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Sphinxa [80]
3 years ago
14

The spring-mounted 0.84-kg collar A oscillates along the horizontal rod, which is rotating at the constant angular rate rad/s. A

t a certain instant, r is increasing at the rate of 610 mm/s.
If the coefficient of kinetic friction between the collar and the rod is 0.61, calculate the friction force F exerted by the rod on the collar at this instant.
Physics
1 answer:
vfiekz [6]3 years ago
8 0

Answer:

Friction = 5.027 N

Explanation:

Kinetic friction does not change with speed or acceleration. The only thing that is required is to state that the object is moving.

Solving the force equation for friction we get:

Friction = coefficient of friction* normal force exerted by the object

Friction = u * R

here R is = 0.84 * Gravity = 0.84 * 9.81 = 8.2404 N

u = 0.61 (as stated in the question)

So frictional force will be:

Friction = 0.61 * 8.2404

Friction = 5.027 N

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The purpose of the scapula to move during arm elevation is increase the range of elevation of the arm.

<h3>What is the importance of movement of the scapula during arm elevation?</h3>

The scapula is an important bone which is found in the shoulder and back region of the body.

The scapula enables and increases the range of motion of the arm with its motions.

During arm elevation, the scapula undergoes an upward rotational motion.

Therefore, the  purpose of the scapula to move during arm elevation is increase the range of elevation of the arm.

Learn more about scapula motion at: brainly.com/question/5133017

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2 years ago
A box has a mass of 150kg and the surface area of the bottom of the box is :
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Explanation:

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is the rate at which ____________________ changes. (Velocity or Acceleration, velocity or acceleration)
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2 years ago
The speed of molecules is indirectly related to a measurement called _____.
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Read 2 more answers
A 65.0-kg runner has a speed of 5.20 m/s at one instant dur- ing a long-distance event. (a) What is the runner’s kinetic energy
vladimir2022 [97]

Answer:

a)KE=878.8 J

b)W=2636.4 J      

Explanation:

Given that

mass ,m = 65 kg

Initial speed ,u = 5.2 m/s

a)

We know that kinetic energy KE is given as follows

KE=\dfrac{1}{2}mu^2

m=mass

u=velocity

Now by putting the values in the above equation we get

KE=\dfrac{1}{2}\times 65\times 5.2^2\ J

KE=878.8 J

b)

We know that

Work done by all forces = Change in the kinetic energy

The final velocity , v= 2 u = 2 x 5.2 m/s

v= 10.4 m/s

W=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2

Now by putting the values in the above equation we get

W=\dfrac{1}{2}\times 65\times 10.4^2-\dfrac{1}{2}\times 65\times 5.2^2\ J

W=2636.4 J

a)KE=878.8 J

b)W=2636.4 J

8 0
3 years ago
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