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otez555 [7]
2 years ago
9

10 kg mass sliding along a horizontal rough surface at speed 30 m/s and rests in 6 seconds. what is the coefficient of friction

Physics
1 answer:
Aleks [24]2 years ago
7 0

The coefficient of friction is 0.5.

<h3>What is Coefficient of friction?</h3>

The ratio of friction force to normal force is known as the coefficient of friction (COF), which has no dimensions. Those materials are said to as lubricous if their COF is less than 0.1. Surface roughness and COF are dependent on the composition of the materials.

Initial velocity, u=30m/s

time taken = 6s

Final velocity, v=0

Acceleration of friction, a=μg=10μms⁻²

Apply first kinematic equation of motion,

v - u = at

0 - 30 = (-10μms⁻²) × 6

-30 = -60μ

μ = 1/2

μ = 0..5  

Hence, coefficient of friction is 0.5

to learn more about Coefficient of friction go to - brainly.com/question/14121363

#SPJ4

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A 1 kg particle moves upward from the origin to (23) m. Wit is work done by the force of gravity which is in - y direction s B.
sveta [45]

Answer:

Explanation:

mass, m = 1 kg

Position (2, 3 ) m

height, h = 2 m

acceleration due to gravity, g = 9.8 m/s^2

Here, no force is acting in horizontal direction, the force of gravity is acting in vertical direction, so the work done by the gravitational force is to be calculated.

Force  mass x acceleration due to gravity

F = 1 x 9.8 = 9.8 N

Work = force x displacement x CosФ

Where, Ф be the angle between force vector and the displacement vector.

Here the value of Ф is 180° as the force acting vertically downward and the displacement is upward

So, W = 9.8 x 2 x Cos 180°

W = - 19.6 J

Thus, option (A) is correct.

4 0
3 years ago
A star produces 2x10^26 watts. how much energy does it lose every minutes
Step2247 [10]

Answer:

Energy loss per minute will be 120\times 10^{26}j

Explanation:

We have given the star produces power of 2\times 10^{26}W

We know that 1 W = 1 J/sec

So 2\times 10^{26}W=2\times 10^{26}J/sec

Given time = 1 minute = 60 sec

So the energy loss per minute =2\times 10^{26}\times 60=120\times 10^{26}j

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7 0
3 years ago
Starting from rest, a 2.3x10-4 kg flea springs straight upward. While the flea is pushing off from the ground, the ground exerts
Harman [31]

Answer:

3.13 m/s

Explanation:

From the question,

Since the flea spring started from rest,

Ek = W................... Equation 1

Where Ek = Kinetic Energy of the flea spring, W = work done on the flea spring.

But,

Ek = 1/2mv²............ Equation 2

Where m = mass of the flea spring, v = flea's speed when it leaves the ground.

substitute equation 2 into equation 1

1/2mv² = W.................... Equation 3

make v the subject of the equation

v = √(2W/m)................. Equation 4

Given: W = 3.6×10⁻⁴ J, m = 2.3×10⁻⁴ kg

Substitute into equation 4

v = √[2×3.6×10⁻⁴ )/2.3×10⁻⁴]

v = 7.2/2.3

v = 3.13 m/s

Hence the flea's speed when it leaves the ground  = 3.13 m/s

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