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andrew-mc [135]
3 years ago
8

Get it right and win a brilliant

Physics
1 answer:
BlackZzzverrR [31]3 years ago
3 0

In Newton's third law, the action and reaction forces D.)act on different objects

Explanation:

Newton's third law of motion states that:

<em>"When an object A exerts a force on object B (action force), then action B exerts an equal and  opposite force (reaction force) on object A"</em>

It is important to note from the statement above that the action force and the reaction force always act on different objects. Let's take an example: a man pushing a box. We have:

  • Action force: the force applied by the man on the box, forward
  • Reaction force: the force applied by the box on the man, backward

As we can see from this example, the action force is applied on the box, while the reaction force is applied on the man: this means that the two forces do not act on the same object. This implies that whenever we draw the free-body diagram of the forces acting on an object, the action and reaction forces never appear in the same diagram, since they act on different objects.

Learn more about Newton's third law of motion:

brainly.com/question/11411375

#LearnwithBrainly

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The bones of the forearm (radius and ulna) are hinged to the humerus at the elbow. The biceps muscle connects to the bones of th
sertanlavr [38]

Answer:

The force exerted by the biceps is 143.8 kgf.

Explanation:

To calculate the force exerted by the biceps, we calculate the momentum in the elbow.

This momentum has to be zero so that her forearm remains motionless.

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W: mass weight (6.15 kg)

d_W= distance to the mass weight (0.425 m)

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d_A: distance to the center of mass of the forearm (0.425/2=0.2125 m)

H: force exerted by the biceps

d_H: distance to the point of connection of the biceps (0.0215 m)

The momemtum is:

H*d_H-A*d_A-W*d_W=0\\\\H=(A*d_A+W*d_W)/d_H\\\\H=(2.25*0.2125+6.15*0.425)/0.0215\\\\H=(0.478125+2.61375)/0.0215\\\\H=3.091875/0.0215=143.8

The force exerted by the biceps is 143.8 kgf.

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