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anonymous

  • 5 years ago

A building is being knocked down with a wrecking ball, which is a big metal sphere that swings on a 12-m-long cable. You are (unwisely!) standing directly beneath the point from which the wrecking ball is hung when you notice that the ball has just been released and is swinging directly toward you. How much time do you have to move out of the way?

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  1. anonymous
    • 5 years ago
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    is that all the information you have?

  2. anonymous
    • 5 years ago
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    unfortunately

  3. anonymous
    • 5 years ago
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    i know that \[T= 2pi sqrt{m/k}\]

  4. anonymous
    • 5 years ago
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    \[T=2\pi \sqrt{m/k}\]

  5. anonymous
    • 5 years ago
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    so, \[T = 2\pi \sqrt{L/g} \] where L = Lenght g = gravity

  6. anonymous
    • 5 years ago
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    T- period?

  7. anonymous
    • 5 years ago
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    T- period in seconds i mean?

  8. anonymous
    • 5 years ago
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    yes, because you know that gravity is in m/s^2 so, period will be measured in seconds. You know the cable is 12m long and gravity is 9.81 solve for T (period) \[2\pi \sqrt{12/9.81} = 6.94922\]

  9. anonymous
    • 5 years ago
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    ok cool i got 6.95 as well

  10. anonymous
    • 5 years ago
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    but, The release through the first vertical point is ¼ of the total cycle, so you have ¼ of the period to get out of the way. \[6.95 * (1/4) = ?\]

  11. anonymous
    • 5 years ago
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    but that is in hertz correct?

  12. anonymous
    • 5 years ago
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    period is in seconds i believe

  13. anonymous
    • 5 years ago
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    ah ok. i ended up with 1.7 seconds

  14. anonymous
    • 5 years ago
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    sounds right to me.

  15. anonymous
    • 5 years ago
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    im still a little confused as to why we multiplied by 1/4

  16. anonymous
    • 5 years ago
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    Period is the time that the ball takes to complete one cycle, as you are standing right beneath the ball (first vertical point) that is 1/4 of the cycle

  17. anonymous
    • 5 years ago
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    ah ok

  18. anonymous
    • 5 years ago
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    thanks! :)

  19. anonymous
    • 5 years ago
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    np (:

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