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    (Original post by Questioner1234)
    Was the capacitance in the first question 8.2 microcoulombs????
    That's what I got (rounded of course).

    And it would be in farads, not coulombs
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    (Original post by tanyapotter)
    I used the equation too. Do you remember what you got for your capacitance?
    I done E=1/2(C)(V)^2 for one voltage then again for another and subtracted
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    (Original post by ombtom)
    Wasn't it V = -mv/N?
    No because they're in opposite direction. So NV - mv = 0


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    (Original post by DesignPredator)
    Here are my answers to Written, the question parts might not be correct.

    Written answers
    1ai) 8.26 (will be a range)
    ii) 2.6x10^-4

    bi)28
    ii)5.9x10^4
    iii) bottom box ticked

    2ai) Derivation
    ii) 6.06x10^6 (3SF only)

    b) Downward Curve staring at 9.81
    Minimum at zero at a point slightly offset from halfway closer to Venus
    Rises up to 8.87

    3a) V=mv/N

    b) Derivation

    ci) 1.15x10^-19 kgms-1 OR Ns

    cii) Anti electron neutrino released also so Beta minus particles have range of Kinetic Energies upon release.

    4a) Small amplitude oscillation as only valid for small angular amplitudes.

    b) -Set up light string with Bob on the end,
    attach top of string to clamp stand boss.
    -Measure length from top to centre of Bob.-Release the Bob so it oscillates with small amplitude.
    -Use stop clock to measure time taken for 20 oscillations.
    -Divide this by 20 for mean T
    -Plot T^2 against L
    -Calc Gradient
    g=(4pi^2)/gradient
    -Repeat procedure

    c) -Student value 4x true value
    -Time period half so all values for T^2 will be 1/4 the true
    -Gradient is 4x lower
    -State equation or show that g is inversely proportional to gradient.

    5a) Lenz's Law: Direction of induced EMF/current is always in a direction that opposes the change that caused it.
    Faraday's Law: Induced EMF is proportional to rate of change of flux linkage.

    b) -Change in flux linkage when current flows.
    -EMF induced in wheel (or whatever it was called)
    -Current induced in the wheel as good conductor.
    -Current carrying conductor (wheel) opposes the field due to coil, therefore there is a force against the direction of motion.

    c) Wheel not in contact with electromagnet so no wear, but in brake pads there is wear due to friction.

    More energy used or less effective or heating.
    Mostly the same
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    (Original post by Jonsmith98)
    I think both answers for increased power drain and electrical failure will be valid. Although my view is that it would be unreasonable to use a powered braking system, because if you park your car on a hill your battery will eventually go flat and then your car will have f***Ed off down the hill by the time you come back with your shopping.
    You must shop for months on end.
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    (Original post by ombtom)
    Wasn't it V = -mv/N?
    Nope. Because V is a magnitude.

    0 = mv - NV

    So:

    NV = mv

    V = mv/N

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    (Original post by shuu00)
    That's what I got (rounded of course).

    And it would be in farads, not coulombs
    Yeah thanks I was just remembering from the y axis (will edit in a second) - did it give the units or did we have to write them?
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    (Original post by duncant)
    what did people say for the effect when that guy recorded only half of T? was it (1/4)g?
    i said they measured 4g
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    (Original post by kingaaran)
    What was the unit - I worked it out to be C^2 kg^(-2)?
    Was it A ?
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    (Original post by Yo12345)
    No because they're in opposite direction. So NV - mv = 0
    (Original post by Euclidean)
    Nope. Because V is a magnitude.
    But they're velocities not speeds, so V would have to be -v x something. I'm dead rn so you're probably right
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    did people get the resuklt for g would be 4x larger not less
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    (Original post by ombtom)
    Wasn't it V = -mv/N?
    Arrow was pointing left for V and arrow for v was right.
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    (Original post by CourtlyCanter)
    Was it A ?
    yeah
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    How many marks would I lose on the first proof momentum one for thinking N is the original nucleus mass? (3 marker)
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    (Original post by Jordenwilder1998)
    did people get the resuklt for g would be 4x larger not less
    yeah
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    (Original post by Jordenwilder1998)
    did people get the resuklt for g would be 4x larger not less
    because g iversly proportional to t and t is 4 ti,mes smaler
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    (Original post by ombtom)
    Wasn't it V = -mv/N?
    Yeah -mv because it was a velocity so had to be going in the opposite direction to conserve momentum!
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    (Original post by shuu00)
    That's what I got (rounded of course).

    And it would be in farads, not coulombs
    i got 8microfards but i have feeling i might be out of range
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    (Original post by CourtlyCanter)
    Okay assuming I'd completed the question thinking N was that of the original, would I only lose marks on the 4-marker proof question or any of the ones following? Do you guys remember what the follow-up ones were?
    For some dumb ass reason I thought N was the mass of the original nucleus too (think I saw a similar question before and went into autopilot). I think marks will be lost for that equation question and obviously the proof as you can't do it with the incorrect equation. Maybe get some method marks for correct expressions?

    The next part was working out the kinetic energy of the alpha particle I think, and you'll lose marks if you sub N in as the mass of the original nucleus. Surprisingly, I subbed in the mass of the daughter nucleus for N so god knows why I didn't go back and realise why I couldn't do the proof lol.
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    (Original post by TSRPAV)
    i got 8microfards but i have feeling i might be out of range
    Mine was 8.1666666 (6 recurring) microfarads.
 
 
 
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