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

but there wasn't anything on transformers and E.M though

Oh really, could you upload it again or link me to the page your uploaded to. Thanks. They have not done circular motion, momentum, transformers and E.M Induction in a while so hoping these topics are kinda certain to come up in section B I think.


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here it is

Thanks so much


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Do you need an A* in the ISA for an overall A*?

No, you just need 90% average UMS at A2 and 80% total UMS for both years. so 270 UMS this year, which could possibly be achieved with an E in the EMPA/ISA if you got full UMS in the written papers!

I'd love to know if anyone has achieved that.


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Please can somebody explain question 11 to me? Thanks. Also, for an A* in Physics you need to average 90% in the papers AND the coursework? That's a bit of a downer, thought it was just the papers.

http://filestore.aqa.org.uk/subjects/AQA-PHYA4-1-W-QP-JAN11.PDF

For an object to to appear weightless at the equator the speed of the rotation of the earth around itself must increase (therefore time period decrease), to such an extent that rotating faster would result in the object be lifted up the surface of the earth as the object is changing direction. At this time the centripetal acceleration is equal to the free fall acceleration of the earth.



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Thanks, I didn't know that. So at the point where objects appear weightless, the centripetal force is the same as the force of weight due to gravity on the object? However I thought both forces act in the same direction, which is confusing me a little.

Has anyone ever heard of the formula:

$[br]B = \dfrac{{\mu_0}NI}{L}[br]$

For magnetic field strength?

Came across it in a really old paper yesterday.


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Thanks, I didn't know that. So at the point where objects appear weightless, the centripetal force is the same as the force of weight due to gravity on the object? However I thought both forces act in the same direction, which is confusing me a little.

Do we need to know the total energy formula for a system undergoing shm? $E_T=E_K+E_P=\frac{1}{2}mA^2\omega^2$

Edit: Also what about $E_P$ for a mass on a spring? : $E_P=\frac{1}{2}kx^2$ (k=stiffness constant)

No worries. That's very good question. We've been told that always the centripetal force is acting towards the centre of the rotation, that's not the correct way of saying it. Rather, for an object to be in a circular motion the centripetal force "must act towards the centre". If not the the object will not remain in circular motion. So in the case of rotating an object with a string the tension in the string is certainly acting towards the centre of the rotation because you are holding the string. If the string snaps then the object will not keep rotating in a circle.
In the case of an object rotating at the equator the centripetal force must act towards the centre which is essentially the weight of the object. But as the rotation gets faster the centripetal force , which would be the "inertia" of the object, overcomes its weight and this time there is no such a thing as the centripetal force


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It's in the "advanced physics for you" book which covers the old a level syllabus. I think it's called Ampere's law, there's a nice experiment about it that I came across a while ago: http://www.cyberphysics.co.uk/topics/magnetsm/electro/Electromagnetism%20experiment.htm

Do we need to know the total energy formula for a system undergoing shm? $E_T=E_K+E_P=\frac{1}{2}mA^2\omega^2$

Edit: Also what about $E_P$ for a mass on a spring? : $E_P=\frac{1}{2}kx^2$ (k=stiffness constant)