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    (Original post by Vikingninja)
    You had that power of the quasar was some number x power of the sun and intensity of the sun was a number x intensity of the quasar. I originally put intensity of the sun = number x intensity of the quasar/4pi *(AU in metres)^2. I then rearranged the intensities so that I had intensity of the quasar = number x intensity of the sun. I then put all numbers under the fraction and that total was the distance squared.
    Are you some kind of genius?? Hahaha i never would've thought of doing that, oh well its over now, it was only 3 marks!
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    (Original post by -jordan-)
    My turning points answers:

    2
    a) 6 marker. Increasing potential difference increases work that needs to be done against the potential difference for electrons to move across the gap, as you increase the potential difference the current will increase. Therefore the kinetic energy of the electrons will be reduced. At the point where current is zero is the stopping potential. Can be used to find the maximum kinetic energy by calculating eV. Einstein's equation is hf = thi + ekmax where ekmax can be replaced by eV and used to find the work function or the frequency.
    1000
    2a) Doesn't the current decrease to zero, as the electrons can't move across the gap?
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    for the first two astrophysics questions, was the distance between the objective and eyepiece 0.9 and the image distance 0.2m?
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    What did people get as the star with the furthest distance?
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    (Original post by micycle)
    2a) Doesn't the current decrease to zero, as the electrons can't move across the gap?
    Yes that's what I meant, they loose all their kinetic energy as it isn't sufficient enough to overcome the potential difference across the plates so it doesn't cross.
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    (Original post by Vikingninja)
    You had that power of the quasar was some number x power of the sun and intensity of the sun was a number x intensity of the quasar. I originally put intensity of the sun = number x intensity of the quasar/4pi *(AU in metres)^2. I then rearranged the intensities so that I had intensity of the quasar = number x intensity of the sun. I then put all numbers under the fraction and that total was the distance squared.
    I did this too but nobody is saying what they got! What was your value?! Was it around 6.(something)x10^13?
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    (Original post by -jordan-)
    My turning points answers:

    1)
    a) Cathode rays are electrons, specific charge was 1800x that of a H+ ion, so mass is very light.
    b) Electron beam collides with the gas atoms, causes ionisation, the gas gets excited and then de-excites, emiting photos in the visible light spectrum.
    c) 9.1x10^10 for the calculated specific charge.
    2
    a) 6 marker. Increasing potential difference increases work that needs to be done against the potential difference for electrons to move across the gap, as you increase the potential difference the current will increase. Therefore the kinetic energy of the electrons will be reduced. At the point where current is zero is the stopping potential. Can be used to find the maximum kinetic energy by calculating eV. Einstein's equation is hf = thi + ekmax where ekmax can be replaced by eV and used to find the work function or the frequency.
    b) Work function - can't remember.
    c) Can't remember what I got here either - 21 or 25?

    3)
    a Electric field and magnetic field, 180 degrees in phase
    b) Increasing the distance means that it reflects back at a different point in the wave cycle, so a different number of wavelengths is reflected back, the phase difference is different and therefore they destructively interfere rather than constructively interfere
    c) 0.7m
    3)c) I got 0.0375m, as I got the wavelength as 0.15m and it asked for the distance between a maximum and a minimum recording, i.e. the distance between an antinode and a node, which is a quarter of a wavelength (0.15/4) = 0.0375m

    Anyone else do it this way?
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    (Original post by ErwinJ)
    I did this too but nobody is saying what they got! What was your value?! Was it around 6.(something)x10^13?
    10^13 sounds familiar.
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    Will 45/75 get a B?
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    In section A I got the volume of gas as 4.4x10^-4 by finding the number of moles and using pV = nRT. Then the next question said to find the number of moles. Not very reassuring.
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    (Original post by a4567)
    3)c) I got 0.0375m, as I got the wavelength as 0.15m and it asked for the distance between a maximum and a minimum recording, i.e. the distance between an antinode and a node, which is a quarter of a wavelength (0.15/4) = 0.0375m

    Anyone else do it this way?
    I think that is right, the 0.7m came from halving my wavelength but it should be a quarter, I got 0.14m for the wavelength though not 0.15
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    (Original post by -jordan-)
    I think that is right, the 0.7m came from halving my wavelength but it should be a quarter, I got 0.14m for the wavelength though not 0.15
    0.034m was the answer. Wavelength = 0.136m


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    (Original post by ombtom)
    In section A I got the volume of gas as 4.4x10^-4 by finding the number of moles and using pV = nRT. Then the next question said to find the number of moles. Not very reassuring.
    I got 4.4 as well. I used p1v1/t1 = p2v2/t2. Still correct in working because you do have the same amount of moles in each case.
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    (Original post by tanyapotter)
    for the first two astrophysics questions, was the distance between the objective and eyepiece 0.9 and the image distance 0.2m?

    Yeah I think that's what I got
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    (Original post by tanyapotter)
    for the first two astrophysics questions, was the distance between the objective and eyepiece 0.9 and the image distance 0.2m?
    I got 0.9 for the image distance and 100% thats right. But I couldnt work out the one after, ended up attempted 1/f=1/u+1/v and got 0.3m. No idea if its right.
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    (Original post by GeorgiaAnne)
    I did! Thought it was v low considering water is about 4900
    There was a past paper where you were told ice was 2100. If i hadn't have done it, I probably would've panicked when I sae my answer to that question

    (Original post by Yo12345)
    55 for count rate?


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    I either got 55 or 56
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    what did people put for why the nuclear fusion reaction is more powerful, and why at a larger distance from a source you should measure the count rate longer?
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    (Original post by Vikingninja)
    10^13 sounds familiar.
    the answer was 2.4 times 10^14, most of the other Physicists in my year got this
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    For the distance question in turning points I just did wavelength/2. What were you supposed to do?
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    (Original post by tanyapotter)
    what did people put for why the nuclear fusion reaction is more powerful, and why at a larger distance from a source you should measure the count rate longer?
    Fusion average binding energy increases sharply, so more energy is released.

    Count rate would be very low, so statically unreliable due to inverse square law. More time to get larger count


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