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AQA PHYA4 ~ 13th June 2013 ~ A2 Physics

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Reply 720

gandanmo

the first Q hen it said units of impulse, it was kg ms-1 right??

Reply 721

The H

Original post by gandanmo

the first Q hen it said units of impulse, it was kg ms-1 right??

Yeah, impulse is change in momentum, and unit of momentum is kgms^-1

Reply 722

TeddyBasherz

Original post by gandanmo

the first Q hen it said units of impulse, it was kg ms-1 right??

yep

Reply 723

Dirtybit

Original post by cooldudeman

Yh amd Blv. I did that in 5secs because in the book it mentioned that

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Surely it's vertical and blv as the rod needs to be moving perpendicular to the magnetic field lines

Ahh nvm I think I fell for the trap.

Reply 724

cooldudeman

Original post by Dirtybit

Surely it's vertical and blv as the rod needs to be moving perpendicular to the magnetic field lines

I can't remember which way the lines were going

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Reply 725

Qwertish

Original post by gandanmo

im sure its 150 and 5pi/6 :smile:

60 and 5pi/6. 150 is the same as 5pi/6...

Reply 726

Gernick

Original post by fizzbizz

I actually found the brightness one quite easy;

The time between the sparks would increase since a larger capacitance means it takes longer to charge (greater time constant)
And since a larger capacitance means greater charge stored per unit potential difference, for each discharge there is more charge to get rid of

So why are the sparks brighter? I thought there are just more sparks of the same brightness :/

Reply 727

The H

Original post by Dirtybit

Surely it's vertical and blv as the rod needs to be moving perpendicular to the magnetic field lines

The current was vertical, the magnetic field was to the right (I think)
therefore the movement is into or out of the plane which is horizontal because it was a side view

Reply 728

gandanmo

HEY GUYS! you know the graph of e.m.f in Q5 i drew the the reflection of the graph they should reflected on y=0 so basically flipped it over making it upside down( or right way up) did the graph start from 0 or a value? so me reflecting it making a cos graph, would i drop both marks ? or just 1 ?

Reply 729

Stellios

Original post by Dirtybit

They gave the values in the question so I presume you can use them to work out the value of g . I did this and the sun had a larger value than the moon

I worked it out too roughly, got E>S>M

Reply 730

MSI_10

Original post by Qwertish

60 and 5pi/6. 150 is the same as 5pi/6...

Would I get anything for putting 2.62 radians?

http://www.mathinary.com/calculation.jsp?calcname=Grader&grader=&pi=Pi&radianer=2.62&

Damn silly mistakes

Reply 731

Jack93o

Original post by cooldudeman

Yh amd Blv. I did that in 5secs because in the book it mentioned that

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yeah but the thing that stumped me, is that it showed the magnetic field in the horizontal direction :confused:

Reply 732

Muppet Science

Having just realised I have completely forgotten everything I wrote in the exam I then scan the thread and result in just petrifying myself... well done me

Reply 733

Jack93o

Original post by Stellios

I worked it out too roughly, got E>S>M

was that answer B?

Reply 734

fizzbizz

Original post by gandanmo

the first Q hen it said units of impulse, it was kg ms-1 right??

DAMN I put kgms^-2

I was thinking of F and not Ft :angry:

Reply 735

Qwertish

Original post by gandanmo

The graph started from 0, with a peak of your peak emf from the previous part and the same frequency as the other graph which was already there.

Reply 736

Jack93o

Original post by The H

The current was vertical, the magnetic field was to the right (I think)
therefore the movement is into or out of the plane which is horizontal because it was a side view

so you put the answer as 'horizontal, BLV'?

Reply 737

S1L3NTPR3Y

I worked pretty hard for this paper, was getting A's in all the past papers but I think I messed this one up quite a bit, I hope the A boundary is reasonable :s-smilie: