The drag on the hull of a ship depends in part on the height of the water waves produced by the hull. the potential energy associated with these waves therefore depends on the acceleration of gravity g. hence, we can state that the wave drag on the hull is d= f(rho[infinity],v[infinity], c, g) where c is a length scale associated with the hull, say, the maximum width of the hull. derive the non-dimensional groups for this problem, one of which is non-dimensional drag. hint: choose rho[infinity],v[infinity], c as the basis variables.

He can play two games. 13/4=3.253.25*2=6.5

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

Work is reduced while time is decreased

Lexy is correct because angle r is between the incident wave,  n  = 0, and a diffracted wave.