Physics

If a 100-g mass was placed at the 25-cm mark, and a 20-g mass at the 10-cm mark, where should a 500-g mass be placed to balance the system?

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asked by Lila
  1. 100*(25- distance from fulcrum) + 20*(40- distance from fulcrum)= 500(x- distance from fulcrum)

    3300=500x Solve for x. then add to 50 because it's in the opposite side of the fulcrum.

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  2. 100*(25- distance from fulcrum) + 20*(40- distance from fulcrum)= 500(x- distance from fulcrum)

    3300=500x Solve for x. then add to 50 because it's in the opposite side of the fulcrum.

    At the 50.66 cm mark the 500g mass should be placed

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    posted by Spencer
  3. the answer should be 56.6 cm because 3300/500 is 6.6.

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    posted by Pam
  4. Statisticly I think the answer is either 8===D or 69 or 666

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  5. Ya know, I think my calculations were slightly off. You should still copy my answers though. Because if 666=666, then 69= Mushroom on the rocks which is 8===D

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