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4 years ago

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

Physics

1 answer:

Alexeev081 [22]4 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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3 years ago

A ball is thrown with a speed of 20 m/s at an angle of 40o above the horizontal from the top of a 22-m tall building.

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Answer:

(a). The distance is 49.79 m.

(b). The speed of the ball is 24.39 m/s.

Explanation:

Given that,

Speed = 20 m/s

Angle = 40°

Height = 22 m

Time = 3.25 sec

(a). We need to calculate the distance

Using formula of distance

$d=u\cos\theta\times t$

Put the value into the formula

$d=20\cos40\times3.25$

$d=49.79\ m$

(b). We need to calculate the horizontal velocity

Using formula of velocity

$v_{x}=u\cos\theta$

Put the value into the formula

$v_{x}=20\times\cos40$

$v_{x}=15.3\ m/s$

We need to calculate the vertical velocity

Using equation of motion

$v_{y}=u\sin\theta-gt$

Put the value into the formula

$v_{y}=20\sin40-9.8\times3.25$

$v_{y}=-19\ m/s$

Negative sign shows the opposite direction.

We need to calculate the speed of ball

Using formula of speed

$v=\sqrt{v_{x}^2+v_{y}^2}$

$v=\sqrt{(15.3)^2+(19)^2}$

$v=24.39\ m/s$

Hence, (a). The distance is 49.79 m.

(b). The speed of the ball is 24.39 m/s.

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3 years ago

An Olympic track runner starts from rest and has an acceleration of 2.4 m/s2 for 3.6 s, then has zero acceleration for the remai

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Answer:

The runner's speed at the following times would remain 8.64 m/s.

Explanation:

Acceleration definition: Acceleration is rate of change in velocity of an object with respect to time.

In this case, after 3.6 seconds the acceleration is zero, it means that the velocity of the runner after 3.6 seconds is not changing and it will remain constant for the remainder of the race. Now, we have to find the velocity of the runner that he had after 3.6 seconds and that would be the runner's speed for the remainder of the race. For this we use first equation of motion.

First equation of motion: Vf = Vi + a×t

Vf stands for final velocity

Vi stands for initial velocity

a stands for acceleration

t stands for time

In the question, it is mentioned that the runner starts from rest so its initial velocity (Vi) will be 0 m/s.

The acceleration (a) is given as 2.4 m/s²

The time (t) is given as 3.6 s

Now put the values of Vi, a and t in first equation of motion

Vf = Vi + a×t

Vf = 0 + 2.4×3.6

Vf = 2.4×3.6

Vf = 8.64 m/s

So,the runner's speed at the following times would remain 8.64 m/s.

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3 years ago

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