Answer:
The tension in the cable when the craft was being lowered to the seafloor is 4700 N.
Explanation:
Given that,
When the craft was stationary, the tension in the cable was 6500 N.
When the craft was lowered or raised at a steady rate, the motion through the water added an 1800 N.
The drag force of 1800 N will act in the upward direction. As it was lowered or raised at a steady rate, so its acceleration is 0. As a result, net force is 0. So,
T + F = W
Here, T is tension
F = 1800 N
W = 6500 N
Tension becomes :

So, the tension in the cable when the craft was being lowered to the seafloor is 4700 N.
8 wave units I guess I tried it should be the answer though
F = 2820.1 N
Explanation:
Let the (+)x-axis be up along the slope. The component of the weight of the crate along the slope is -mgsin15° (pointing down the slope). The force that keeps the crate from sliding is F. Therefore, we can write Newton's 2nd law along the x-axis as
Fnet = ma = 0 (a = 0 no sliding)
= F - mgsin15°
= 0
or
F = mgsin15°
= (120 kg)(9.8 m/s^2)sin15°
= 2820.1 N
Kinetic energy is the answer to your question.
Answer:
Thats the third law______