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3 years ago

A monkey hangs motionless by one arm from a tree branch. Draw the free-body diagram.

Physics

1 answer:

Alexeev081 [22]3 years ago

7 0

Find the free-body diagram in attachment.

Explanation:

The free-body diagram of an object is a diagram representing all the forces acting on a body. Every force is represented using an arrow, where:

- The length of the arrow is proportional to the magnitude of the force

- The direction of the arrow shows the direction of the force

For the monkey in this problem, we have two forces acting on it:

- The weight of the monkey, labelled as (mg), acting downward, where

m is the mass of the monkey

g is the acceleration of gravity

- The tension in the tree branch, labelled as T, which acts upward

The monkey is motionless: this means that its acceleration is zero, so the net force actin on it is zero - this means that the forces acting on the monkey, the weight and the tension, are balanced, therefore they are shown as two arrows of same length, but opposite direction.

Learn more about forces and weight here:

brainly.com/question/8459017

brainly.com/question/11292757

brainly.com/question/12978926

#LearnwithBrainly

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A car is moving at 19 m/s along a curve on a horizontal plane with radius of curvature 49m.

JulsSmile [24]

Answer:

$\mu =0.75$

Explanation:

<u>Frictional Force </u>

When the car is moving along the curve, it receives a force that tries to take it from the road. It's called centripetal force and the formula to compute it is:

$F_c=m.a_c$

The centripetal acceleration a_c is computed as

$\displaystyle a_c=\frac{v^2}{r}$

Where v is the tangent speed of the car and r is the radius of curvature. Replacing the formula into the first one

$F_c=m.\frac{v^2}{r}$

For the car to keep on the track, the friction must have the exact same value of the centripetal force and balance the forces. The friction force is computed as

$F_r=\mu N$

The normal force N is equal to the weight of the car, thus

$F_r=\mu .m.g$

Equating both forces

$\displaystyle \mu .m.g=m.\frac{v^2}{r}$

Simplifying

$\displaystyle \mu =\frac{v^2}{rg}$

Substituting the values

$\displaystyle \mu =\frac{19^2}{(49)(9.8)}$

$\boxed{\mu =0.75}$

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3 years ago

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saveliy_v [14]

Answer:

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Explanation:

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adell [148]

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Rasek [7]

Answer:

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