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zaphod

  • 3 years ago

Help please :)

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  1. zaphod
    • 3 years ago
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  2. zaphod
    • 3 years ago
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    Show that the gain in potential energy of the block is approximately 0.01J?

  3. zaphod
    • 3 years ago
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    @Coolsector

  4. zaphod
    • 3 years ago
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    @sauravshakya @experimentX

  5. rvgupta
    • 3 years ago
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    |dw:1350213222437:dw| now subtract them and get the answer this difference is due to the increase in hight of second square due to being on its edge

  6. zaphod
    • 3 years ago
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    why is it 0.16

  7. rvgupta
    • 3 years ago
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    it is an approximation due to finding the height of the triangle

  8. zaphod
    • 3 years ago
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    can u show the calculation?

  9. zaphod
    • 3 years ago
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    how did u get mgh = 0.15?

  10. rvgupta
    • 3 years ago
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    as m=150/1000 kg g=10 m/s^2 h=10/100 m multiply them and get the ans

  11. zaphod
    • 3 years ago
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    wt bt the other one 0.16

  12. rvgupta
    • 3 years ago
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    |dw:1350213752879:dw| here h is slightly greater than the previous one thats why the ans will be 0.16 approx.

  13. zaphod
    • 3 years ago
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    its 14.14 ? the height

  14. rvgupta
    • 3 years ago
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    how did u find the h

  15. zaphod
    • 3 years ago
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    pythagoras....10^2 + 10^2 = x^2 x = 14.14

  16. rvgupta
    • 3 years ago
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    right, but how will u get the solu.

  17. zaphod
    • 3 years ago
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    thats what i posted this question..

  18. rvgupta
    • 3 years ago
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    ok

  19. zaphod
    • 3 years ago
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    how can the height be 11?

  20. rvgupta
    • 3 years ago
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    it was an appro

  21. zaphod
    • 3 years ago
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    from 14?

  22. rvgupta
    • 3 years ago
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    was my method right

  23. zaphod
    • 3 years ago
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    how did u get 11? i mean frm wt did u approximatre

  24. rvgupta
    • 3 years ago
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    it was just to get the ans dude

  25. zaphod
    • 3 years ago
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    hmm...do we actually consider the height from the horizontal to cntre of gravity or as a whole?

  26. rvgupta
    • 3 years ago
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    can u give the solu

  27. zaphod
    • 3 years ago
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    i dont knw how to solve it, coz the answer doesnt equal to 0.01 j, maybe theres a problem in the question. anyway thanks for helping

  28. rvgupta
    • 3 years ago
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    were u talking about GPE or only about PE

  29. zaphod
    • 3 years ago
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    gain in potential energy?

  30. rvgupta
    • 3 years ago
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    i think the H should be taken from the cnt of gravity

  31. zaphod
    • 3 years ago
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    so its 5 cm

  32. Vincent-Lyon.Fr
    • 3 years ago
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    Gain in potential energy is m g Δh , where Δh is gain in altitude of the centre of mass of the body.

  33. zaphod
    • 3 years ago
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    yes :) so i got the answer as 0.03j not 0.01 can u chck where iwent wrong

  34. Vincent-Lyon.Fr
    • 3 years ago
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    Δh = 5√2 - 5 = 2.07 cm = 0.0207 m g = 9.81 m/s² m = 0.150 kg Hence gain = 0.03 J

  35. zaphod
    • 3 years ago
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    so i am right :)

  36. Vincent-Lyon.Fr
    • 3 years ago
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    You are!

  37. zaphod
    • 3 years ago
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    i hve a second part in the question, cn i post it here..

  38. rvgupta
    • 3 years ago
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    go ahead

  39. zaphod
    • 3 years ago
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    The collision is perfectly elastic and, without sliding, the block turns about the corner as shown. The block is able to reach the position as shown in the image. 25% of the kinetic energy of the ball is transferred to the block. Calculate 1. the kinetic enrgy of the balll just before it strikes the block

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