1. a sonar pulse is sent to the ocean bottom which is 25◦ celsius. it takes 4.2 seconds for the wave echo to return to the ship. how deep is the ocean bottom? medium m/s air 0○c 331 air 20○c 343 helium 0○c 972 water 25○c 1,493 seawater 25○c 1,533 copper 25○c 3,560 iron 25○c 5,130

The sonar pulse takes t=4.2 s to reach the ocean bottom and then to travel back tothe ship. If we call d the depth of the ocean bottom, this means that the sonar pulse covered a distance equal to 2d in a time of 4.2 s.

The speed of the sonar pulse is the speed of sound in water at 25C, so looking at the table is v=1493 m/s. Using the relationship between space and time for a uniform motion, and keeping in mind that the total distance covered by the pulse is 2d, we can write:

which becomes

so, the ocean bottom is 3135 m deep.

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