Recent content by Philip Robotic

  1. Philip Robotic

    I don't know how I got "mysterious energy" in a Carnot heat pump

    So I will be able to heat up a certain substance more by using a Carnot Heat Pump than by theoretically converting all the available energy to heat? Do I understand it correctly?
  2. Philip Robotic

    I don't know how I got "mysterious energy" in a Carnot heat pump

    So first I transformed the equation no 2 like this: $$|Q_L|=K\cdot|W|$$ And then I transformed the first equation to find ##|Q_Z|## $$|Q_L|=|Q_H|-|W|$$ Plugging the result into the first equation $$|Q_H|=K\cdot |W|+|W|$$ $$|Q_H|=|W|\cdot (K+1)$$ We know that the efficiency coefficient K is...
  3. Philip Robotic

    Prime numbers and divisibility by 12

    Homework Statement Prove that if ##p## is a prime number and if ##p>5## then ##p^2-37## is divisible by ##12## Homework EquationsThe Attempt at a Solution So I think that the number ##p^2-37## should be expressed in a way that we can clearly see that it is divisible by 3 and by 2 twice...
  4. Philip Robotic

    Not understanding the behaviour of the vis-viva equation

    So what I tried doing here is to solve a two-body problem. I attempted at simplifying it by giving the barycentre a "mass" so I can model the body's orbit around the barycentre as if it were a static body in space with a mass of ##M##. As it is static from out point of reference it has no...
  5. Philip Robotic

    Not understanding the behaviour of the vis-viva equation

    Sorry, I meant angular momentum there. I have corrected it already. As I understand the solution would look the same if we had two objects where one has a mass ##m## and the other ##\frac{4}{5}m## and we were looking for ##d_p## where the second body is the point of reference. But I am not...
  6. Philip Robotic

    Not understanding the behaviour of the vis-viva equation

    [Moderator's note: Thread moved from a technical forum hence no formatting template] Hi everyone! I came here to seek some explanation. As I was doing some orbital mechanics I came up with a problem that in short looks like this: Having the velocity at apoapsis and its distance from the...
  7. Philip Robotic

    How to get from tan(a)/tan(b) to sin(a)/sin(b) or something simmilar

    Alright... Again, thank you all very much for help :smile: I've really learned a lot in this discussion from you and I have finally managed to find an alternative solution that doesn't require converting a fraction of tangents into a fraction of sines, but still it was a lot of fun trying to...
  8. Philip Robotic

    How to get from tan(a)/tan(b) to sin(a)/sin(b) or something simmilar

    Still working on it, but that's a great tip :smile: I am still new to computer calculations
  9. Philip Robotic

    How to get from tan(a)/tan(b) to sin(a)/sin(b) or something simmilar

    The thing is I can't post the entire task. It is a part of an ongoing physics contest. But thanks a lot for asking :wink:
  10. Philip Robotic

    How to get from tan(a)/tan(b) to sin(a)/sin(b) or something simmilar

    I tried now it and unfortunately I can only find approximations with spreadsheet (referring back to post #4). The software didn't find a function that matches both of the fractions well enough for the task. Seems that I will have to try a different approach on the "big" problem itself. I would...
  11. Philip Robotic

    How to get from tan(a)/tan(b) to sin(a)/sin(b) or something simmilar

    If I understand the question correctly, it would be good if it were true, because a general relationship between these two fractions could be derived. I am not sure if is true or false, but since you asked this question it is false, right? Unfortunately the answer they are asking for is that...
  12. Philip Robotic

    How to get from tan(a)/tan(b) to sin(a)/sin(b) or something simmilar

    Yes, that is it. Please note this is just a part of a bigger problem and that is as far as I got. I managed to find the ratio of tangents, but I need to have a ratio of sines.
  13. Philip Robotic

    How to get from tan(a)/tan(b) to sin(a)/sin(b) or something simmilar

    If I am not mistaken that is ##\frac{1}{cos^{2}a}## Thus I could cancel out the ##cos^2a## at the bottom of my equation and then I would have$$ \frac{(sina+sinb)\cdot (sina-sinb)}{sin^{2}b}+tan^2a\cdot\frac{(sina+sinb)\cdot (sina-sinb)}{sin^{2}b}+1$$ But I don't know what next should I do...
  14. Philip Robotic

    How to get from tan(a)/tan(b) to sin(a)/sin(b) or something simmilar

    I just spotted I made a mistake in my attempt at solution in post #1. It is correct now (but still unsolved :cry:). I also graphed both of the functions and it seems that they nearly perfectly match when the bottom angle (b) is negative and the top stays positive. You can check it here...
  15. Philip Robotic

    How to get from tan(a)/tan(b) to sin(a)/sin(b) or something simmilar

    Referring to kuruman's first post, unfortunately the angles in this task are rather large, between 30 to 80 degrees (approximately). But they are smaller than 90 deg.
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