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Levart [38]
3 years ago
6

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 7000 N {\rm N}. When the craft was lowered or raised at a steady rate, the motion through the water added an 1800 N {\rm N} drag force.
Part A

What was the tension in the cable when the craft was being lowered to the seafloor?

Express your answer to two significant figures and include the appropriate units.

Part B

What was the tension in the cable when the craft was being raised from the seafloor?

Express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
Talja [164]3 years ago
4 0

Answer:

A) 5.2 x 10³ N

B) 8.8 x 10³ N

Explanation:

Part A)

F_{g} = weight of the craft in downward direction = tension force in the cable when stationary = 7000 N

T = Tension force in upward direction

F_{d} = Drag force in upward direction = 1800 N

Force equation for the motion of craft is given as

F_{g} - F_{d} - T = 0

7000 - 1800 - T = 0

T = 5200 N

T = 5.2 x 10³ N

Part B)

F_{g} = weight of the craft in downward direction = tension force in the cable when stationary = 7000 N

T = Tension force in upward direction

F_{d} = Drag force in downward direction = 1800 N

Force equation for the motion of craft is given as

T  - F_{g} - F_{d} = 0

T - 7000 - 1800  = 0

T = 8800 N

T = 8.8 x 10³ N

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