D.3. a.4. sorry this is all i know hope this helps
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a. <span>FM GmMmr2
</span>= 6.67 x 10-11N.m2kg27 .35 x 1022 kg 70 kg 3.78 x 108 m2
<span>= 2.40 x 10-3 N
b. </span><span>FE GmEmr2
= 6.67 x 10-11 N.m2kg 25 .97 x 1034 kg (70kg) 6.38 x 106 m2
=685 N
FMFE 2.40 x 10-3N685 N= 0.0004%</span>
Explanation:
For this problem, use the first law of thermodynamics. The change in energy equals the increase in heat energy minus the work done.
ΔU=Q−W
We are not given a value for work, but we can solve for it using the force and distance. Work is the product of force and displacement.
W=FΔx
W=3N×2m
W=6J
Now that we have the value of work done and the value for heat added, we can solve for the total change in energy.
ΔU=Q−W
ΔU=10J−6J
ΔU=4J
Answer is 4J
i think this may help you very much