An early submersible craft for deep-sea exploration was raised and lowered by a cable from a ship. When the craft was stationary, the tension in the cable was 6000 N. When the craft was lowered or raised at a steady rate, the motion through the water added an 1800 N drag force. What was the tension in the cable when the craft was being lowered to the seafloor?

T =  4200N

Explanation:

When the submersible craft is at rest, the tension in the cable is 6000N.

With this information you can calculate the weight of the craft by summing the forces (the summation of the force is zero because the craft is at rest):

When the craft is going down with a constant speed, there is a drag force of 1800N. Then, by using the second Newton law you have:

  (1)

Fd: drag force

The summation of the forces is zero because the craft moves with constant velocity, that is, there is no acceleration.

You calculate the new tension on the cable by solving the equation (1) for T:

hence, the tension is 4200N

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer