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3 years ago

An object has a coefficient of kinetic friction of 0.2 and a normal force of 30N. Find the force of kinetic friction.

Physics

1 answer:

shutvik [7]3 years ago

6 0

Answer:

Force of kinetic friction, Fk = 6N

Explanation:

Given the following data;

Coefficient of kinetic friction, μ = 0.2

Normal force, Fn=30N

To find the Force of kinetic friction;

Mathematically, the force of kinetic friction is given by the formula;

Fk = μFn

Where;

Fk represents the force of kinetic friction.

μ represents the coefficient of friction.

Fn represents the normal force.

Substituting into the equation, we have;

Fk = 0.2*30

Fk = 6N

Therefore, the force of kinetic friction is 6N.

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3 years ago

An electron is released from rest at the negative plate of a parallel-plate capacitor. If the distance across the plate is 2.0 m

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Answer:

Velocity of electron will be $1.325\times 10^6m/sec$

Explanation:

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4 years ago

A 2.0-kg block slides on a rough horizontal surface. A force (magnitude P = 4.0 N) acting parallel to the surface is applied to

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When the applied force increases to 5 N, the magnitude of the block's acceleration is 1.7 m/s².

<h3>Frictional force between the block and the horizontal surface</h3>

The frictional force between the block and the horizontal surface is determined by applying Newton's law;

∑F = ma

F - Ff = ma

Ff = F - ma

Ff = 4 - 2(1.2)

Ff = 4 - 2.4

Ff = 1.6 N

When the applied force increases to 5 N, the magnitude of the block's acceleration is calculated as follows;

F - Ff = ma

5 - 1.6 = 2a

3.4 = 2a

a = 3.4/2

a = 1.7 m/s²

Thus, when the applied force increases to 5 N, the magnitude of the block's acceleration is 1.7 m/s².

Learn more about frictional force here: brainly.com/question/4618599

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2 years ago

Which of the following is example of convection check all that apply

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3 years ago

A young diver is practicing his skills before an important team competition. Use the diagram below in order to analyze the energ

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Mass of the diver (m) = 90 Kg.
Height of the board from the ground (h) = 10 m.
Acceleration due to gravity (g) = 9.8 m/s^2.
Height of the diver from the ground when he reaches point C (x) = 5m
Initial velocity (u) = 0 m/s
We know, gravitational potential energy of a body = mass × acceleration due to gravity × height.
Therefore, the gravitational potential energy of the diver when he reaches point C (GPE) = mg(h - x)
or, GPE = [90 × 9.8 × (10-5)] J
or, GPE = [90 × 9.8 × 5] J
or, GPE = 4410 J
For a freely falling body,
v^2 - u^2 = 2gh
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We know, kinetic energy of a body = 1/2 mv^2
Therefore, kinetic energy of the diver when he reaches point C (KE) = 1/2 m(2gx)
Here, 2gx = (2 × 9.8 × 5) = 98 (m/s)^2
We have already seen v^2 = 2gh
or, v = √2gh
So, the velocity of the diver = √2gx = √98 m/s = 9.9 m/s

Answers:

The gravitational potential energy of the diver when he reaches point C is 4410 J.

The velocity of the diver is 9.9 m/s.

Hope you could get an idea from here.

Doubt clarification - use comment section.

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2 years ago

Read 2 more answers