Press two pieces of tape onto the surface and write a B (for bottom) on them. Then press another tape on top of each B tape and label it T (for top). Pull each pair of tapes off the surface as a unit. After they are off the surface, separate the T and B tapes. Hang one of the T tapes and one of the B tapes from a support. Describe the interaction between the following pairs of tape when they are brought near one another. a. Two T Tapes
b. Two B Tapes
c. a T and a B tape

This is the answer of your question

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

It was crowded with sewer like smells. bad sewer systems