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2016 Official AQA New Spec AS Level Physics Paper 1 - 24th of May 2016 watch

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    (Original post by kierans99)
    What did you guys get for the stopping potential of the photoelectrons. I got like 1.4*10^-19J for the one before it asked for the stopping potential and then on that question it was only mark so I just put 1.4v?
    for the stopping potential I got 0.88v
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    For the stopping potential. You had to use maximum kinetic energy= eV
    1.4*10-19/1.6x10-19= 0.875
    Two sf 0.90volts
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    (Original post by 11234)
    What did people get for the percentage change in diameter. Did anyone get 0.79%
    I got 21% decrease
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    (Original post by haza01)
    I measured 59mm how was it 11.8cm???

    Posted from TSR Mobile
    59 X 2 = 118 You measure the distance from one node to the other which is only one half of a wavelength I think
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    (Original post by Hariss)
    for the stopping potential I got 0.88v

    Was it one sig fig or 2 i first had 0.875 then put it to 0.90
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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
    Is there much difference between the old and new spec?
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    What did everyone get for the reduction in diameter?? 25%??
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    (Original post by Pra99)
    Does anyone have the paper?
    yeah, in my pocket.

    let me get it out
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    (Original post by HasanRaza1)
    What did everyone get for the reduction in diameter?? 25%??
    I got 0.79
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    [QUOTE=Pra99;65115641]Is there much difference between the old and new spec?[/QUOTE

    Quite a handful of difference, they are more strict on sig figs and the the content is more about application i think, my teacher but apart from that ....
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    What did people get for the last electricity where the battery had internal r
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    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)
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    To work out decelaration did you have to find the resultant force using the risistive forces and force already known then use this in f= ma, which gives you -10. Is this right?
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    (Original post by Pouh)
    For the stopping potential. You had to use maximum kinetic energy= eV
    1.4*10-19/1.6x10-19= 0.875
    Two sf 0.90volts
    two sf is 0.88.
    the first 0 doesn't count
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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)
    Yes, what did you get for the one with the voltmeter placed between A and B
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    For the choloclate bar question it melted at each antinode
    you have to remember the stationary wave is unpolarised
    the wavelength was i believe 11.7/11.8
    I made a dumb mistake and wrote it as 10^-3 instead of 10^-2 when converting into metres. Anyone know how many marks i would lose????
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    (Original post by Pouh)
    To work out decelaration did you have to find the resultant force using the risistive forces and force already known then use this in f= ma, which gives you -10. Is this right?
    I got 2.7
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    [QUOTE=Pouh;65115755]
    (Original post by Pra99)
    Is there much difference between the old and new spec?[/QUOTE

    Quite a handful of difference, they are more strict on sig figs and the the content is more about application i think, my teacher but apart from that ....
    Is it harder though?
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    (Original post by Laceypr82)
    I forgot that....I got wavelength as 120mm. How many marks would I lose
    The answer is 120 not 80 because you have the distance between the antibodies not the the nodes. The distance between 2 nodes is half a wavelength and the distance between 2 anti nodes is the same as between 2 nodes so it's also half a wavelength[/QUOTE]

    It's alright being a Jack of all trades, but not if you're a master of none
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    Does anyone reckon boundaries will go up
 
 
 
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