AQA Physics PHYA4 - Thursday 11th June 2015 [Exam Discussion Thread]

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Reply 4060

fruity97

Original post by wat a wizard

0.19 for induced emf?

I got 0.19!! Was you flux linkage like 0.0605 wb or something like that also?

Reply 4061

000alex

Original post by chizz1889

Average I got 0.0475. Worked out what I was at 90• (0.0950) then divided by 2 ☺️ (Think I may have needed to divide by root2 though as I think RMS applies

It said average emf induced, i don't know if that means rms or not

Reply 4062

k9000

Original post by king cobra

It was to the power of -6 not -5

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-5 i think you must've done 2.78x10-9 in your calculations instead of 2.78x10-8 if using whole values

Reply 4063

IWantSomeMushu

I enjoyed the MC quite a bit, but section B was a nightmare for me...

Reply 4064

Disney0702

Grade boundary predictions?

Reply 4065

Fred Cantoni

Original post by king cobra

0.12 was right I think

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I put 0.12v!

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Reply 4066

QueNNch

Alright guys I drew my graph like this

I think in probably wrong because the initial current (y axis) would be lower.

Thoughts?

Reply 4067

DanTheMan358

Original post by chizz1889

I really liked that paper, I know I got the one about which statement isn't correct about the gravitational potential wrong. Also all the damping but wrong and the curve as I forgot to use a shallower gradient but pretty sure I got the rest of the marks. What was the answer to number of electrons? I got 2.2*10^10, not sure if that's right 😊

EDIT: I also dropped a mark on the electric field strength one as my final answer wasn't negative 😓

Think i done the same thing, also i got 4.4x10^10 😓

Reply 4068

Humbers

Original post by DannySmith420

What did people get fir the hight lifted by the capacitor motor? Becuase i got 1.9m using mgh and the energy stored

It gave you weight instead of mass, maybe you went wrong there?

I think the energy stored was roughly 4.8J = mgh = Wh
H = 4.8/3.5 = 1.4

Reply 4069

QueNNch

Original post by Fred Cantoni

I put 0.12v!

Same I got 0.12 V as the emf.

Reply 4070

Sbarron

Original post by johnny147

Here's my answers to the objective test:

DAADABCBABACDCACCDBABACBC[/QUOT

I got 15 mathching

Reply 4071

king cobra

Original post by k9000

-5 i think you must've done 2.78x10-9 in your calculations instead of 2.78x10-8 if using whole values

Wasn't the next one based on this one ?

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Reply 4072

wat a wizard

Original post by fruity97

I got 0.19!! Was you flux linkage like 0.0605 wb or something like that also?

Yes, i got that

Reply 4073

Disney0702

Original post by MSB47

you reckon they'll penalise me for putting the field strength to two sig figs (1.6x10^3)?

I do not think they'll penalise you.
They usually have 2 sf written in the mark scheme

Reply 4074

Sbarron

[QUOTE="johnny147;56877845"]Here's my answers to the objective test:

DAADABCBABACDCACCDBABACBC[/QUOTE

I got 15 mathching

Reply 4075

Fred Cantoni

Original post by QueNNch

Alright guys I drew my graph like this

I think in probably wrong because the initial current (y axis) would be lower.

Thoughts?

I believe your current has to be half or 1/2000 depending if it was a typo or not, the your gradient should be shallower a time constant increases!

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Reply 4076

Will177

Original post by QueNNch

Alright guys I drew my graph like this

I think in probably wrong because the initial current (y axis) would be lower.

Thoughts?

That's what I did but I wasn't sure if the initial current changed

Reply 4077

000alex

Original post by QueNNch

Alright guys I drew my graph like this

I think in probably wrong because the initial current (y axis) would be lower.

Thoughts?

Well the question said 150 ohms in the first part and 300k ohms in the second, It might be a mistake so AQA might remove the mark for intercept?

Reply 4078

fruity97

Original post by imyimy

1.4m for energy stored by capacitor= mgh?
0.06 Wb Turns for a question about magnetic fields, and the emf as 0.12V for second part (think thats wrong)

circular motion Qs
130 degrees
T=15N (think thats wrong)
2 revolutions

6 marker
talked about constant speed but accelerating ..change in direction so velocity etc.. so F=MA newtons second law = mv2/r, then some stuff about resolving tension horizontally and vertically with MG acting downwards etc, and newtons laws about opposite and equal force and no resultant force=0 if nothing else acting on it at rest/constant velocity...

I got 15N at first for tension! But I remember changing it towards the end to 12N because I didn't include the weight I think? Stil not sure which is right but they said horizontal so I changed it

Reply 4079

Fred Cantoni

Original post by DanTheMan358

Think i done the same thing, also i got 4.4x10^10 😓

I put 2.2x10^10 cause the total number of electrons would be 4.4x10^10 but half would already be on that plate, and the other half moves over from the other plate when they're is a potential difference? Anyone confirm or deny this?

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