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    i got the time constant RC to be 35.4s, the time constant of a discharging capacitor is the time it drops to 63% its original value, since the time to drop 80% was 45s, 1% = 45/80 = 0.5625 .... 63% = 0.5625 x 63 = 35.4 seconds. but no one on this thread seems to have done that, even though i knew i could have used the equation, but this method seemed more legit in my mind. how many marks will i have lost?
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    (Original post by tiggins)
    i got the time constant RC to be 35.4s, the time constant of a discharging capacitor is the time it drops to 63% its original value, since the time to drop 80% was 45s, 1% = 45/80 = 0.5625 .... 63% = 0.5625 x 63 = 35.4 seconds. but no one on this thread seems to have done that, even though i knew i could have used the equation, but this method seemed more legit in my mind. how many marks will i have lost?
    Not sure if it would be worth any marks because it doesn't really show the equation being used or understanding that V=0.2Vo. Sorry
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    (Original post by particlestudent)
    I got -3Q/8 for electric potential, probably wrong
    And zero for that square one
    Do you still remember the question to -3Q/8 thing, both -3Q and -5Q look familiar to me somehow
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    (Original post by Vikingninja)
    Did anyone get the question where you had to show that the energy of the alpha particle was the total energy x some fraction, no one I've talked to at college has got it.
    I did.

    You had to do E = E (alpha) + E (nucleus)

    = 1/2mv^2 + 1/2m^v^2/N
    = 12mv^2 (1 + m/N)

    Take the brackets over to the other side ----> E/(1+m/N) = 1/2mv^2

    Combine the bracketed terms into one fraction.

    E/((N+m)/N) = E (N/m+N) = 1/2 mv^2

    As 1/2mv^2 = E (alpha), E (N/m+N) also equals E (alpha).

    Thought this was a solid paper in general. Guessed around 7 MCQ's (desperately ran out of time). Hopefully no more and hopefully dropped no more than 7/8 in section B.
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    (Original post by hahaox768)
    Do you still remember the question to -3Q/8 thing, both -3Q and -5Q look familiar to me somehow
    I don't unfortunately. I am 100% sure the answer was -3Q/8pi(e)(r) though. My guess is people that got -5Q forgot to make the denominators the same when adding.
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    (Original post by tiggins)
    i got the time constant RC to be 35.4s, the time constant of a discharging capacitor is the time it drops to 63% its original value, since the time to drop 80% was 45s, 1% = 45/80 = 0.5625 .... 63% = 0.5625 x 63 = 35.4 seconds. but no one on this thread seems to have done that, even though i knew i could have used the equation, but this method seemed more legit in my mind. how many marks will i have lost?
    I did this lol, 3 marks I think.
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    (Original post by Simple Harmonic)
    I did.

    You had to do E = E (alpha) + E (nucleus)

    = 1/2mv^2 + 1/2m^v^2/N
    = 12mv^2 (1 + m/N)

    Take the brackets over to the other side ----> E/(1+m/N) = 1/2mv^2

    Combine the bracketed terms into one fraction.

    E/((N+m)/N) = E (N/m+N) = 1/2 mv^2

    As 1/2mv^2 = E (alpha), E (N/m+N) also equals E (alpha).

    Thought this was a solid paper in general. Guessed around 7 MCQ's (desperately ran out of time). Hopefully no more and hopefully dropped no more than 7/8 in section B.
    How many marks do you think I got for doing you first 2 steps? 2 marks?


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    in one of the capacitance questions did I have to work out area under the graph between two points (area of trapezium)??
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    (Original post by perpetua)
    So what u reckon for the gb
    Defo less than June 2014, perhaps 67 for max ums...
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    (Original post by tiggins)
    i got the time constant RC to be 35.4s, the time constant of a discharging capacitor is the time it drops to 63% its original value, since the time to drop 80% was 45s, 1% = 45/80 = 0.5625 .... 63% = 0.5625 x 63 = 35.4 seconds. but no one on this thread seems to have done that, even though i knew i could have used the equation, but this method seemed more legit in my mind. how many marks will i have lost?
    Not sure if you'd get any sorry. The decay is exponential whereas you're working shows you've presumed it to be linear

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    (Original post by Questioner1234)
    Do you remember how many marks this and the one before it was? I got the final answer but accidently used the symbol M instead of N and just changed it at the end and I didn't have time to change all my working? Not sure if I'll lose all the marks or just a few or none?
    4 marks for the proof, 2 for the momentum one.

    Damn, just read that it was 80% LOST in the capacitance question, there's another 2/3 marks gone down the drain.
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    (Original post by Yo12345)
    How many marks do you think I got for doing you first 2 steps? 2 marks?


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    2 at least.
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    (Original post by Sharpify)
    Shouldn't have needed to, the answer was in micro joules IIRC.
    You are right. I remember converting my answer back to 8.1 as the unit given on the line was in micro already.
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    Hey is there an unofficial mark scheme up yet? x
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    Any chance that the gradient range would stretch to 8.4? Not sure why mine was 0.2 out
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    (Original post by tiggins)
    i got the time constant RC to be 35.4s, the time constant of a discharging capacitor is the time it drops to 63% its original value, since the time to drop 80% was 45s, 1% = 45/80 = 0.5625 .... 63% = 0.5625 x 63 = 35.4 seconds. but no one on this thread seems to have done that, even though i knew i could have used the equation, but this method seemed more legit in my mind. how many marks will i have lost?
    None unfortunately, the time constant is the time taken to drop to 37% of the initial value!
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    (Original post by ¡Muy bien!)
    same for me apart from s4 and chemistry. Also I'm doing turning points. Good luck!
    You too!
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    (Original post by cowie)
    in one of the capacitance questions did I have to work out area under the graph between two points (area of trapezium)??
    Guys was the energy stored the area of the triangle or trapezium?


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    would dropping 20 marks give me an A?
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    (Original post by Yo12345)
    Guys was the energy stored the area of the triangle or trapezium?


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    Well, if the graph started at (0,0) you'd work out the area of the larger triangle (the one at 12 V) and subtract from that the area of the smaller one (9V) and the remainder is a trapezium, right?
 
 
 
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