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2 years ago

6

The triceps muscle in the back of the upper arm is primarily used to extend the forearm. Suppose this muscle in a professional b

oxer exerts a force of 1.9 × 103 N with an effective perpendicular lever arm of 2.85 cm, producing an angular acceleration of the forearm of 150 rad/s2?

1 answer:

Romashka-Z-Leto [24]2 years ago

4 0

Answer:

I=0.0361 kg.m^2

Explanation:

Torque is the rotational equivalent of a force

Torque= perpendicular distance r X Force F

Torque T = I(moment of inertia) X α (angular acceleration)

T= Iα

r= 0.0285m

F= 1.9 x 10^3

T=0.0285 x 1.9 x 10^3

T= 54.15Nm

I=T/α

I=54.15/150

I=0.361 kg.m^2

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2 years ago

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Answer:

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From the question we are told that

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The coefficient of kinetic friction of the first block is $\mu_1 = 0.205$

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The coefficient of kinetic friction of the second block is $\mu _2 = 0.105$

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The force acting on the mass $M_1$ is mathematically represented as

$F_1 = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1$

=> $M_1 a = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1$

Where T is the tension on the rope

The force acting on the mass $M_2$ is mathematically represented as

$F_2 = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

$M_2 a = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

At equilibrium

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So

$T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1 =M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

making a the subject of the formula

$a = \frac{M_2 g sin \theta_2 - M_1 g sin \theta_1 - \mu_1 M_1g cos \theta - \mu_2 M_2 g cos \theta_2 }{M_1 +M_2}$

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=> $a = 0.7156 m/s^2$

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3 years ago

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alexira [117]

Answer:

The minimum inductance needed is 2.78 H

Explanation:

Given;

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