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Zigmanuir [339]
3 years ago
11

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 AWhat 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 BWhat 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:
brilliants [131]3 years ago
3 0

Answer:

Explanation:

When the craft was stationary , weight will be balanced by tension

T = mg

T = 7000 N

A)

when the craft was being lowered to the seafloor

drag force will act in upper direction , so

T₁ + 1800 = mg

T₁  = mg - 1800

= 7000 - 1800

= 5200 N

52 X 10² N

B)

when the craft was being raised from the seafloor , Tension will act in downward direction

T₂ = mg+ 1800

T₂  = 7000 - 1800

= 8800N

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Read 2 more answers
A spring with spring constant 15.5 N/m hangs from the ceiling. A ball is suspended from the spring and allowed to come to rest.
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Answer:

A) 138.8g

B)73.97 cm/s

Explanation:

K = 15.5 Kn/m

A = 7 cm

N = 37 oscillations

tn = 20 seconds

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ω² = k/m and m = k/ω²

Also, angular frequency (ω) = 2π/T

Now, T is the time it takes to complete one oscillation.

So from the question, we can calculate T as;

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v(t) = vmax Sin (ωt + Φ)

It is from the derivative of;

v(t) = -Aω Sin (ωt + Φ)

So comparing the two equations of v(t), we can see that ;

vmax = Aω

Vmax = 7 x 10.5672 = 73.97 cm/s

6 0
3 years ago
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