The force applied by the competitor is littler than the heaviness of the barbell. At the point when the barbell quickens upward, the power applied by the competitor is more prominent than the heaviness of the barbell. When it decelerates upward, the power applied by the competitor is littler than the heaviness of the barbell.
Answer:
Answer for A
Explanation:
F1=GmM/r1^2
If r2 becomes r2=5r
F2=GmM/(25r^2)
Multiply with 25 gives to maintain the same force
I.e.,25F2=F1
F2=G(25m)M/25r^2=F1
By the factor 25 would change to increase to same.
The engine's efficiency is (35J)/125J) = 28% .
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Answer:
14.8 m
Explanation:
S= ut + a
where u = initial velocity
S= (0 )(2) + (7.4)(2)
S= (7.4)(2)
S=14.8 m