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OCR Physics B G495 Field and Particle Pictures June 21st 2011 Exam Thread watch

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    (Original post by Swindan)
    Hmm it probably is What I was trying to say though, is that only low-risk patients will get a scan. So low-risk patients, getting a very small amount of radiation.

    Anywho, I think we've had enough confirmation on here that it's 4.35. If that's too low then blame OCR
    OCR doesn't wanna scare students so kindly said that it's 4.35 that's wt i thought.. anyway i should hv put 4/5 instead ?
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    (Original post by wki)
    got ignore anyone know the phasors one in section c? n also how to find the mass of uranium per year or anything?
    Mass of U-235 per year:

    Multiply Watts/power given by number of seconds in a year, get total energy used per year.

    Divide total energy by energy released per fission to get no of nuclei that decay.

    Get actual mass of one U-235 nucleus by multiplying 235 by u to get mass in kg

    Multiply no of nuclei by the mass...result?
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    (Original post by Revolution is my Name)
    Well, it asked to comment on the amplitudes, so I said that of OX (or whichever it was where they were removing the air) would increase because less was being refracted and so more reflected, and there'd be no change for OY. Then it said to mention the detector signal, and I think I guessed at their being more minima.
    Hmmm... is this similar to my answer :confused:
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    (Original post by jimmeh)
    Nah, definitely meant to get a positive root :/

    What equation did you end up with after you rearranged to find V?

    I think (can someone confirm this?) it was meant to be v = \sqrt{c^{2}(1-\gamma ^{2})}
    Damn yeah i get that, must have messed something up in the exam
    I got pretty stressed on that question
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    (Original post by wki)
    same here how to find the mass of uranium per year or anything?
    Divide the wattage given by the number of J per fission, to give the fissions per second. Mulitply that by seconds in a year, to give the number of fissions per year. Then multiply that by the amount of uranium per fission (mass number x number of u per kg)
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    (Original post by Revolution is my Name)
    Divide the wattage given by the number of J per fission, to give the fissions per second. Mulitply that by seconds in a year, to give the number of fissions per year. Then multiply that by the amount of uranium per fission (mass number x number of u per kg)
    Same I got 980Kg
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    Oh another thing,
    Radioactive question:
    something like find N to be 9x10^something (can't remember))
    I got 9.89 .. and so did like alllll my friends .. is 9.89 supposed to be ~ 9? :s
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    (Original post by arianex)
    Mass of U-235 per year:

    Multiply Watts/power given by number of seconds in a year, get total energy used per year.

    Divide total energy by energy released per fission to get no of nuclei that decay.

    Get actual mass of one U-235 nucleus by multiplying 235 by u to get mass in kg

    Multiply no of nuclei by the mass...result?
    I can't remember my exact method, but I think I got just over 1000kg for this question. Anyone else get something similar?

    Edit: Different answer to Summerdays, hence wrong
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    (Original post by Summerdays)
    They were, before the air was put in front of one of the mirrors, I think?
    air was being removed not put in front of the mirror? I dunno what the answer would be but the wavelength would be refracted less as the refractive index would be less.
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    (Original post by ejw93)
    air was being removed not put in front of the mirror? I dunno what the answer would be but the wavelength would be refracted less as the refractive index would be less.
    I put that. I am not even sure about my answer, I just guessed.
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    (Original post by Swindan)

    Edit: Different answer to Summerdays, hence wrong
    I imagine there's different potential answers, depending on whether you used 3x10-11 from the previous question, or the exact answer that you found.
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    (Original post by Swindan)
    I can't remember my exact method, but I think I got just over 1000kg for this question. Anyone else get something similar?

    Edit: Different answer to Summerdays, hence wrong

    (Original post by Summerdays)
    Same I got 980Kg
    Feckload of uranium, that is. I might've gone wrong, divvied it by 1000 because I thought I got it in g. Oops. XD

    EDIT// I did get an answer that was over 1000 though. Before the divvying. XD
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    (Original post by Revolution is my Name)
    I imagine there's different potential answers, depending on whether you used 3x10-11 from the previous question, or the exact answer that you found.
    Yeah. I actually foun the average binding energy and then went from there, like it doe in the CPG book

    I was talking bout the air and mirror question
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    (Original post by arianex)
    Mass of U-235 per year:

    Multiply Watts/power given by number of seconds in a year, get total energy used per year.

    Divide total energy by energy released per fission to get no of nuclei that decay.

    Get actual mass of one U-235 nucleus by multiplying 235 by u to get mass in kg

    Multiply no of nuclei by the mass...result?
    i did that, it seems correct.
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    (Original post by Swindan)
    I can't remember my exact method, but I think I got just over 1000kg for this question. Anyone else get something similar?

    Edit: Different answer to Summerdays, hence wrong
    i got this!
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    (Original post by arianex)
    Feckload of uranium, that is. I might've gone wrong, divvied it by 1000 because I thought I got it in g. Oops. XD

    EDIT// I did get an answer that was over 1000 though. Before the divvying. XD
    I used MY answer that's why, which was 3.410^-11J
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    (Original post by Summerdays)
    Yeah. I actually foun the average binding energy and then went from there, like it doe in the CPG book

    I was talking bout the air and mirror question
    Well since it gets refracted less, wouldn't the speed (and hence the wavelength) of said beam change, thus rendering the phasor longer/shorter and thus unable to totally cancel out the other one? I totally blagged that Q xD
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    (Original post by Summerdays)
    I used MY answer that's why, which was 3.410^-11J
    Now I'm questioning my answer to the previous question too I ended up with something like 2.95^-11
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    (Original post by arianex)
    Well since it gets refracted less, wouldn't the speed (and hence the wavelength) of said beam change, thus rendering the phasor longer/shorter and thus unable to totally cancel out the other one? I totally blagged that Q xD
    I blagged it as well
    I really didn't understand the question
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    Grade boundaries people! Similar to last June? (75 for A, 67 for B, 83 for 90)
 
 
 
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