anonymous
  • anonymous
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?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
is that all the information you have?
anonymous
  • anonymous
unfortunately
anonymous
  • anonymous
i know that \[T= 2pi sqrt{m/k}\]

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anonymous
  • anonymous
\[T=2\pi \sqrt{m/k}\]
anonymous
  • anonymous
so, \[T = 2\pi \sqrt{L/g} \] where L = Lenght g = gravity
anonymous
  • anonymous
T- period?
anonymous
  • anonymous
T- period in seconds i mean?
anonymous
  • anonymous
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) \[2\pi \sqrt{12/9.81} = 6.94922\]
anonymous
  • anonymous
ok cool i got 6.95 as well
anonymous
  • anonymous
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. \[6.95 * (1/4) = ?\]
anonymous
  • anonymous
but that is in hertz correct?
anonymous
  • anonymous
period is in seconds i believe
anonymous
  • anonymous
ah ok. i ended up with 1.7 seconds
anonymous
  • anonymous
sounds right to me.
anonymous
  • anonymous
im still a little confused as to why we multiplied by 1/4
anonymous
  • anonymous
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
anonymous
  • anonymous
ah ok
anonymous
  • anonymous
thanks! :)
anonymous
  • anonymous
np (:

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