biggrin:
Original post by The H
I put 8F ( If i'm thinking of the right question) and I also guessed the Mars one, but I think I should have put 25MJ


I'm fairly sure the Mars question was -13MJ? Because...

The Earth has -63MJ and for Mars the mass is 10 times smaller so that would become -6.3MJ. However the diameter (and therefore the radius) is twice as small so you're dividing by a number twice as small (as potential is proportional to 1/R) so you multiply by 2 which gives you -12.6 = -13MJ (2sf)
Original post by Lay-Z
I thought it was really good, only question I didn't really like was the brightness one.. if only I did better in my isa's might have been able to get A* :sigh:

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that one is so hard.the unit of the brightness is W m-2.
Thus B=IV/area, IV is unchanged , area is increased, thus the brightness is decreased. i just realized that last second!!!i even thought brightness=IV!!!
i really think none can get it!!!!
Reply 773
Original post by cooldudeman
This is ridiculous. Jan13 was the hardest one ive ever done but it had the highest boundaries out of the ones in the past. In order of the latest ones, the full ums has been 63, 63, 62, 62, 62, 60, 62. Im guessing its gonna be 63 this time but what does everyone else think????

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wait, I thought jan this year had full ums at 65 :s-smilie:#

edit, you're right
(edited 10 years ago)
Reply 774
Original post by marcolliwright
I'm fairly sure the Mars question was -13MJ? Because...

The Earth has -63MJ and for Mars the mass is 10 times smaller so that would become -6.3MJ. However the diameter (and therefore the radius) is twice as small so you're dividing by a number twice as small (as potential is proportional to 1/R) so you multiply by 2 which gives you -12.6 = -13MJ (2sf)

I hope you're right, because that's the one I guessed :tongue:
Original post by cottonsocks0_0
that one is so hard.the unit of the brightness is W m-2.
Thus B=IV/area, IV is unchanged , area is increased, thus the brightness is decreased. i just realized that last second!!!i even thought brightness=IV!!!
i really think none can get it!!!!


Pretty sure brightness is just to do with the energy stored. If you increase C you increase the energy stored, because it can store more charge, so it can release more energy so it is brighter. There was a similar question on the specimen paper with a camera
Reply 776
Original post by Jack93o
wait, I thought jan this year had full ums at 65 :s-smilie:#

edit, you're right


This means a mark of 65/72 = 100 ums?
Original post by cottonsocks0_0
that one is so hard.the unit of the brightness is W m-2.
Thus B=IV/area, IV is unchanged , area is increased, thus the brightness is decreased. i just realized that last second!!!i even thought brightness=IV!!!
i really think none can get it!!!!


Lmao... Think you've over complicated it there
Reply 778
Was time for the semi circle dee Q 6.8 X 10^ -8 seconds

And used circumference is equal to PI R instead of 2 PI R because it is a semi circle
Reply 779
Original post by MSI_10
Was time for the semi circle dee Q 6.8 X 10^ -8 seconds

And used circumference is equal to PI R instead of 2 PI R because it is a semi circle

Yeah, i got that except I think i used 3 sig figs in the time

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