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  1. histephenson007

    A2 Physics | Post your doubts here

    Well, guys, I'm going to sleep now. Need to be fresh for tomorrow's exam! Thank you all for participating in this forum and teaching me some valuable tips to get me as ready as I can be for tomorrow's exam. May one and all do well in the exam and achieve your desired results! (y):p
  2. histephenson007

    A2 Physics | Post your doubts here

    I had the exact same problem. But, I suppose they would allow for small mistakes like that.
  3. histephenson007

    A2 Physics | Post your doubts here

    Complete (y)(y)
  4. histephenson007

    A2 Physics | Post your doubts here

    Nah, I don't have any reason for that one. I just know that if anything like that comes up in the test, I will answer that the half-life is too large o_O
  5. histephenson007

    A2 Physics | Post your doubts here

    I do not quite get ur question here. The dV is positive. And, because W = -P (dV). The total work done is negative.
  6. histephenson007

    A2 Physics | Post your doubts here

    Q3 b) iii) I think what you say is right. Work done on the system should be positive. Although I explain this from that perspective, I will try to make some sense by explaining it in an alternative way. If you just ignore the words for a moment, We know the formula W = -P (dV) , where dV is...
  7. histephenson007

    A2 Physics | Post your doubts here

    In an earlier question, it asked the total energy change in PQRP which is zero. So, the sum of the third column should be zero. Since we already know that the first two are -360and +720, -360+720+ x = 0 So, x = -360 Then u = q + w -360 = 480 + x so, x = -360 - 480 = -840
  8. histephenson007

    A2 Physics | Post your doubts here

    look at my previous answer again, I edited it (y)
  9. histephenson007

    A2 Physics | Post your doubts here

    GOT IT!!!!!!!!!! In an earlier question, it asked the total energy change in PQRP which is zero. So, the sum of the third column should be zero. Since we already know that the first two are -360and +720, -360+720+ x = 0 So, x = -360 Then u = q + w -360 = 480 + x so, x = -360 - 480 = -840
  10. histephenson007

    A2 Physics | Post your doubts here

    I believe the xtremepapers' paper is the correct one. But I am still trying to find out why the answer is -840J .
  11. histephenson007

    A2 Physics | Post your doubts here

    the one with the lowest SF
  12. histephenson007

    A2 Physics | Post your doubts here

    yea, just check the ms in xtremepapers. something is not right
  13. histephenson007

    A2 Physics | Post your doubts here

    Yea, I know how that feels. But they always say they accept 1 extra significant figure. And more good news!! My teacher has attended a conference this year and found out that they are going to be less strict about significant figures from this year. So don't worry, and always use one extra...
  14. histephenson007

    A2 Physics | Post your doubts here

    hmmm...how come different papers?? there is a different paper in xtremempapers with the same year same name paper!!
  15. histephenson007

    A2 Physics | Post your doubts here

    dV is delta Volume P -> Q : P = 4x10^5 , dV = -6x10^-4 ............. W = - p(dV) = - (4x10^5)(-6x10^-4) = 240 J (first column) ............. u = q + w = -600 + 240 = -360 Q -> R : +720J R -> P : omg, how'd they get -840? .............. by reverse calculation, 840 = (1/2)*P*(dV) which is...
  16. histephenson007

    A2 Physics | Post your doubts here

    For 1) The amplitude is constant, because it is frequency modulation. 2) Definition of frequency modulation : carrier frequency varies with the displacement of the signal wave. Maximum frequency = Carrier wave frequency + Maximum displacement * frequency deviation of carrier wave Max...
  17. histephenson007

    A2 Physics | Post your doubts here

    Just use W = -P (ΔV) to get first column values And add heat supplied to get the third column
  18. histephenson007

    A2 Physics | Post your doubts here

    For O/N 2008 Q5, Use energy conservation : K1+P1 = K2+P2 P1 is zero, because they are far from each other K2 is zero, because they just come into contact P2 = Q1Q2 / 4(pi)(E) * d (d is the diameter which equals to the sum of both the radiuses of the deutirium nuclei)
  19. histephenson007

    A2 Physics | Post your doubts here

    no problem, the answer is in page 44
  20. histephenson007

    A2 Physics | Post your doubts here

    p41/42/43 ??? I'm guessing 41 & 42. cuz 43 doesn't have a 2 iii lol So, why is internal energy directly proportional to the temperature? u = k.e. + p.e. but we assume that there are no intermolecular forces in ideal gases, hence p.e. = 0 u = k.e but according to ideal gas equations, we can...
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