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    (Original post by Pouh)
    Was it one sig fig or 2 i first had 0.875 then put it to 0.90
    I put it as 0.9 as well... I hope that's right
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    (Original post by champ_mc99)
    Are we all sure the grade boundaries will be lower than last year? Last year was the lowest it's been.

    It probably would be significantly higher- compare it to the grade boundaries when the old spec was brought in
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    (Original post by SaltandSugar)
    Why zero? wasnt there 4 resistors in total or something and they were difference along each parallel branch?
    I dunno, i said zero cos there were no components in parallel with the voltmeter to begin with at that point
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    (Original post by LivToBe)
    It probably would be significantly higher- compare it to the grade boundaries when the old spec was brought in
    Uhhh... except the old spec is 100x easier.
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    (Original post by BainesyA)
    Hey guys, first post in this thread, quick question...

    Did anyone else find the whole paper relatively forgiving except for the electricity questions?

    (Also, ticking a box for a final question made me smile)
    Same. It was easier than the specimen papers, at least
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    Does anyone know when an unofficail mark scheme is on its way?
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    (Original post by Pra99)
    Is the paper out of 60?
    70
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    (Original post by studentabcde)
    Same. It was easier than the specimen papers, at least
    I just can't seem to get my head around electricity. Or is that because I revised the whole of the electricity unit in 3 hour last night between 9pm and midnight? :congrats:

    But I think it went okay, the worst part is when you come out of the exam and then people are screaming their answers left, right and centre. "Did you get this?" "What did you get for question 1,7,99, 3million and 29?" They just lower my confidence
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    (Original post by o.s.s15)
    Did anyone get 2.6 for deceleration
    It was 2.7ms^-2, using F=ma

    And the question after it told us the speed when it touched the runway was 68ms^-1, you know a and you know it goes until its at rest
    So

    s = s

u = 68ms^{-1}

v = 0ms^{-1}

a = -2.71ms^{-2}

t = t

    You can ignore t here, it's irrelevant

    Using v^{2} = u^{2} + 2as

     0 = 68^{2} + 2(-2.71)s

     \therefore 5.42s = 68^{2}

     s = \dfrac{68^{2}}{5.42} = 853.13 = 853m
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    (Original post by Gurkin)
    I dunno, i said zero cos there were no components in parallel with the voltmeter to begin with at that point
    Was the set up something like this? http://i.imgur.com/bE1Yv5V.png

    Wouldnt there be a potential difference as the resistors were different?

    I'm not sure tbh I guessed too
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    (Original post by BainesyA)
    I just can't seem to get my head around electricity. Or is that because I revised the whole of the electricity unit in 3 hour last night between 9pm and midnight? :congrats:

    But I think it went okay, the worst part is when you come out of the exam and then people are screaming their answers left, right and centre. "Did you get this?" "What did you get for question 1,7,99, 3million and 29?" They just lower my confidence
    Yeah same. probably shouldn't be going on TSR right now lol
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    (Original post by SaltandSugar)
    Was the set up something like this? http://i.imgur.com/bE1Yv5V.png

    Wouldnt there be a potential difference as the resistors were different?

    I'm not sure tbh I guessed too
    Yes, the only reason i think the answer may not be as simple as saying the pd is zero is because it mentioned the resistance of one of the components in the question
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    (Original post by studentabcde)
    Yeah same. probably shouldn't be going on TSR right now lol
    I must admit, the majority of TSR answers ring a bell in my head and are similar to what I put, my mates where trying to tell me that the chocolate melted at the nodes

    I thought, that surely can't be right, and now TSR have confirmed that it was indeed the antinodes where the chocolate melts.
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    (Original post by edothero)
    It was 2.7ms^-2, using F=ma

    And the question after it told us the speed when it touched the runway was 68ms^-1, you know a and you know it goes until its at rest
    So

    s = s

u = 68ms^{-1}

v = 0ms^{-1}

a = -2.71ms^{-2}

t = t

    You can ignore t here, it's irrelevant

    Using v^{2} = u^{2} + 2as

     0 = 68^{2} + 2(-2.71)s

     \therefore 5.42s = 68^{2}

     s = \dfrac{68^{2}}{5.42} = 853.13 = 853m
    I got a rounding error how many marks would I lose. I got 856
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    (Original post by BainesyA)
    I must admit, the majority of TSR answers ring a bell in my head and are similar to what I put, my mates where trying to tell me that the chocolate melted at the nodes

    I thought, that surely can't be right, and now TSR have confirmed that it was indeed the antinodes where the chocolate melts.
    I crossed out nodes and changed it to antinodes when i was checking through. so glad now!
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    (Original post by 11234)
    I got a rounding error how many marks would I lose. I got 856
    Not sure bro
    Probably 1?
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    (Original post by edothero)
    It was 2.7ms^-2, using F=ma

    And the question after it told us the speed when it touched the runway was 68ms^-1, you know a and you know it goes until its at rest
    So

    s = s

u = 68ms^{-1}

v = 0ms^{-1}

a = -2.71ms^{-2}

t = t

    You can ignore t here, it's irrelevant

    Using v^{2} = u^{2} + 2as

     0 = 68^{2} + 2(-2.71)s

     \therefore 5.42s = 68^{2}

     s = \dfrac{68^{2}}{5.42} = 853.13 = 853m
    Can I ask why you haven't rounded the answer to 850, given that the value of 'u' was to two significant figures?
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    The max sig fig in that question was 2 so i left my answer as 850
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    (Original post by studentabcde)
    I crossed out nodes and changed it to antinodes when i was checking through. so glad now!
    Antinodes make sense, by all means it could be nodes, but that would mean the majority of TSR is wrong...
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    (Original post by haes)
    I don't remember seeing that here so will those be for the second paper?
    well its supposed to be every paper. Here I copied this from the paper 1 specimen mark scheme

    1. Significant figure penalties An A-level paper may contain up to 2 marks (1 mark for AS) that are contingent on thecandidate quoting the final answer in a calculation to a specified number of significantfigures (sf). This will generally be assessed to be the number of sf of the datum with theleast number of sf from which the answer is determined. The mark scheme will give therange of sf that are acceptable but this will normally be the sf of the datum (or this sf -1).The need for a consideration will be indicated in the question by the use of ‘Give youranswer to an appropriate number of significant figures’. An answer in surd form cannotgain the sf mark. An incorrect calculation following some working can gain the sf mark.
 
 
 
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