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    (Original post by PawanAviator)
    i did take one value because it was the away i did it back when i done GCSE maths

    yeah , you are probably right because it is taught at the start of the book.
    ahhh i think i see where you are coming from now
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    (Original post by james1221212)
    your absolutely wrong mate, it is 22.5 because it was linear and straight at the end and had the steepest gradient, 10-8=2 and 170-125=45 so as your 'a* ' would tell you that equates to a speed of 22.5 m/s
    Well , it is weird how i can't get 22.5m/s and ended up with 17m/s :/

    Update!!!!! : I found out something.
    You are wrong!

    170 / 10 = 17m/s

    125 / 8 = 15.625m/s.

    ITS NOT LINEAR.
    THE SPEED TELLS YOU THE GRADIENT. IF THE THE SPEED IS DIFFERENT THE GRADIENT IS DIFFERENT. THEREFORE IT IS NOT LINEAR. GOOD TRICK AQA!
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    (Original post by Positivedani)
    WAS There a question that asked for the units of resistance?
    Yes
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    (Original post by ridaboo99)
    okay thank you i think i dropped around 12-14 marks on p2 would that be an A*?
    I would say it would be on the boundary between A and A* so may be a high A, may be a low A* 😊
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    (Original post by Matt1278)
    It was a diagram so probably only 1, can't remember though
    Thanks anyway
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    (Original post by hannah4475)
    Ah right ok thanks I had a bit of a blank moment and for some reason just found the speed at the end of the line (top right) so got 17m/s but I'm pretty sure that's wrong now...

    Can you remember how many marks it was?
    no haha probably 1 maybe max 2
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    (Original post by WhiteScythe)
    What do you think of these revised grade boundaries?

    P2: 48/49 A* (54 = 100 UMS)
    P3: 44/45 A* (51 = 100 UMS)
    C2: 48/49 A* (54 = 100 UMS)
    C3: 45/46 A* (49 = 100 UMS)
    B2: 43/44 A* (49 = 100 UMS)
    B3: 44/45 A* (49 = 100 UMS)
    P2 is not going to be that high, are you kidding? P3 seems right tho.
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    (Original post by Positivedani)
    There was a primary coil in the battery! So an alternating pd enters the primary coil creating an alternating magnetic field in the iron bar; and this is induced by the secondary coil to create an akternating current in the secondary coil in the brush x
    Is that the right answer?
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    (Original post by WhiteScythe)
    The graph gave you information about the change in focal length -that was the whole point.
    No it wasn't though :P
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    (Original post by WhiteScythe)
    What do you think of these revised grade boundaries?

    P2: 48/49 A* (54 = 100 UMS)
    P3: 44/45 A* (51 = 100 UMS)
    C2: 48/49 A* (54 = 100 UMS)
    C3: 45/46 A* (49 = 100 UMS)
    B2: 43/44 A* (49 = 100 UMS)
    B3: 44/45 A* (49 = 100 UMS)
    I dont think chemistry will be that high because a lot of people found it quite hard but other than that i think those boundaries would be good
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    (Original post by PawanAviator)
    Well , it is weird how i can't get 22.5m/s and ended up with 17m/s :/

    Update!!!!! : I found out something.
    You are wrong!

    170 / 10 = 17m/s

    125 / 8 = 15.625m/s.

    ITS NOT LINEAR.
    THE SPEED TELLS YOU THE GRADIENT. IF THE THE SPEED IS DIFFERENT THE GRADIENT IS DIFFERENT. THEREFORE IT IS NOT LINEAR. GOOD TRICK AQA!
    accept your wrong? we all make mistakes from time to time
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    (Original post by Conor1998)
    Is that the right answer?
    Erm I think so as it was similar to a transformer!
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    Doesn't wire A have the greatest resistance because it had the highest voltage, I always had the thought of increasing the voltage would increase the resistance?
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    (Original post by isabellagrace)
    Lewis, do you think everybody on this thread is wrong and not you? As I am pretty sure that everyone on here thinks charge is measured in Coulombs and you are the only one who thinks otherwise..... Are we all going to get Us then?
    C'mon now, there's no need to use everyone else's opinion to gang up on him.
    I'm sure we will all get As and A*s so lets all be happy!
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    (Original post by Decklan)
    Doesn't wire A have the greatest resistance because it had the highest voltage, I always had the thought of increasing the voltage would increase the resistance?
    They all had the same voltage (at the end). But they had different values of current.
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    I'm pretty sure you had to subtract 0.1 from 0.5 or did I just do an extra step and lose a mark lol
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    (Original post by Pra99)
    I bet my house that the 6 marker in p2 will be about stars
    I don't think a single student out of the 100,000+ that sat it would have bet it was on an atom LOL
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    (Original post by james1221212)
    accept your wrong? we all make mistakes from time to time
    why are you doing170/10 that is not the gradient you a* student.

    it is the CHANGE y divided by the CHANGE in x

    deliberately 8 seconds was made to have 125m and 10 seconds 170 and the gradient is steepest and is a straight line so your genius a* would tell you that is 22.5.
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    (Original post by neil20143)
    No it wasn't though :P
    I'm sorry Neil 0.5 is the right answer, if I can't persuade you otherwise then I apologise but try not to be obstinate and deny it because you got a different answer.
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    (Original post by lewis99taylor)
    Yep and if you could read you would realise that i corrected myself as i found out i was wrong. You are the reason for low grade boundaries.
    Yeah okay I don't want to argue, sorry
 
 
 

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