During a retrograde loop of mars, would you expect mars to be brighter than usual in the sky, about average in brightness, or fainter than usual in the sky? explain.

The objects are acted upon by only one force. gravity. mass times the acceleration of gravity (-9.81meters/second squared).

W= fd
where f is an applied force and d is the distance over which it is applied. when a parachuting skydiver is at constant velocity, he is experiencing "terminal velocity", the point at which his acceleration down is canceled out by air resistance. that means that the resisting force of the air is equal to his weight, but in the opposite direction. with this, we can solve for the work done by the force.
w = fd
w = (92.0)(9.81 m/s^2) (325)
w= 2.93 x 10^5 j
the value for work in this case is positive because it is pushing up on the skydiver, who has a downward force of gravity.
therefore the answer is 2.93 x 10^5 j

The answer is b on edgenuity

B.air resistancehoped i   can i get brainliest