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

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    (Original post by 000alex)
    Equation for electric potential in a uniform field?
    V = q /(4*pi*epsilon*r)

    It's in the data sheet.
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    (Original post by lawson_g)
    Does everyone reckon SHM for the 6 marker?
    I hope not. What would it be if it was about SHM? Like damping and stuff?
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    (Original post by MsFahima)
    V = q /(4*pi*epsilon*r)

    It's in the data sheet.
    Isn't that only for a radial field?


    Tried that equation for this question and it didn't work
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    (Original post by JackRowbotham)
    june 13 question 1.c) ... help?
    also june 14 question 3 on the multiple choice?
    pleasssssse:-)


    any tips for the 6 marker?
    June 14 Q3: everything is in terms of seconds. Calculate the mass of water leaving per second using density formula (mass ÷ volume) as you know the volume and density. M works out to be 0.2 kg/s. Now to work out momentum you meed toe velocity of the water leaving per scond. You know the volume and the cross-area so use this to work out the length of water leaving per second which is equal to the speed ( volume ÷ area ) then the speed is 0.27. Mv is momentum so 0.27 x 0.2 = 0.055555 i.e 5.6×10^-2
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    (Original post by ivyb)
    GPE inscreases (i.e. becomes less negative) since you have to put work in against the gravitational force (gains energy) to move it further from the planet.
    Gpe does increase but does not become less negative, potential becomes less negatve the further away you go from a planet/body. This increase in gpe is equivalent to work done by an external force
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    (Original post by Mehrdad jafari)
    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
    It's not really about whether you are doing maths or not. I'm doing a level maths too but i still haven't applied that maths anywhere in physics. The maths does help but that's where we have an idea of what's going on and not where we have no idea, lol


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    Trust me if you try and start a physics A level not knowing ANY maths it makes a HUGE difference! It's full of maths... maybe what some would call 'basic' but I started not even knowing how to rearrange one equation!! And then there's exponentials, logs, the list goes on.... I've had to get a private maths tutor once a week just for the 'maths' part! But hopefully I'm mostly there now!
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    (Original post by JackRowbotham)
    june 13 question 1.c) ... help?
    also june 14 question 3 on the multiple choice?
    pleasssssse:-)


    any tips for the 6 marker?
    And also question 1c. There is 2 ways of doing it dependent on which time period you use. I.e you can use the smaller time period (1.9) and divide this by theire difference (0.1) to get 19 swings until they are in phase. Then 19 swings times the period of the larger time period (2.0) equals 38 seconds. Or the other way is if you use the larger period initially (2.0) and divide this by their difference (0.1) to get 20 swings, then times this by the smaller time period (1.9) to again get 38 seconds
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    (Original post by DannySmith420)
    June 14 Q3: everything is in terms of seconds. Calculate the mass of water leaving per second using density formula (mass ÷ volume) as you know the volume and density. M works out to be 0.2 kg/s. Now to work out momentum you meed toe velocity of the water leaving per scond. You know the volume and the cross-area so use this to work out the length of water leaving per second which is equal to the speed ( volume ÷ area ) then the speed is 0.27. Mv is momentum so 0.27 x 0.2 = 0.055555 i.e 5.6×10^-2
    you're a star!
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    (Original post by 000alex)
    Isn't that only for a radial field?


    Tried that equation for this question and it didn't work
    Potential is constant in a uniform field.
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    How are the graphs related for induced emf and flux linkage?
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    (Original post by MsFahima)
    E = V/D

    And it's an electron so it will be attracted to the 50v plate.
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    (Original post by donutellme)
    That's field strength...

    Potential is constant in a uniform field

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    oops. I understand what I've done wrong!
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    Can anyone please explain how a transformer works please?
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    (Original post by MsFahima)
    oops. I understand what I've done wrong!
    Oh sorry... :P it looked similar to another question asked earlier....

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    (Original post by donutellme)
    Oh sorry... :P it looked similar to another question asked earlier....

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    No no. I edited it. My mistake.

    Btw, that question.. what would the potential be? 50v?
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    (Original post by MsFahima)
    Potential is constant in a uniform field.
    Why is it 25?
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    (Original post by EloiseFH)
    How are the graphs related for induced emf and flux linkage?
    90 degree out if phase i think
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    (Original post by 000alex)
    Why is it 25?
    I'm not sure.. :/

    donutellme help please?
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    could someone help me on the q17 and q12 on the jan 2013 paper
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    (Original post by 000alex)
    Why is it 25?
    Sorry, I thought it was a different question.

    Field strength is the gradient of the potential. So in a uniform field, in this case 50, the potential goes from 50 to 0. That means half way along it's gonna be 25.

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