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    Protoxylic Question 13b. http://filestore.aqa.org.uk/subjects...W-QP-JUN10.PDF

    I'm feeling rather silly and stupid now, and the mark scheme isn't helping. Why this time do we need to times it by 4 at 25% efficiency when the previous one we times'd it by 10 for 10% efficiency?

    Edit: I found energy at 25% efficiency and then divided that by the energy released each fission. I then got 6.25*10^27 times it needs to fission. Then what?
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    Section A is okay, turning points is fairly straight forward but too many people and theories to memorise!
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    (Original post by rich1334)
    Section A is okay, turning points is fairly straight forward but too many people and theories to memorise!
    June 2015 special relativity stuff was horrible. Otherwise, it's been ok


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    Do we ACTUALLY need to know that disgusting gas equation proof? I've never even bothered with it. Apparently we just "need to be aware of it". But then what's the point of it being in the book?
    (Original post by treeporn)
    For those doing the Turning Points module, could someone tell me whether the the proper value of time dilation t0 is in the moving or rest frame? From my research (which is contradicting the mark schemes) I've been going by the fact that 't' is the proper time in the rest frame and 't0' is the time in the moving frame because the time dilated in the moving frame is greater?So I've paired up L0 (length is always greatest in the rest frame) with t in the rest frame and L with t0 in the moving frame but the mark schemes indicate that L0 is paired up with t0 meaning time dilated is greatest in the rest frame so I'm dead confused.I kind of get the concepts but I look at these questions more from ratio point of view so if anyone could clear up my confusion i'd be grateful.
    For some reason,

    t= rest frame.

    T0= proper time

    However,


    L0= rest frame.

    And

    L0= proper length


    It makes no sense but I've learnt to accept it. It gets me the right answer afterall
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    (Original post by lucabrasi98)
    Do we ACTUALLY need to know that disgusting gas equation proof? I've never even bothered with it. Apparently we just "need to be aware of it". But then what's the point of it being in the book?
    For some reason,

    t= rest frame.

    T0= proper time

    However,


    L0= rest frame.

    And

    L0= proper length


    It makes no sense but I've learnt to accept it. It gets me the right answer afterall
    It does make sense as if t0 is proper time, that frame has the smallest time interval. X=ct, so for the travelling frame it will see a shortest space interval and thus everything travelling relative to the x direction will be length contracted. It's just that you call proper length the length in the rest frame because this is the unmodified length and you call proper time the time interval in the travelling frame (traveling at u) which is the smallest time interval possible, all other frames experience at larger time interval.
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    (Original post by particlestudent)
    Protoxylic Question 13b. http://filestore.aqa.org.uk/subjects...W-QP-JUN10.PDF

    I'm feeling rather silly and stupid now, and the mark scheme isn't helping. Why this time do we need to times it by 4 at 25% efficiency when the previous one we times'd it by 10 for 10% efficiency?

    Edit: I found energy at 25% efficiency and then divided that by the energy released each fission. I then got 6.25*10^27 times it needs to fission. Then what?
    Ah I think you linked an astro paper?
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    Could anyone give some tips on the specific heat capacity questions? I always seem to miss a calculation out when there are several energy transfers.
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    (Original post by Protoxylic)
    It does make sense as if t0 is proper time, that frame has the smallest time interval. X=ct, so for the travelling frame it will see a shortest space interval and thus everything travelling relative to the x direction will be length contracted. It's just that you call proper length the length in the rest frame because this is the unmodified length and you call proper time the time interval in the travelling frame (traveling at u) which is the smallest time interval possible, all other frames experience at larger time interval.

    None of that makes any sense to me though. I'm not even sure what you mean by shortest time interval and largest. And i would've thought t0 was the biggest time.

    But instead of worrying about it, i realised it doesn't need to make sense for me to ace the paper. It's fine as long as I can apply it, which I can. I just practiced time dilation questions until i wasn't dropping any marks.

    I would ask my teacher to explain the theory but she says she doesn't get it. She claims it's because she only did a chemistry degree lol

    The only thing I'm worried about now is that gas equation proof
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    (Original post by lucabrasi98)
    None of that makes any sense to me though. I'm not even sure what you mean by shortest time interval and largest. And i would've thought t0 was the biggest time.

    But instead of worrying about it, i realised it doesn't need to make sense for me to ace the paper. It's fine as long as I can apply it, which I can. I just practiced time dilation questions until i wasn't dropping any marks.

    I would ask my teacher to explain the theory but she says she doesn't get it. She claims it's because she only did a chemistry degree lol

    The only thing I'm worried about now is that gas equation proof
    Has the gas equation proof ever come up? Ive just learned it and im wondering if ive wasted my time....
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    (Original post by Protoxylic)
    Ah I think you linked an astro paper?
    My bad this one 13b: https://e4cf8bb391554b7c9d8e0fc42269...r%20Energy.pdf

    It's not so much the calculations confusing me it's what I am actually doing during the last steps.

