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

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    http://www.tomred.org/uploads/7/7/8/...b_specimen.pdf
    http://www.tomred.org/uploads/7/7/8/...4_specimen.pdf

    Can anyone explain 2b please, how do you know what the phase difference is from the frequency?
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    (Original post by aprocrastinator)
    Can someone please explain this question - the answer is T/2 but I thought the acceleration was in the same direction at this time?
    ...... when t is T/2 , the mass is accelerating upwards towards the centre, the mass in the circle will be at the top, acceleration is downwards towards the centre
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    For SHM 6 marker, what context would be asked??
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    Could someone help me with this please?

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    OMG just done june 2011 paper!! If only it is as wonderful tomorrow!!! I WISH! That's gota be the easiest paper I've done but I don't want to get my hopes up now!!
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    (Original post by _Caz_)
    Could someone help me with this please?

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    Get rid of the constant (4 pi) in the force formula and it should become clearer that the answer is D

    All terms have to be in the same place as normal.


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    (Original post by huniibehi)
    This is the hardest momentum question I've seen, any ideas??

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    mg = 5N on earth, g = 10 so Mass is 0.5

    On the moon, W(orF) = mv/t

    mv/t = mg
    which gives
    v = tg
    v = 3 x (10/6) = 5
    momentum is mass x velocity so 5 (v) x 5 (m) = 25

    Thats what I got
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    (Original post by Sbarron)
    OMG just done june 2011 paper!! If only it is as wonderful tomorrow!!! I WISH! That's gota be the easiest paper I've done but I don't want to get my hopes up now!!
    Same I hit an A* on that paper haha, it was so so nice - but its really tough to say if itll be that easy
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    (Original post by _Caz_)
    Could someone help me with this please?

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    obviously D....just use the formula.... force between charges
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    The electrostatic force of repulsion will be acting in opposite directions to one another. However, this is regarding the electric fields strength. Both charges are positive therefore their electric fields go outwards radially, away from the charge, rather than into the charge creating electric fields.Where the two field lines meet, the the fields line don not intersect and repel each other. The E field strength that the charge possess is constant on all planes. The 4 charge is left because the 6 charge is to the left of it and we require the midpoint
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    Anyone got any tips for the multiple choice questions? Mainly the ones that require algebra manipulation? I often get stuck on those and end up doing the "Whatever looks closet to my answer" or "Makes the most sense" approach.
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    (Original post by a.a.k)
    Emf=Blv
    Emf=emf zero*sin( 2*pi*f*t)

    These are not in the data sheet.

    Anyother plz share

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    Emf = BANwSin(wt) is the exact same as
    Emf = emfzeroSin(2pift)

    Because emfzero is the max EMF induced. max emf is when the Sin function = 1 so that would leave Eo=Banw


    (Original post by CD223)
    Just wanna say thanks to everyone for making this such a lively and buzzing thread since the beginning. It's definitely helped me prepare and I hope it has for everyone else!

    Let's all smash this paper!


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    You're a god. Have put up with me so many times in this thread I can't be more thankful. All the best to you with you offers!
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    what are your offers for where if you don't mind me asking
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    Question 3 on june 2014 multiple choice help please?
    http://filestore.aqa.org.uk/subjects...1-QP-JUN14.PDF
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    (Original post by dominicwild)
    Anyone got any tips for the multiple choice questions? Mainly the ones that require algebra manipulation? I often get stuck on those and end up doing the "Whatever looks closet to my answer" or "Makes the most sense" approach.
    Very good question. If it says something like 'radius is doubled' etc, then formulate an equation and sub in 2r or 2m etc, and then you will find the overall equation is being multiplied by a single constant. so if F= 2m / (2r)sq, then F is being multiplied by x 2/2sq = 2/4 / 1/2 so the new force will be 1/2F. Hope this helps but these questions are the toughest
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    (Original post by AR_95)
    mg = 5N on earth, g = 10 so Mass is 0.5

    On the moon, W(orF) = mv/t

    mv/t = mg
    which gives
    v = tg
    v = 3 x (10/6) = 5
    momentum is mass x velocity so 5 (v) x 5 (m) = 25

    Thats what I got
    Pretty sure it's A ( 2.5kg ).

