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jessica01479
10.03.2020 •
Physics
You are holding a 5-kg dumbbell straight out at arm’s length. Assuming that your arm is 0.70 m long, and that your shoulder acts as a pivot, what torque is the dumbbell exerting?
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Ответ:
34.3 N.
Explanation:
Torque: This is a force that tend to cause rotation or twisting in a system. The S.I unit of torque is Nm.
From the question,
Torque = Force×distance from the point of pivot
T = F×R equation 1
Where GT = torque, F = force or weight of the dumbbell, R = length of the arm
But,
F = mg Equation 2
Where m = mass of the dumbbell, g = acceleration due to gravity.
Substitute equation 2 into equation 1
T = mg×R Equation 3
Given: m = 5 kg, g = 9.8 m/s², R = 0.70 m.
Substitute into equation 3
T = 5(9.8)(0.7)
T = 34.3 N.
Ответ:
use
Explanation:google