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    Question 7c was clearly 4 ohms
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    (Original post by swagadon)
    i got the same as you, i dont think they care about the significant figures unless they tell you so
    which question are you guys talking about please
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    Anyone remember how many marks there were for the question about changing the x/y gain on the oscilloscope?
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    (Original post by Felix Ivers)
    Anyone remember how many marks there were for the question about changing the x/y gain on the oscilloscope?
    there were 3
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    (Original post by Goldn)
    Question 7c was clearly 4 ohms
    Last one was 30000 ohms or 30 kilo ohms
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    (Original post by Goods)
    there were 3
    I forgot to mention keep y the same only talked about x how much do you think i'll get?
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    For the question on conservation of energy, what did everyone do, i added the powers, 50.4 i think i got and as p=IV, and current was 4.2, i divided 50.4/4.2, which gave me 12v, the amount which the cell was supplying, is this right?


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    (Original post by Qari)
    I forgot to mention keep y the same only talked about x how much do you think i'll get?
    It definitely wouldn't cost you more than a mark, so i'd say 2 out of 3?
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    (Original post by BayHarborButcher)
    That's what I put but I'm the only person I know that did It may reduce their energy but they're still being emitted surely? (Until it's below the threshold)
    ive put the photons will have less energy, and therfore the electrons will have less kinetic energy
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    (Original post by zedd01)
    For the question on conservation of energy, what did everyone do, i added the powers, 50.4 i think i got and as p=IV, and current was 4.2, i divided 50.4/4.2, which gave me 12v, the amount which the cell was supplying, is this right?


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    I did something like that, I added the others and I did 12*4.2
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    (Original post by swagadon)
    ive put the photons will have less energy, and therfore the electrons will have less kinetic energy

    I said the maximum kinetic energy of the photoelectrons decreases, but the number of photoelectrons emitted remains the same (as the intenstity remains the same).
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    (Original post by swagadon)
    ive put the photons will have less energy, and therfore the electrons will have less kinetic energy
    I've put lower frequency so less will be emitted due to less Ek
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    (Original post by Qari)
    It does the kinetic energy would be lower
    doubling the intensity means doubling the number of photons given out per meter squared,so as there are double the amount of photons( the photons frequency has not changed), there will be double the amount of photoelectrons emitted
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    (Original post by zedd01)
    For the question on conservation of energy, what did everyone do, i added the powers, 50.4 i think i got and as p=IV, and current was 4.2, i divided 50.4/4.2, which gave me 12v, the amount which the cell was supplying, is this right?


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    Yeah, I added the powers too and I got 50.4W. Then, I used P=I^2V and got 50.4 again. This showed that the energy transferred was conserved.
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    (Original post by BayHarborButcher)
    That's what I put but I'm the only person I know that did It may reduce their energy but they're still being emitted surely? (Until it's below the threshold)
    The lower the frequency, the lower amount of kinetic energy each electron has until the frequency goes below the threshold frequency, where no electrons are then emitted, but the amount of electrons emitted remains the same. Check question 4a) on June 2009, it's a similar question but the opposite way (asking about an increase).
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    (Original post by Zakee)
    I said the maximum kinetic energy of the photoelectrons decreases, but the number of photoelectrons emitted remains the same (as the intenstity remains the same).
    No but the ones deeper in the metal might not get emitted due to lower Ek
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    (Original post by Qari)
    I forgot to mention keep y the same only talked about x how much do you think i'll get?
    The question did not asked for what to keep same, but what to change. you should get 3/3.
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    (Original post by swagadon)
    doubling the intensity means doubling the number of photons given out per meter squared,so as there are double the amount of photons( the photons frequency has not changed), there will be double the amount of photoelectrons emitted
    I wrote increase first, but then realised they've said doubled, so I wrote number of emitted electrons will be double
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    Are you guys still talking about intensity and frequency?
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    Any worked solutions PDF ting up anywhere yet?


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