IP Force to Hold a Baseball A person holds a 1.42-N baseball in his hand, a distance of 34.0 cm from the elbow joint, as shown i n the figure(Figure 1) . The biceps, attached at a distance of 2.75 cm from the elbow, exerts an upward force of 12.6 N on the forearm. Consider the forearm and hand to be a uniform rod with a mass of 1.50 kg . What is the net torque?
2 answers:
We will assume that the CM of the arm is at "L"
from the elbow, and the ball is at 34cm. Then the net torque is computed
by:
= 48.28 N*cm + 1.50kg * 9.8m/s² * 34 cm/2 – 12.6 N*2.75cm
= 48.28 N*cm + 499.8 – 34.65 N*cm
τ = 513.43 N*cm or
5.1343 N*m
Answer:
Explanation:
Net τ = 1.42 N * 34 cm + 1.50 kg * 9.8m/s² * 34 cm/2 – 12.6 N*2.75cm
= 48.28 N*cm + 1.50kg * 9.8m/s² * 34 cm/2 – 12.6 N*2.75cm
= 48.28 N*cm + 499.8 – 34.65 N*cm
τ = 513.43 N*cm or 5.1343 N*m
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Explanation:
Given data
time= 1 minute= 6 seconds
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P= I^2R
P/R= I^2
2/12= I^2
I^2= 0.166
I= √0.166
I= 0.4 amps
We know also that
Q= It
substitute
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what are u asking there isnt a question
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