Asked by Lila
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?
Answers
Answered by
James Lee
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.
3300=500x Solve for x. then add to 50 because it's in the opposite side of the fulcrum.
Answered by
Spencer
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
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
Answered by
Pam
the answer should be 56.6 cm because 3300/500 is 6.6.
Answered by
Chloe
why ? 40- distance from fulcrum
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