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

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Original post by CD223
Get well soon! Core 4 wasn't great - spent my day drowning my sorrows and being unproductive haha. Good luck for Chem!


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Thanks! (I think this is legitimately the worst cold I've ever had though why). At least core 4 boundaries are usually really low, and you did so well on core 3 it'll probably balance out alright :smile:
Original post by NEWT0N
Yeah I have trouble understanding this as well. After giving it some thought all I've come up with is that the weight of the object acts in the same direction as the centripetal force (towards the center of the planet), so if centripetal force=weight of object this simply means that there is a double centripetal force! Definitely not weightless. An explanation would be much appreciated. (Or does the weight act outwards when an object is spinning... but I thought centrifugal force doesn't exist? :s )


I always get stuck on the whole water in bucket on a rope thing where the weight of the bucket acts upwards along with the tension so centripetal force is mg PLUS t!!!!
Having a little trouble with the following q:
If an ideal transformer has twice as many turns in the secondary coil as in the primary coil, what will the voltage across the secondary coil be?

I think ideal transformer leads a hint but no clue what that is. Thanks in advance :P

Edit: Oh crud, I get it now...
(edited 8 years ago)
Original post by Sbarron
I always get stuck on the whole water in bucket on a rope thing where the a of the bucket acts upwards along with the tension so centripetal force is mg PLUS t!!!!


ImageUploadedByStudent Room1433877976.327041.jpg

I forgot about centripetal force equation and just think which way each of the forces are acting. I then equate them so the ones in the direction the the centre are +ve away are -ve then I make them equal to the centripetal force


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Original post by gabbons
Having a little trouble with the following q:
If an ideal transformer has twice as many turns in the secondary coil as in the primary coil, what will the voltage across the secondary coil be?

I think ideal transformer leads a hint but no clue what that is. Thanks in advance :P

Edit: Oh crud, I get it now...


More coils = bigger voltage

Power must be conserved so current must be smaller

As p=iv

So current is smaller in secondary coil


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Original post by GANGA_SLAYER


Hey, it's just Hooke's Law I think so her mass (mg) provides the force for F = kx so x = mg/k so (58 x 9.81)/54 = 10.5m
Original post by GANGA_SLAYER


Here is your desired thought process in terms of how AQA like to word questions:
First clue - 'behaves like a spring' automatically F=Kx should spring to mind (see what I did there lol)

Second clue - 'Hooke's law' > just reinforces my initial point about F=Kx.

Now you're given the force (9.81 x 58) and you're also given X the spring constant K. Rearrange and there you have it sir :smile:
Original post by NEWT0N
Well I definitely remember it having something to do with the flux through the secondary coil.

Ah yes, it's coming back: A changing magnetic field is needed in the secondary coil because (by Faraday's law) the rate of change of flux linkage needs to be nonzero to induce an emf in the secondary coil. A direct current will produce a steady field so rate of change of flux linkage = 0 --> no induced emf, which is bad. Great, thanks!


Hey Isaac,

Can you explain the key points which need to be mentioned when answering how a transformer operates, please

Thanks
I'm confused as ****, aqa shows the exam as wednesday, my friend says wednesday, i got thursday in the back of my book

AQA A2 Physics unit 4, someone confirm for me pls
Original post by ima fail
I'm confused as ****, aqa shows the exam as wednesday, my friend says wednesday, i got thursday in the back of my book

AQA A2 Physics unit 4, someone confirm for me pls


Thursday the 11th of june.

He's messing
ty babes
(edited 8 years ago)
Original post by NEWT0N
Can any smart person inform me as to why an object feels weightless if it is on a planet rotating with a centripetal force equal to the object's weight?


Not entirely sure but this may help

F = mv^2/r F=mg

mg = mv^2/r

g = v^2/r

- could you tell me what question you're talking about plez <3
I cannot fxcking believe im asking this - how on earth do you do Q3
http://filestore.aqa.org.uk/subjects/AQA-PHYA41-QP-JUN14.PDF
Original post by Kennethm
I cannot fxcking believe im asking this - how on earth do you do Q3
http://filestore.aqa.org.uk/subjects/AQA-PHYA41-QP-JUN14.PDF


Mass of water = pv
so 1000 x (2x10^-4) = 0.2 kg


Speed = vol/area

7.2x10^-4 / 2x10^-4 = 3.6m/s


Momentum = 0.2 x 3.6
=0.72


IGNORE THIS
(edited 8 years ago)
Original post by Adangu
Mass of water = pv
so 1000 x (2x10^-4) = 0.2 kg


Speed = vol/area

7.2x10^-4 / 2x10^-4 = 3.6m/s

Momentum = 0.2 x 3.6
=0.72


but the answer is B
Original post by Kennethm
but the answer is B


I now feel incredibly stupid. My teacher marked it right too. My school is just terrible haha.

Apologies.
Original post by Master Sam
Hey Isaac,

Can you explain the key points which need to be mentioned when answering how a transformer operates, please

Thanks


Hey, I know the q wasn't directed to me but this really helped:
http://www.cyberphysics.co.uk/topics/magnetsm/electro/Transfromer/trnsfrmr.htm

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