A 20-kg child is on top of a slide was pushed down by his brother giving him an initial speed of 2 m/s down the slide. What is the child's speed at the bottom of the slide if the slide is 5 m high and 10 m long and the force of friction between the child and the slide is 10 N?​

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

When seismic waves hit a fault, the waves are reflected off the fault. seismograph can detect these reflected seismic waves. geologists then use these data to map the fault's length and depth

Ray 2 should cross through the focal point on the right side of the lens