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

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Why do I not include the radius of the earth in this? I thought newtons law was the graviational force of attraction the centre of 2 masses??

Thanks

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I think radius does mean from centre to centre

Reply 2661

gcsestuff

Original post by sykik

I think radius does mean from centre to centre

Just a bit confusing how it says "from the surface of the earth" and then radius "around the earth"

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Reply 2662

sykik

Original post by gcsestuff

Just a bit confusing how it says "from the surface of the earth" and then radius "around the earth"

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A picture i found on internet. So i think when they say Orbital radius then you assume tat it's from centre to centre.

Reply 2663

_Caz_

Original post by CD223

An A hopefully! Need AAA for uni. You? :smile:

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I need an A*A in either Maths or physics (any order) and then another A in one of my other subjects. I am worried about this exam though, my mock was definitely not good for unit 4.

I'm hoping they lay off the fields a bit this year because for two years in a row they've had section b being made primarily of fields. I don't mind gravitational fields but electric and magnetic really confuse me and I can't visualise the context at all.

I'll be surprised if you don't get an A* tbh (no pressure haha) because you seem to know so much when you're helping people out in this thread so you deserve one IMO.

Reply 2664

Disney0702

Original post by Mehrdad jafari

To be honest i don't know why the ratio doesn't hold for energy
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EDIT: for questions like this one it's better not to use the ratio as you are not given any voltage, but only by how much it's changed so the ratios are exact multiples and can easily be figured out thinking about it. If you were given inexact values if voltage then you will have to use ratio as you can't figure out by logic.

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Thank you for this. The answer is 72ms.

I'm still a little confused on how you had

E = \dfrac{CV^2}{2}

and then concluded that E ∝ t V². How did you get the t?

And from I t = C V how did you get t ∝ C, I assumed you couldn't do that because I and V are not constant.

Reply 2665

CD223

Original post by _Caz_

That's very kind of you, thank you! I just don't think an A* is possible due to my EMPA :/

I agree! What do you think the 6 marker will be? I reckon SHM.

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Reply 2666

Disney0702

Original post by CD223

[br]P = \dfrac{\Delta E}{\Delta t}[br]

[br]\therefore P = \dfrac{0.5 \times 50 \times 10^{-6} \times (30 \times 10^{3})^2}{5 \times 10^{-3}}[br]

[br]\therefore P = 4.5 \text{MW}[br]

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Thank you very much. :smile:

Original post by CD223

Sorry - didn't realise someone had already answered!I can tell you why this doesn't work though - you've worked out the average current - the capacitor discharge current falls exponentially and therefore is not a constant value :smile:

Edit: and I've just realised this was answered too - I'm not having a good day 😂Posted from TSR Mobile

I really do appreciate your explanation, it is better for my understanding when I get different explanations for the same thing, it makes it easier for me to tackle questions later on.

Reply 2667

CD223

Original post by Disney0702

Thank you for this. The answer is 72ms.

I'm still a little confused on how you had

E = \dfrac{CV^2}{2}

and then concluded that E ∝ t V². How did you get the t?

And from I t = C V how did you get t ∝ C, I assumed you couldn't do that because I and V are not constant.

What question is this? I remember being told a really simple way by my teacher!

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Reply 2668

Disney0702

Original post by CD223

What question is this? I remember being told a really simple way by my teacher!

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It is Q9 Jun '07 and you can find it in this link.

Reply 2669

CD223

Original post by Disney0702

It is Q9 Jun '07 and you can find it in this link.

Thanks!

Essentially you know it always takes 36ms for V to halve in value. This is because, for exponential decay the same decrease is observed over equal time intervals.

As E is proportional to V squared, this means that when E becomes E/16, V must have halved again to become V/4.

This means that the time for V to halve twice from its initial value and therefore become V/4 is:

36ms + 36ms = 72 ms

Does that help?

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Reply 2670

Disney0702

Original post by CD223

Oh my! I had that penny drop moment! :u:

Thank you so much! That's a good way of thinking!

I was on a similar track when I was doing this question but I thought I had to halve 36 ms, but your explanation made so much sense.

Thank you very much!

Reply 2671

getback339

https://08c6cd28b8bef288858e878e57451f5685184876.googledrive.com/host/0B1ZiqBksUHNYQ2p4Y29lcHhMOXM/From-AQA-Set-A/4.1.3%20SHM.pdf

could someone also explain question number 13? thanks

Reply 2672

CD223

Original post by Disney0702

Oh my! I had that penny drop moment! :u:

Haha! Glad it helped.

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Reply 2673

_Caz_

Original post by CD223

That's very kind of you, thank you! I just don't think an A* is possible due to my EMPA :/

I agree! What do you think the 6 marker will be? I reckon SHM.

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Yeah there's not been a SHM six marker yet has there? It'll be that or something to do with magnetic fields I think. My teacher made an excel spreadsheet of all the topics of the past paper questions. I've attached it here if you want it for reference. It's useful to find the six markers quickly.
The black cells are six markers and the numbers in the cells are the question number that they appeared in the exam in.

Don't worry about your EMPA it's not worth as much as the exams. :smile:

A2 Past paper question type log.xls43.0KB

Reply 2674

CD223

Original post by _Caz_

Thank you!

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Reply 2675

gcsestuff

Original post by _Caz_

Omg no transformer question since 2012:/

Defo need to nail this now!!

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Reply 2676

Disney0702

Original post by getback339

https://08c6cd28b8bef288858e878e57451f5685184876.googledrive.com/host/0B1ZiqBksUHNYQ2p4Y29lcHhMOXM/From-AQA-Set-A/4.1.3%20SHM.pdf

could someone also explain question number 13? thanks

The answer is B.

Lightly damped systems have a very sharp resonance peak. Their amplitude only increases drastically when the driving force is very close to its natural frequency.

Heavily damped systems have a flatter response. Their amplitude doesn't increase very much near the natural frequency and they aren't as sensitive to the driving frequency.

I hope that explanation helps. :smile:

Reply 2677

getback339

Original post by Disney0702

yeah thanks for the reply, im not too sure i get the difference between A and B though

Reply 2678

Jed-Singh

Can anyone help me with the multiple choice questions that involve 'when potential or field strength etc' equals 0. I never get them correct :frown:

Reply 2679

Absent Agent

Original post by Disney0702

Thank you for this. The answer is 72ms.

I'm still a little confused on how you had

E = \dfrac{CV^2}{2}

and then concluded that E ∝ t V². How did you get the t?

And from I t = C V how did you get t ∝ C, I assumed you couldn't do that because I and V are not constant.

Oh, sorry, i meant to write 72ms as i worked with ms in the question.
And yeah, you're correct in saying that you cannot assume the proportionality of t with C as C is constant, but this proportionality has a different sense. t is proportional to c means that the greater the capacitance the longer the time the capacitor takes to discharge as it can contain more charge. I needed that proportionality to see how the voltage varies with time. If, for example, the greater the capacitance the less time it would take to discharge then proportionally would differ. It would become :
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