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

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I'm talking about when you read off a graph, say. Because they only put one value on the ms

If you're reading off a graph then technically you shouldn't quote it to any more precision than the scale itself. So, if each grid line was 0.1, you could only answer with 1d.p.

Reply 2201

SuperMushroom

Original post by FN510

Do you prefer unit 5/ find it easier?

Yep for some reason i find it alot easier, especially the applied physics option that i am doing, i know that is a bit unusual haha.

How about you ?

Reply 2202

CD223

Original post by JJBinn

Not understanding how to get the efficiency on 25, whenever I do it it comes out as about 68%, really weird. I am also confused over 10,12 and 13 although I managed to get them right by judgement, I don't really understand them. If anybody could explain some/all of these questions that'd be awesome. Thanks

http://filestore.aqa.org.uk/subjects/AQA-PHYA4-1-QP-JAN13.PDF

Question 25:
ImageUploadedByStudent Room1433016111.069745.jpg

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Reply 2203

CD223

Original post by JJBinn

Question 13:
ImageUploadedByStudent Room1433016144.073153.jpg

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Reply 2204

CD223

Original post by 000alex

Can someone explain what these equations find?

Are the emf equations average emf induced and is the 2nd average flux linkage?
Also when should I use each of them (e.g. for a fixed rotation or continuous rotation?)

The top equation is related to "average emf" induced in any situation. This may apply to rotating coils, but equally could apply to a plane cutting the earth's magnetic flux.

The bottom two are specific to a rotating coil. (Could you specify what you mean by fixed or continuous?)

The bottom two assume the coil is in continuous rotary motion if that's what you mean to say?

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Reply 2205

Crash1234

If my answer is 2.88x10^4 and the mark scheme says 2.87x10^4, do you think I would get the marks??

Reply 2206

Absent Agent

Original post by Crash1234

If my answer is 2.88x10^4 and the mark scheme says 2.87x10^4, do you think I would get the marks??

Unfortunately not the accuracy mark. Is the second 8 in a bracket?

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Reply 2207

jordan_bennett14

Just wondering does anyone know what the closest past paper to the exam is likely to be? I'm guessing it will be last year's June 2014 paper?

Also, I'm hovering on the A/B borderline - my last few past papers have only been 1 or 2 marks from an A, bearing in mind that these are 2013 to 2011 papers. Any tips on how to improve my grade? Just more past papers or should I revisit the material? It's usually the explanation and 6 markers that I fall down on! :s-smilie:

Thanks!

Reply 2208

Lau14

Original post by Crash1234

If my answer is 2.88x10^4 and the mark scheme says 2.87x10^4, do you think I would get the marks??

It probably depends on why - if you didn't round any numbers (used the answer button on your calculator throughout) then you should. If you've actually rounded wrong then you probably will drop a mark, and if you rounded too much/too early on in a series of calculations you may also drop a mark? The mark schemes aren't always very clear on it though :/

Reply 2209

Lau14

Original post by jordan_bennett14

Thanks!

Go back over questions you've got wrong and make sure you've learnt them. Redoing the six markers is a good idea if you have time, particularly if there's any you did a while ago. Double check that you know all your definitions and things like that :smile:

Reply 2210

jordan_bennett14

Original post by Lau14

Okay thanks for the quick response at such a late time! :smile:

I have to study for 5 other exams so I'll try and fit it all in :s-smilie:

Hope your exams go well! And good luck to everyone else! :smile:

Reply 2211

000alex

Original post by CD223

Yeah by fixed I meant the coil is turned by a certain angle, e.g. 90 degrees only
And yes, continuous rotary motion is what I meant by continuous.

So if a coil is rotated 90 degrees in a certain amount of time, you would use the first equation?
Why doesn't using the third equation work if you knew BAN and t and find omega as 90/t (in radians) and put them into the equation?

Reply 2212

gcsestuff

Why does a positive charge flow in the same direction as current using Flemings left hand rule, but a negative charge is in the opposite direction?

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Reply 2213

Absent Agent

Original post by gcsestuff

Why does a positive charge flow in the same direction as current using Flemings left hand rule, but a negative charge is in the opposite direction?

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First because that was the best way Fleming could get away with the direction of the force, field, and current. And second because the magnetic field induced when a positive charge is in relative motion has field lines in opposite direction to the field lines when a negative charge is in relative motion :smile:

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Reply 2214

bwr19

Original post by gcsestuff

Why does a positive charge flow in the same direction as current using Flemings left hand rule, but a negative charge is in the opposite direction?

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(Conventional) Current is defined at the flow of positive charge.

So if an electron is moving from right-to-left, then that is kinda the same as a positive charge moving left-to-right.

Reply 2215

CD223

Original post by 000alex

All three equations can apply to a rotating coil. You just have to select the most appropriate one to use based on the information they give you.

However, the top one can also be applied to other applications.

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Reply 2216

gcsestuff

Original post by bwr19

(Conventional) Current is defined at the flow of positive charge.

So if an electron is moving from right-to-left, then that is kinda the same as a positive charge moving left-to-right.

Thanks for all your help!! I get it now

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Reply 2217

gcsestuff

ImageUploadedByStudent Room1433066786.488903.jpg

Can someone explain why at the surface of the moon it goes up then comes back down to its final point?

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Reply 2218

CD223

Original post by gcsestuff

Can someone explain why at the surface of the moon it goes up then comes back down to its final point?

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Potential is a scalar quantity.

[br]\displaystyle V = (-) \dfrac{GM}{r}[br]

The minus sign explains why the graph is negative.

All the way along the graph, as we travel from the earth to the moon, the total potential is the sum of the earth and the moon's potential:

[br]\displaystyle {V_{Total}} = {V_{Earth}} + {V_{Moon}}[br]

V \propto M

, and the mass of the earth is much greater than that of the moon, the shape of the graph will largely follow the trend of the earth's potential as the distance increases.

However, as we get closer to the moon, there is a point X where the resultant gravitational field strength is zero between earth and the moon, and the potential for earth reaches a maximum on graph you see.

At this point, the potential due to the moon starts to take over and is greater than the potential due to the earth, causing a dip in the graph you see.

ImageUploadedByStudent Room1433067403.732519.jpg