    On the last line of your calculation you've used 5 instead of 0.5.
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    (Original post by CD223)
    I will try to write down as many answers as I can after the exam. The MC will be hard to.

    Do you know if you can get a hold of a copy of the MC paper?

    Failing that does anyone know if they can? It'll mean we can make a proper mark scheme.


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    I don't think we're allowed to take the MC paper, unfortunately. But we'll remember as much as we can. No pressure because we need to concentrate on the exam instead. Btw, how long do you take doing the MC?
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    (Original post by Adangu)
    Question 3 on june 2014 multiple choice help please?
    http://filestore.aqa.org.uk/subjects...1-QP-JUN14.PDF
    To find momentum, you need mass and velocity.

    Mass is density x volume
    Velocity is distance per second, so in this case the length of a tube of water leaving the hose each second.
    Look at the question and you have everything you need
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    (Original post by Adangu)
    Question 3 on june 2014 multiple choice help please?
    http://filestore.aqa.org.uk/subjects...1-QP-JUN14.PDF
    MOMENTUM = (densityxvolume) X (volume/CSA)
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    [
    (Original post by Mehrdad jafari)
    To be honest i don't know that eitherYeah mathematically that proves it but if you think about it the the change in flux in the coil is no different when it passes horizontally through the magnetic field lines than it passes vertically. In fact if you think about it, the emf induced would have to be greater when the coil passes vertically through the magnetic field lines than when it passes horizontally through the magnetic field lines. But the truth is otherwise, that is the emf induced is greatest when the coil passes horizontally to the magnetic field lines and minimum when it passes vertically through the magnetic field lines. I have read explanations that the change in flux is greatest when the coil passes horizontally through the field lines and minimum when it passes vertically through the field lines. But this cannot be true as the change in flux in the coil is as equal when the coil passes horizontally as when it passes vertically through the field lines.Yeah I totally get the part where emf is max at 90 to field/flux lines as they are perpendicular and more are being cut.... But to me that is the same as maximum flux linkage as it's where the most flux is being cut!But I obviously can't see the difference between maximum flux linkage and maximum flux apart from the number of loops are considered!???!And maths explanations won't help I don't do alevel maths and didn't do GCSEs for years so Ihave only leant the basic physics maths skills requiredPosted from TSR Mobile
    QUOTE=Mehrdad jafari;56851639]To be honest i don't know that eitherYeah mathematically that proves it but if you think about it the the change in flux in the coil is no different when it passes horizontally through the magnetic field lines than it passes vertically. In fact if you think about it, the emf induced would have to be greater when the coil passes vertically through the magnetic field lines than when it passes horizontally through the magnetic field lines. But the truth is otherwise, that is the emf induced is greatest when the coil passes horizontally to the magnetic field lines and minimum when it passes vertically through the magnetic field lines. I have read explanations that the change in flux is greatest when the coil passes horizontally through the field lines and minimum when it passes vertically through the field lines. But this cannot be true as the change in flux in the coil is as equal when the coil passes horizontally as when it passes vertically through the field lines.Yeah I totally get the part where emf is max at 90 to field/flux lines as they are perpendicular and more are being cut.... But to me that is the same as maximum flux linkage as it's where the most flux is being cut!But I obviously can't see the difference between maximum flux linkage and maximum flux apart from the number of loops are considered!???!And maths explanations won't help I don't do alevel maths and didn't do GCSEs for years so Ihave only leant the basic physics maths skills requiredPosted from TSR Mobile[/QUOTE]
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    (Original post by Plasmapause)
    Pretty sure it's A ( 2.5kg ).

    On the last line of your calculation you've used 5 instead of 0.5.
    yeah it's A, just a silly error from my half but the working was fine
 
 
 
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