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    (Original post by Danny_L)
    Yeah, nah last couple years they've kept 100%und in low 50's or there about, I reckon today was no exception, I presume that's where they're trying to set the bar. I guess 53 for 100% ums
    Fair enough, lower the amount of marks required the better! :P
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    (Original post by President J)
    I got power of 41.
    lol same, seems impossible tho
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    (Original post by Jm098)
    also when talking about the energy conversions, I said GPE into KE and frictional forces in the tube and between pellets?
    Between pellets wouldn't be right I don't think, but heat loss to the tube, air resistance, anything like that should be fine.
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    Anyone remember was the the initial height the thing was dropped from in question 1.
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    http://i.imgur.com/Ogi2A0R.jpg Here is the workings for the S2 mass question if people want to see it =p
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    (Original post by Jackhawkins21)
    Starts at Max displacement 30 up and is and opposite curve to the PE
    i think ur probs right, but wat doesnt make sense is that from ur graph the glider wud never have zero kinetic energy, and i dont think thats right, thats why i drew it down to zero for max displacement.
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    Someone make a poll please.

    WHAT I DONT understand, why 2 questions about gravitational fields, and they missed almost HALF of the whole specification....They want us to work with NASA or what..
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    (Original post by tobybes)
    For such a mathematical paper I actually think that would be fair!
    For the people who take Maths and Further Maths it makes it easier for them so that might cancel out?
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    For sketching graph I got this
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    here are the workings for the S2 question
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    (Original post by BriO)
    Anyone remember was the the initial height the thing was dropped from in question 1.
    1.7m bro
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    (Original post by BriO)
    Anyone remember was the the initial height the thing was dropped from in question 1.
    1.7m
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    Teeeeeeaaaaaachheeer cooooool? Wheeeerreeeee aaaaaaarreeeee yoooouuuuu????
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    Anyone do simultaneous equations to find out R1 and R2? I thought I was in a maths mechanics exam rather than a physics exam..
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    eZZZZ
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    (Original post by Zonothene)
    here are the workings for the S2 question
    I got 1.73*10^33
    M = (4pi^2*radius^3)/(G * T^2)
    = (4pi^2*(1.2*10^12)^3)/((6.67*10^-11)*(86400*365*4)^2)
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    (Original post by Howshotmyexamis)
    Guys can u tell me why is it 1070?
    (92+8.25sin55)×9.81 then i got 969
    Because the force from the hose is actually from Newtons 2nd law. So F=mv/t with t=1 m=8.25 v=25. Then you resolve vertically and times by sin55. Then add the weight of the man.
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    http://www.thestudentroom.co.uk/show....php?t=4176590

    Made a poll guys, votevotevote
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    (Original post by BriO)
    Anyone remember was the the initial height the thing was dropped from in question 1.
    I think 1.7m?
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    (Original post by Danny_L)
    I got 1.73*10^33
    M = (4pi^2*radius^3)/(G * T^2)
    = (4pi^2*(1.2*10^12)^3)/((6.67*10^-11)*(86400*365*4)^2)
    As its a binary system you have to use this if you wish to do it using kepler's 3rd law (i.e adding the two masses together in the equation).
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Updated: August 23, 2016
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