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  1. G

    HSC Physics Marathon 2013-2015 Archive

    re: HSC Physics Marathon Archive * Transportation can use superconducting coils in applications like Maglev trains. This is far more efficient that current usage of electromagnets in places like Shanghai and Japan (No resistance and heat loss in wiring). This is also because of the Meissner...
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    HSC Physics Marathon 2013-2015 Archive

    re: HSC Physics Marathon Archive Ahhh thank you, so this is assuming that it will always gain U - 13000 m/s in the example - worth of speed?
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    Greatest term/coefficient binomial

    Yea, if guess and check is a valid method, I don't see why this isn't :)
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    HSC 2015 MX1 Marathon (archive)

    Re: HSC 2015 3U Marathon yes!
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    HSC 2015 MX1 Marathon (archive)

    Re: HSC 2015 3U Marathon http://puu.sh/kUxZc/9389879c8a.JPG A particle is performing Simple Harmonic Motion in a straight line. In 1 minute of it's motion, it completes exactly 15 oscillations and travels exactly 120 metres. What is the amplitude of the motion?
  6. G

    Tricky perms and combs questio

    Ah yes! Ughhh, I hate permutations and combinations, everything makes sense and seems like its right until you realise it's wrong.
  7. G

    Tricky perms and combs questio

    Ah yes, it's just the maximum possible mixed double pairings without restrictions. Ok I see where I went wrong now, I should have used permutations so 5P2*5P2
  8. G

    Tricky perms and combs questio

    Ahh I see, thanks! So it would be 5*5*4*4
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    HSC 2015 MX1 Marathon (archive)

    Re: HSC 2015 3U Marathon let roots= (a-x), a, (a+x) sum of roots => 3a = 36/8 a = 3/2 product of roots=> a^3 - ax^2 = -21/8 but since a = 3/2 ; x=2/-2 roots are -1/2, 3/2, 7/2
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    Help 3uniy intergrand

    Positive as in third quadrant?
  11. G

    Tricky perms and combs questio

    Hey, not sure if I'm reading the question right, but shouldn't the answer be 80 considering that it is mixed doubles, wouldn't the maximum possibility be 5C2*5C2 = 100 I think the stipulation states that one specific man has to be in the game with his wife (uh I think this is wrong, but I'm...
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    HSC Physics Marathon 2013-2015 Archive

    re: HSC Physics Marathon Archive I don't understand, why is there no change in speed from the perspective of the observer on the planet?
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    HSC Physics Marathon 2013-2015 Archive

    re: HSC Physics Marathon Archive Uh, I think it's because electrons and holes travel in opposite directions the first place.
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    HSC Physics Marathon 2013-2015 Archive

    re: HSC Physics Marathon Archive This is my assumption. To be carrying a current, it necessitates that electrons flow in opposite directions to the holes. From the right hand palm rule, the flow of holes- conventional current- would be downwards when subjected to the magnetic field...
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    What would happen if you use a superconducting iron core in a transformer

    Wait, isn't the whole point of a soft iron core to allow for more efficient propagation of magnetic field to allow for better electromagnetic induction. If it were a super conductor, wouldn't this meant that the Meissner effect would exclude the field, making it counter-productive to transformation?
  16. G

    Maths Extension 2 thoughts

    either on the right or on the left and symmetrical probability of both?
  17. G

    HSC 2015 MX2 Marathon (archive)

    Re: HSC 2015 4U Marathon Ahh cool, so from -1/z = r.cis(-2pi/3), you could get -1/z = r.[cos(-2pi/3) + isin(-2pi/3)] which would net you -1/z = -[r(1+ sqrt3)]/2 and re-arrange to get rz =2/(1+ sqrt 3) Is that and using that fact that we know x= 1 to find r and y correct? Thank you so much btw!!
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    HSC 2015 MX2 Marathon (archive)

    Re: HSC 2015 4U Marathon Just wondering if this is remotely correct: arg( \frac{-1}{z}) = \frac{-2 \pi}{3} from this -1/z = -r (1+ isqrt 3) [where r is modulus of -1/z] rz = 1/ (1+ isqrt3) r(x+ iy) = (1- isqrt3)/4 but since we know that x =1 can we assume that r = 1/4...
  19. G

    Argand proofs

    haha oops, I've always thought that 3 and 4 quadrant would be -θ
  20. G

    Argand proofs

    1) Consider the values of sin and cos around the different quadrants of the unit circle (A-S-T-C) It's obvious that cosθ = cos-θ and sin-θ = sin-θ (for mod arg form, θ should be in the first 2 quadrants) 2) sin (pi/2-θ)= cos θ and similarly, cos (pi/2-θ) = sin θ (sum and difference of two angles)
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