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

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    (Original post by Sbarron)
    [
    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]


    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
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    Would you still get full marks for this question if you did not include any of the words in brackets in your answers?
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    r=mv/bq is the same as r=p/bq

    If p(momentum) is doubled, does that mean r=2m2v/bq or r=2mv/bq???
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    (Original post by palaseum9)
    For SHM 6 marker, what context would be asked??
    could be resonance and damping, or maybe some experiment with a spring/mass or pendulum, either to estimate g or estimate the length of the pendulum (just a guess)
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    (Original post by IWantSomeMushu)
    Would you still get full marks for this question if you did not include any of the words in brackets in your answers?
    yeah you would get full marks
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    (Original post by Lau14)
    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
    (Original post by Dante991)
    MOMENTUM = (densityxvolume) X (volume/CSA)
    Thanks you guys, sorry if this is dumb but what is the volume?

    When I did it before i had 0.72 but the answer is 0.2
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    (Original post by AR_95)
    r=mv/bq is the same as r=p/bq

    If p(momentum) is doubled, does that mean r=2m2v/bq or r=2mv/bq???
    r = 2mv/bq, momentum = mv therefore 2 x momentum = 2mv
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    (Original post by CD223)
    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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    thanks for your help I see it now! Think I'm getting a little tired

    (Original post by DCMed96)
    obviously D....just use the formula.... force between charges
    I'm glad you feel so comfortable doing this exam. What's 'obvious' to you might not be obvious to other people.
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    (Original post by _Caz_)
    Could someone help me with this please?

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    What paper is it
    • Welcome Squad
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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
    Hi.

    Basically we need to find momentum which is mass x velocity.

    We can find mass as we have density and volume.

    Then we have the Area. We can use to find the length of the pipe to as A X L = V

    The length gives use the speed of the water coming out.

    Hence, v * m gives us the answer.
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    (Original post by Adangu)
    Thanks you guys, sorry if this is dumb but what is the volume?

    When I did it before i had 0.72 but the answer is 0.2
    Volume is actually just 2x10^-4. You get a hint from the units, its m^3 PER SECOND. and the answer they want is per second
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    2p =2(mv)=2mv
    [QUTE=AR_95;56860843]r=mv/bq is the same as r=p/bq

    If p(momentum) is doubled, does that mean r=2m2v/bq or r=2mv/bq???[/QUOTE]
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    (Original post by moeeahmed)
    What paper is it
    It's june 10. I've managed it now though
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    (Original post by _Caz_)
    thanks for your help I see it now! Think I'm getting a little tired



    I'm glad you feel so comfortable doing this exam that you feel like you need to talk down at others. What's 'obvious' to you might not be obvious to other people so sit down please. This is a study help thread not a study brag thread.
    amen
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    Hey guys does anyone know how to calculate the time period for this:
    January 2011



    The correct answer is B but my calculations can never get that answer.
    Apparently we are supposed to use ω2r = 9.81

    Thanks
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    (Original post by thedon96)
    for SHM (the left one), acceleration always acts towards equilibrium and for circular motion, acceleration acts towards the centre (centripetal acceleration). At the initial time therefore, acceleration for the mass-spring system is downwards and the particle in circular motion is upwards because there's where the centre is. After half an oscillation, the particle in circular motion will be at the top and the mass on the spring will be at the bottom so opposite accelerations. Hope this helps
    Thank you that explanation was great but I realised I read the question wrong I read it as the mass was at its lowest point for some reason??? Which is why I was confused, but you explained it great anyway!
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    Can anyone please help me with question's 19, 20 & 21 on the June 14 paper. Thanks
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    Can someone pls help me on this have no clue
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    (Original post by saad97)
    Can anyone please help me with question's 19, 20 & 21 on the June 14 paper. Thanks
    yeah me too, 19 was just like what
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    (Original post by MsFahima)
    Hi.

    Basically we need to find momentum which is mass x velocity.

    We can find mass as we have density and volume.

    Then we have the Area. We can use to find the length of the pipe to as A X L = V

    The length gives use the speed of the water coming out.

    Hence, v * m gives us the answer.
    (Original post by Dante991)
    Volume is actually just 2x10^-4. You get a hint from the units, its m^3 PER SECOND. and the answer they want is per second
    I was being dumber. I kept getting 0.056 which is the right answer but for some reason I convinced myself 0.2 was the answer. Sorry to have wasted your time haha! my brain is just mush
 
 
 
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