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    (Original post by tetra-s)
    for the plot the rms voltage on the graph..did you have to plot it when the voltage was negative also? seems it would be the case as it was worth 2 marks..forgot to do it lol oops
    Crap, I think you do. I didn't do that. diojsoidjs
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    (Original post by StalkeR47)
    Well no. But I hope you get what you want. Amen.
    Thank you anyway
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    (Original post by GeneralOJB)
    Crap, I think you do. I didn't do that. diojsoidjs
    It was plot DC voltage not rms
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    How many UMS do you guys recon 58 would be ? More than 110 ? Any chance
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    (Original post by Seth James)
    I got that as well, did you do V/I=R then add 5000 to it and then do 90000-Ans
    Yeahh exactly! Because I tried to keep current constant, therefor the total residtance had to add upto 90kilo ohms


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    (Original post by GeneralOJB)
    Crap, I think you do. I didn't do that. diojsoidjs
    I thought of this during the exam and the question definitely said '...draw *a* line on Figure 2...' so I just did the positive one, still not sure though.
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    (Original post by g.k.galloway)
    Yeh S1 was a little weird, but okay
    Good luck! I abandoned History at GCSE, too much writing!!
    Yeah, I thought S1 was weird as well, only 6 questions but I managed to fall for the same mistake I made in January (not recognising the cumulative frequency distribution).

    History is a lot of writing, and I'm always uncertain as to how well I've done, but it's a subject I enjoy the lessons for. Whether I'll continue it next year is another matter.
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    (Original post by Qari)
    It was plot DC voltage not rms
    Isn't rms the equivalent DC voltage? So would one line be ok?
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    (Original post by Qari)
    It was plot DC voltage not rms
    It was plot the DC voltage that would give the same power, which is the rms.
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    (Original post by GeneralOJB)
    Isn't rms the equivalent DC voltage? So would one line be ok?

    It's only one horizontal line as a direct current doesn't reverse direction remember. It just flows in one conventional direction. If it were two horizontal lines, that would mean a current is flowing simultaneously in both directions which, well, is baffling.
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    (Original post by benroxo)
    Not very related but does anyone know the rough raw mark to ums conversion for the ISAs?
    http://www.aqa.org.uk/exams-administ...t-marks-to-ums

    Use this to calculate raw marks/UMS marks.

    UMS marks vary all the time, you can't predict them as there isn't a specific conversion factor
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    (Original post by Semanresu)
    Hi guys,

    The test was okay; didn't expect that 6-marker!

    These are my answers. I'm not sure if they are right.

    Q1
    28 n, 20 p, 18 e
    charge=+3.2x10^-19
    specific charge=4.0x10^6 Ckg^-1

    Q2
    D quark, W+ boson, positron, electron neutrino
    Weak interaction
    B was the exchange boson
    photon has infinite range, W+ has limited range
    Charge conserved, baryon number conserved

    Q3
    6 marker, hadrons experience strong and weak, leptons weak only. hadrons are protons neutrons, leptons are electrons, neutrinos. both experience weak, both have antiparticles, both have charge. hadrons include baryons and mesons etc
    proton, antiproton
    antiparticles have opposite charge sign

    Q4
    E depends on f. 1 electron absorbs 1 photon. KE=hf-work function. Max value. KE is less than max if work is done on deeper electrons.
    work function=1.8eV
    frequency=4.4x10^14 Hz
    frequency reduced, less energy, less kinetic energy
    intensity increased, more photons per second, more electrons per second.

    Q5
    peak to peak=128V
    peak=64V
    RMS=45(.3)V
    frequency=100Hz
    DC= straight ine at 45V
    Oscilloscope, y-gain is same as before, 20Vdiv^-1. Change time base to include 2 cycles. 1 cycle=10ms, 2 cycles=20ms, so 2msdiv^-1.

    Q6
    6.3V
    5.7V across 2 ohm resistor
    2.85A across 2 ohm resistor
    1.35A across unknown resistor
    4.2 ohm resistance
    2.86 ohms total resistance
    power: Internal resistor=26.5W, 2 ohm resistor=16.2W, other resistor=7.7W
    energy conserved: battery power-total power dissipated=0
    50.4-26.5-7.7-16.2=0

    Q7
    6.7x10^-5 A
    1/3 V
    resistance of LDR decreases, so greater proportion of pd across resistor, reading increases
    30000 ohms resistance.

    Tell me if there are any issues.
    Good Luck!

    PS, any ideas for grade boundaries?
    question 4, about the frequency being reduced, surely this has no effect unless the frequency is below the threshold frequency?
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    (Original post by Zakee)
    It's only one horizontal line as a direct current doesn't reverse direction remember. It just flows in one conventional direction. If it were two horizontal lines, that would mean a current is flowing simultaneously in both directions which, well, is baffling.
    True. Not sure why it was 2 marks then, but at least I got em.
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    It shows roughly 55 Raw Marks to get A
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    > I messed up and put Beta minus decay, so W- Boson, Anti-neutrino and electron, how many marks do you think this has cost me?

    ​so annoyed.
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    (Original post by Kayakayakaya)
    question 4, about the frequency being reduced, surely this has no effect unless the frequency is below the threshold frequency?
    For total resistance don't you include internal resistor?
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    (Original post by Wisp_)
    > I messed up and put Beta minus decay, so W- Boson, Anti-neutrino and electron, how many marks do you think this has cost me?

    ​so annoyed.
    At the least 2 of the 3
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    (Original post by GeneralOJB)
    True. Not sure why it was 2 marks then, but at least I got em.
    Which Question?
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    in the specific charge question for number 1, do you have to include the mass of the electrons as their mass is almost negligible? you get the same answer (4.0x10^6) if you don't.
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    (Original post by `God)
    in the specific charge question for number 1, do you have to include the mass of the electrons as their mass is almost negligible? you get the same answer (4.0x10^6) if you don't.
    I got 3.99x10^6
 
 
 
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