Water steam is entering the nozzle at 40 bars of pressure and 400 degrees C with a velocity of 15 m/s. The steam exits the nozzle at 15 bars of pressure, with a temperature of 320 degrees C. Calculate the exit velocity of the steam assuming steady-state conditions, assuming that no work is done inside the nozzle and that no heat exchange is taking place through the walls of the nozzle. The that potential energy change is neglected as well. Keep in mind that enthalpy from the tables is in units of kJ/kg and that 1 kJ/kg

Wind or electric energy force force of friction

Answer(b) 60 miles per hour;

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

That big glowy thing in the sky is called the sun