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

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    How is everyone spending tomorrow revising? It's my first full day before the exam focusing on PHYA4 -,-


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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.
    Wouldn't 21 be B) ? You're increasing flux linkage.
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    (Original post by CD223)
    How is everyone spending tomorrow revising? It's my first full day before the exam focusing on PHYA4 -,-


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    In my chemistry exam :/ which is also in the afternoon, for bonus annoyingness for physics! And I have come down the the cold from hell so hopefully the revision I've done so far will be enough with a quick recap tomorrow evening
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    I'm lost...
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    Power loss stuff that you need to know:

    1

    The bindings around the core need to be of low resistance as possible. As when the large AC current flows a hearing effect in the current occurs. Using thick copper is best. This is most important on the higher current side of the transformer (so in a step up voltage is less on the primary coil therefore current must be more, as power has to be conserved.


    2

    The AC current de magnitizes and magnitizes the core continuously in opposite directions due to the AC current traveling around the core. This means that the metal need to be magnetically soft so that no energy Is lost when the magnetic flow reverses. Iron is used.

    3

    Core design- if there is a large distance between the two coils this can be inefficient and energy wasted. By better design, such as placing the coils on top of each other this can be reduced.

    4

    To prevent the phenomenon of eddy currents the core needs to be laminated. As the magnetic flux passing through the core is changing continuously the core is being cut by flux continuously changes emf. This heats the core. Separating the magnetic core prevent unnecessary flux being cut

    Hope this help! Hope they come up too! Just done this from memory


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    (Original post by IWantSomeMushu)
    I'm lost...
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    (Original post by Whateverisbest)
    Wouldn't 21 be B) ? You're increasing flux linkage.
    Whoops yeah you're right, I didn't read the "increases" and "decreases" parts because I was rushing so I just saw "2.1" and immediately wrote down A. :P
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    (Original post by Lau14)
    In my chemistry exam :/ which is also in the afternoon, for bonus annoyingness for physics! And I have come down the the cold from hell so hopefully the revision I've done so far will be enough with a quick recap tomorrow evening
    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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    (Original post by NEWT0N)
    I think we need to know about light, heavy, critical and overdamping. It's good to know about overdamping even if it's possibly not on the spec (think about polar orbits not being on the spec but you still get questions about them here and there).

    Light: The amplitude of oscillation reduces slowly over time.
    Heavy: Amplitude of oscillation reduces more quickly over time than for a lightly damped system
    Critical: The system doesn't oscillate but returns to its equilibrium position in the shortest possible time (in less than one cycle, due to no oscillation)
    Overdamping: The system doesn't oscillate but also doesn't return to its equilibrium position. This is why you would perhaps not want to overdamp a car suspension system (car won't return to equilibrium, which is bad). You can think of this as a pendulum moving through a thick gooey substance from maximum displacement

    Those are the general ideas anyway.
    Thanks a lot!
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    Thanks!
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    Hey again guys hahah, can someone explain q15 of june 13 section A?! I think i may just being an idiot but I can't figure it out!
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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
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    (Original post by IWantSomeMushu)
    I'm lost...
    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 )
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    Could someone please explain question 3 a part ii from this paper please https://60abffc9b401b1c0936e01291c15...%20Physics.pdf
    I just can't understand it
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    (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!!!!
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    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...
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    (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!!!!
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    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 Sbarron)
    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!!!!
    Well that seems counter-intuitive, are you sure it's true?

    I would think mg acts vertically down and tension acts up, so centripetal force would be t-mg.

    Edit: nvm, see gcsemaths' description
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    (Original post by GANGA_SLAYER)
    Could someone please explain question 3 a part ii from this paper please https://60abffc9b401b1c0936e01291c15...%20Physics.pdf
    I just can't understand it
    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
 
 
 
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