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AQA Physics PHYA4 - 20th June 2016 [Exam Discussion Thread] watch

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    (Original post by CourtlyCanter)
    Ye
    I wrote something along the lines of-
    Advantage-
    The electric-magnet brake thing doesn't wear out as quickly as the pads so it wouldn't need to be replaced as frequently and so less cost on the long run.
    Disadvantage-
    The electric-magnet brake thing uses EMF which means electrical energy from the car, costs more as fuel is used up quicker.

    I am a bit sceptical on the disadvantage one though.
    I put the same for the advantage, and something about the battery being drained more for a disadvantage.
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    (Original post by Mango Milkshake)
    no

    i put the amplitude has to be lower than 10 degrees
    Same. Because sinθ approximates to θ.
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    (Original post by CourtlyCanter)
    Ye
    I wrote something along the lines of-
    Advantage-
    The electric-magnet brake thing doesn't wear out as quickly as the pads so it wouldn't need to be replaced as frequently and so less cost on the long run.
    Disadvantage-
    The electric-magnet brake thing uses EMF which means electrical energy from the car, costs more as fuel is used up quicker.

    I am a bit sceptical on the disadvantage one though.
    My disadvantage was that the system uses electrical power to operate, therefore if the power supply is lost the breaks fail.
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    (Original post by IMZL)
    Same. Because sinθ approximates to θ.
    I don't remember a question about amplitude, was it in section A or B and what was the question
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    No, they were both positive so considering the charge on the left it would move to the left (Y to X)
    (Original post by Barbecuetime123)
    Wouldn't it be X to y, from the bigger charge to the smaller ?
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    (Original post by aelahi23)
    I don't remember a question about amplitude, was it in section A or B and what was the question
    It was for the 6 mark question and the 1 mark question before it (about the restriction).
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    (Original post by C0balt)
    I can't draw but 1/x graph shape down from to 0 near Venus and then up to 8 whatever

    Idk lol maybe?

    I put V across the resistor is small
    But think when charging resistor the V across the resistor is small when it is fully charged so that would be equivalent to the start of the discharge graph which is when the current is greatest. So wouldnt it be when the voltage across the resistor is large- ie when the current is low?
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    How did you guys find the equation for the energy of the alpha particle?
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    (Original post by tanyapotter)
    if pd across the resistor is small then doesn't it mean that the pd across the capacitor is big? so how would the current be greatest at that point? (since V is proportional to I)

    i got that one wrong probably
    When discharging pd across both capacitor and resistor exponentially decrease
    (Original post by Random1357)
    1/x^2
    Yeah whatever lol
    (Original post by Questioner1234)
    But think when charging resistor the V across the resistor is small when it is fully charged so that would be equivalent to the start of the discharge graph which is when the current is greatest. So wouldnt it be when the voltage across the resistor is large- ie when the current is low?
    See above
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    (Original post by jess.olo)
    same here !!!
    I think we wouldn't lose marks on the six marker as we specified the way to find g from our graph, but I think I'd lose all 3 marks?

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    unofficial mark scheme?
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    (Original post by aelahi23)
    I don't remember a question about amplitude, was it in section A or B and what was the question
    Someting about limitations of the equation for time period, section B, 4a.
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    unofficial mark scheme anyone?
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    Got around 38-40 on Section B. How many would I need on the multiple choice to get a mid-A?


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    (Original post by ak33m98)
    How did you guys find the equation for the energy of the alpha particle?
    (Copied from my last post)

    E = KEnucleus + Ealpha

    V = -mv/N --> V^2 = m^2 v^2 / N^2

    v = -VN/m --> v^2 = V^2 N^2 / m^2

    KEnucleus = 1/2 N V^2 = 1/2 N (m^2 v^2 / N^2) = 1/2 m^2 v^2 / N

    Ealpha = 1/2 m v^2 --> v^2 = 2*Ealpha / mE = 1/2 m^2 v^2 / N + Ealpha

    E = 1/2 m^2 (2*Ealpha / m) / N + Ealpha

    E = m*Ealpha / N + Ealpha

    E = Ealpha (m/N + 1)

    Ealpha = E (1 / (m/N + 1))

    Ealpha = (N / (m + N))*E


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    (Original post by Jonsmith98)
    My disadvantage was that the system uses electrical power to operate, therefore if the power supply is lost the breaks fail.
    I suspect there'll be a backup power supply for that ie when your car just shuts down its system or runs out of battery. You may be right though.
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    (Original post by wallpaperpaste)
    I'm sure in previous papers the correct answer for the energy stored in a capacitor was given by the area under the graph BUT WITH V ON THE Y AXIS RATHER THAN ON THE X AXIS. By this logic shouldn't the energy stored in Q1 be given by the area of the trapezium to the left of the line, rather than the area 'under' the graph?
    I think you're right. When the gradient is 1/C, the energy stored is the area between the line, the limits, and the x-axis.

    I just used the equation E=(1/2)(C)(V^2) and then did the difference so deltaE=(1/2)(C)(12^2-9^2)
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    (Original post by Questioner1234)
    But think when charging resistor the V across the resistor is small when it is fully charged so that would be equivalent to the start of the discharge graph which is when the current is greatest. So wouldnt it be when the voltage across the resistor is large- ie when the current is low?
    I agree with you
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    It was definitely X to Y for that question. The answer is A.
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    Probably wrong: but for advantage: I put braking force can be made larger by increasing field strength

    Disadvantage: There would be more heat loss

    Do you think i'd get one mark?


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