    I got the nuclear reactor gives 2*10^17J of energy throughout the year (times'd by 4 because of the 25% efficiency).
    I then divided this by the amount of energy per fission to get the number of fissions that occur, so I got 6.3*10^27. From there I'm stuck
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    Turning points questions

    In the cathode ray experiment, cathodes rays were produced in an evacuated glass tube, what produced the 'rays' if the electrons couldn't collide with anything as it was evacuated?

    Also in the jj Thompson experiment where there was a small amount of hydrogen in the tube, what causes the production of hydrogen ions? is it the electron ray produced by the cathode/anode knocking them off? Are the electrons colliding with the proton of the hydrogen nucleus and creating a neutron? I'm confused as to what causes the production of hydrogen ions? Is it multiple factors

    Also it is said that Thompson concluded that the cathode ray consisted of particles that made up an atom, how did he come to this conclusion based on this experiment?

    Thanks in advance for any help
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    (Original post by particlestudent)
    My bad this one 13b: https://e4cf8bb391554b7c9d8e0fc42269...r%20Energy.pdf

    It's not so much the calculations confusing me it's what I am actually doing during the last steps.

    I got the nuclear reactor gives 2*10^17J of energy throughout the year (times'd by 4 because of the 25% efficiency).
    I then divided this by the amount of energy per fission to get the number of fissions that occur, so I got 6.3*10^27. From there I'm stuck
    Are these made up questions?


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    (Original post by particlestudent)
    My bad this one 13b: https://e4cf8bb391554b7c9d8e0fc42269...r%20Energy.pdf

    It's not so much the calculations confusing me it's what I am actually doing during the last steps.

    I got the nuclear reactor gives 2*10^17J of energy throughout the year (times'd by 4 because of the 25% efficiency).
    I then divided this by the amount of energy per fission to get the number of fissions that occur, so I got 6.3*10^27. From there I'm stuck
    I did this and sort of got the answer, I got 83,080kg which I think is right but just to more SF, once you find the amount of fissions of 235, you times by atomic mass number times by 235 to find mass of U235 required. Then do 97/3 to find how many more U238 there was then find mass of u238 in the same way (mass number times atomic mass unit) then add them and you get 83,080kg.

    There is probably an easier way involving E=MC^2 though
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    (Original post by Yo12345)
    Are these made up questions?


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    Nope. But I don't know which year each are from
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    (Original post by philo-jitsu)
    I did this and sort of got the answer, I got 83,080kg which I think is right but just to more SF, once you find the amount of fissions of 235, you times by atomic mass number times by 235 to find mass of U235 required. Then do 97/3 to find how many more U238 there was then find mass of u238 in the same way (mass number times atomic mass unit) then add them and you get 83,080kg.

    There is probably an easier way involving E=MC^2 though
    I don't understand .
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    (Original post by particlestudent)
    I don't understand .
    which part
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    (Original post by philo-jitsu)
    which part
    I got that it needs to fission 6.3*10^27 times, so I * that by 235... and then I am stuck after that ...
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    (Original post by particlestudent)
    I got that it needs to fission 6.3*10^27 times, so I * that by 235... and then I am stuck after that ...
    235 is number of nucleons, and look on the formula sheet, the atomic mass number, or mass of a single nucleon is 1.661*10^-27kg. (u)

    so 235*u*amount of fission's (you worked that out) gives you mass of all 235 atoms, then there are 97% u-238, so find out how many more u238 there are, do 97/3 then times by how many fissions occur which account for amount of u235 atoms.

    then u238 has 3 extra nucleons so multiply u*238*amount of 238 atoms whihc youve just worked out then add both 235 & 238 masses.
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    (Original post by philo-jitsu)
    235 is number of nucleons, and look on the formula sheet, the atomic mass number, or mass of a single nucleon is 1.661*10^-27kg. (u)

    so 235*u*amount of fission's (you worked that out) gives you mass of all 235 atoms, then there are 97% u-238, so find out how many more u238 there are, do 97/3 then times by how many fissions occur which account for amount of u235 atoms.

    then u238 has 3 extra nucleons so multiply u*238*amount of 238 atoms whihc youve just worked out then add both 235 & 238 masses.
    I got 82,985kg , thanks.

    The part in bold is the main confusing part for me though. Can't we do 97/100* the amount of fissions minus 3*100 * the amount of fissions to find the difference?
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    For Intensity, is the corrected count rate the intensity? If not, what is the corrected count rate in calculation terms. How did you find activity from corrected count rate?


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