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

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    (Original post by NEWT0N)
    In JAN10 4bii why does it use the input power of 3W rather than the secondary power of 2.7W? I thought the power wasted due to the standby mode is 2.7W (from 4bi).

    Edit: Typo (now corrected in bold)
    Anyone? D:
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    (Original post by gcsestuff)
    14 in January lol


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    I got 23 in that one! One of my better ones


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    (Original post by NEWT0N)
    Dang should've revised transformers earlier, boring as hell.
    What exactly do we need to know about them?!


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    (Original post by Kennethm)
    woah all these replies - I did it a long time ago and did it for the first time today basically going into it without remembering what it was before and I must say this is certainly the most difficult MC I've done. The others pale in comparison and I nearly ace the others...

    Im sure its been posted here before but could someone explain Q13 from that paper please? Its here : http://filestore.aqa.org.uk/subjects...1-QP-JUN14.PDF

    Cannot wrap my head around it.
    For convenience let the charges be x and y and let k=\frac{1}{4\pi \varepsilon_0} (because we suspect we won't need the exact value since this is essentially a "ratios" exercise. By Coulomb's law we have F=\frac{kxy}{d^2}. We are also given that 0.5F=\frac{kxy}{20^2} (where all distances are in mm). Therefore substituting the first equation into the second one we get 0.5(\frac{kxy}{d^2})=\frac{kxy}{  20^2}, hence \frac{0.5}{d^2}=\frac{1}{20^2}. Now take square roots and stuff and solve for d.

    EDIT: Whoops silly mistake, now edited.
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    ??
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    (Original post by Fred Cantoni)
    What exactly do we need to know about them?!


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    The jan2010 has everything on power losses in a transformer. Another paper has stuff on power transmission via national grid. Then you just need to know the basics like how a transformer works and that's all I think.
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    (Original post by AR_95)



    ??
    EQ=ma

    Now solve as you know the charge and mass of an electron!


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    (Original post by AR_95)



    ??
    By definition F=Eq=1.5\times 10^5\times 1.6\times 10^{-19}=2.4\times 10^{-14}, hence F=ma=2.4\times 10^{-14}. Since this is an electron m=9.11\times 10^{-31}. Plug in and solve to get a.

    EDIT: Sniped
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    (Original post by NEWT0N)
    The jan2010 has everything on power losses in a transformer. Another paper has stuff on power transmission via national grid. Then you just need to know the basics like how a transformer works and that's all I think.
    Alright cheers, I'll have another look at the power losses!


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    (Original post by NEWT0N)
    For convenience let the charges be x and y and let k=\frac{1}{4\pi \varepsilon_0} (because we suspect we won't need the exact value since this is essentially a "ratios" exercise. By Coulomb's law we have F=\frac{kxy}{d^2}. We are also given that 0.5F=\frac{kxy}{20^2} (where all distances are in mm). Therefore substituting the first equation into the second one we get 0.5(\frac{kxy}{d^2})=\frac{kxy}{  20^2}, hence \frac{0.5}{d^2}=\frac{1}{20^2}. Now take square roots and stuff and solve for d.

    EDIT: Whoops silly mistake, now edited.
    you sir are an angel! However - what part of the question implies that it's d+20x10^3 for the 0.5F part?
    EDIT: nvm lol
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    (Original post by NEWT0N)
    By definition F=Eq=1.5\times 10^5\times 1.6\times 10^{-19}=2.4\times 10^{-14}, hence F=ma=2.4\times 10^{-14}. Since this is an electron m=9.11\times 10^{-31}. Plug in and solve to get a.

    EDIT: Sniped
    (Original post by Jimmy20002012)
    EQ=ma

    Now solve as you know the charge and mass of an electron!


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    FFS that's simple as ****, it's always the basic ones that get me
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    Two pendulums, P and Q, are set up alongside each other. The period of P is 1.90s andthe period of Q is 1.95s. How many oscillations are made by pendulum Q between two consecutive instantswhen P and Q move in phase with each other?

    any quick method for working these out?
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    (Original post by Fred Cantoni)
    Alright cheers, I'll have another look at the power losses!


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    No worries.

    This is what I know about power losses (most of it is from JAN10 paper):

    When a transformer is in operation AC current flows in the primary and secondary coil. Since the coils have some resistance this causes some power to be lost in the coils, which is dissipated to the surroundings as heat. To avoid this you can use thick copper wire for the coils, which has a low resistivity and therefore a low resistance. Thicker wire should be used in the secondary coil than in the primary coil due to the higher current in the secondary coil.

    The AC current in the transformer core continuously magnetises, demagnetises and remagnetises the core in opposite directions. Work is done in building up the magnetic field because energy is needed to line up all the domains of the core's material (technical term). The energy recovered as the magnetic field falls is less than the energy put in. This is known as a hysteresis loss and is due to the core heating up. One way of minimizing this is to use a material for the core that magnetises and demagnetises easily such as soft iron.

