An early submersible craft for deep-sea exploration was raised and lowered by a cable from a ship. When the craft was stationary
1 answer:
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.
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We must place the pivot to keep the meter stick in balance at 90 cm (10 cm from the weight) from the free end.
Answer: Option B
<u>Explanation:</u>
In initial stage, the meter stick’s mass and mass hanged in meter stick at one end are same. Refer figure 1, the mater stick’s weight acts at the stick’s mid-point.
If in case, the meter stick is to be at balanced form, then the acting torques sum would be zero. So,
Taking out ‘mg’ as common and we get
Hence, the stick should be pivoted at a distance of,
So, the stick should be pivoted at a distance of 75 cm at the free end
Now, replace mass with another mass. i.e., four times the initial mass (as given)
If in case, the meter stick is to be at balanced form, then the acting torques sum would be zero. So,
Taking out ‘mg’ as common and we get
Hence, the stick should be pivoted at a distance of,
So, the stick should be pivoted at a distance of 10 cm from the free end.
Therefore, the option B is correct 90 cm (10 cm from the weight).