    (Note: There are many other sources of heat losses such as eddy currents)
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    (Original post by Kennethm)
    you sir are an angel! However - what part of the question implies that it's d+20x10^3 for the 0.5F part?
    EDIT: nvm lol
    Yeah silly mistake, no idea what I was thinking there
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    (Original post by Corectspelling)
    Two pendulums, P and Q, are set up alongside each other. The period of P is 1.90s andthe period of Q is 1.95s. How many oscillations are made by pendulum Q between two consecutive instantswhen P and Q move in phase with each other?

    any quick method for working these out?
    Answered on a previous page but can't find it atm.

    When they are next in phase the faster one must have undergone exactly one more complete oscillation than the slower one (it can't be the other way round for obvious reasons). Therefore if we assume the slower one has undergone X oscillations then the faster one has undergone X+1 oscillations. So X*1.95=(X+1)*1.9, then solve for X to get X=38.
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    if some one could kindly explain





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    (Original post by AR_95)
    if some one could kindly explain





    19) Answer: D.
    By Fleming's left hand rule the force acts at right angles to both the conventional current (opposite the velocity of the electron) and the field. Thus at any rate the force is at right angles to the field.

    21) Answer: A.
    \Phi=BAN\cos \theta.
    At \theta=50^o we have \Phi=(2.8\times 10^{-2})(4.2\times 10^{-3})(50)(\cos 50^o)=3.779... Wb\ turns.
    At \theta=0^o we clearly just have \Phi=BAN, which is simply the previous answer divided by \cos 50^o (since previous answer = BAN\cos 50^o). Therefore \frac{3.779...}{\cos 50^o}=5.88\times 10^{-3}Wb\ turns.
    It follows that \Delta\Phi=5.88\times 10^{-3}-3.779...\times 10^{-3}=2.1\times 10^{-3}Wb\ turns\ (2\ s.f), as claimed.

    24) Answer: Not entirely sure about this one but I think it's A.
    Step up transformer so N_p<N_s. For any transformer at all it's obvious that P_{out}\le P_{in} (with equality if and only if it is 100% efficient). Therefore I_sV_s\le I_pV_p or \frac{I_s}{I_p}\le \frac{V_p}{V_s}=\frac{N_p}{N_s}<  1 where the final inequality follows from the fact that N_p<N_s (see above). We also have \Phi\propto N since B,A and theta are all constant in this situation (and \Phi=BAN\cos \theta). Therefore \frac{\Phi_s}{\Phi_p}=\frac{N_s}  {N_p}>1.
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    (Original post by NEWT0N)
    19) Answer: D.
    By Fleming's left hand rule the force acts at right angles to both the conventional current (opposite the velocity of the electron) and the field. Thus at any rate the force is at right angles to the field.

    21) Answer: A.
    \Phi=BAN\cos \theta.
    At \theta=50^o we have \Phi=(2.8\times 10^{-2})(4.2\times 10^{-3})(50)(\cos 50^o)=3.779... Wb\ turns.
    At \theta=0^o we clearly just have \Phi=BAN, which is simply the previous answer divided by \cos 50^o (since previous answer = BAN\cos 50^o). Therefore \frac{3.779...}{\cos 50^o}=5.88\times 10^{-3}Wb\ turns.
    It follows that \Delta\Phi=5.88\times 10^{-3}-3.779...\times 10^{-3}=2.1\times 10^{-3}Wb\ turns\ (2\ s.f), as claimed.

    24) Answer: Not entirely sure about this one but I think it's A.
    Step up transformer so N_p<N_s. For any transformer at all it's obvious that P_{out}\le P_{in} (with equality if and only if it is 100% efficient). Therefore I_sV_s\le I_pV_p or \frac{I_s}{I_p}\le \frac{V_p}{V_s}=\frac{N_p}{N_s}<  1 where the final inequality follows from the fact that N_p<N_s (see above). We also have \Phi\propto N since B,A and theta are all constant in this situation (and \Phi=BAN\cos \theta). Therefore \frac{\Phi_s}{\Phi_p}=\frac{N_s}  {N_p}>1.
    24 is A yep.

    21) that was my initial approach of doing things but I didn't follow through unfortunately
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    (Original post by Fred Cantoni)
    I got 23 in that one! One of my better ones


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    Well done mate proud of you have a gold star


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    (Original post by Corectspelling)
    Two pendulums, P and Q, are set up alongside each other. The period of P is 1.90s andthe period of Q is 1.95s. How many oscillations are made by pendulum Q between two consecutive instantswhen P and Q move in phase with each other?

    any quick method for working these out?
    The difference between them is 0.05

    If you want to find Q you divide P by 0.05 and that will give you the number

    Don't know how it works though lol


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