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Juliette [100K]
3 years ago
11

You need to move a 122-kg sofa to a different location in the room. It takes a force of 81 N to start it moving. What is the coe

fficient of static friction between the sofa and the carpet?
Physics
1 answer:
jok3333 [9.3K]3 years ago
6 0

Answer:

The coefficient of static friction = 0.0677

Explanation:

Friction: These is the force that tend to oppose two bodies in contact. The force of friction is a contact force.

F = μR

Making μ the subject of the equation above.

μ = F/R ....................... Equation 1

Where F = frictional Force, R = normal reaction, μ = Coefficient of static friction

But, R = mg ................ Equation 2

<em>Given:  m= 122 kg, </em>

<em>Constant: g = 9.8 m/s²</em>

<em>Substituting these values into equation 2</em>

<em>R = 122×9.8</em>

<em>R = 1195.6 N.</em>

<em>Also Given: F = 81 N.</em>

Also substituting these values into equation 1

μ = 81/1195.6

μ = 0.0677

Therefore the coefficient of static friction = 0.0677

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2 years ago
A truck with a mass of 1370 kg and moving with a speed of 12.0 m/s rear-ends a 593 kg car stopped at an intersection. The collis
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Answer:

speed of car after collision, v2 =16.1 m/s and of the truck, v1 = 4.6 m/s

Explanation:

Given:

mass of truck M = 1370 kg

speed of truck = 12.0 m/s

mass of car m = 593 kg

collision is elastic therefore,

Applying law of momentum conservation we have

momentum before collision = momentum after collision

1370×12 + 0( initially car is at rest) = 1370×v1+ 593×v2               ....(i)

Also for a collision to be elastic,

velocity of approach = velocity of separation

12 -0 = v2-v1                  ....(ii)

using (i) and (ii) we have

So speed of car after collision, v2 =16.1 m/s and of the truck, v1 = 4.6 m/s

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2 years ago
The block slides on a horizontal frictionless surface. The block has a mass of 1.0kg and is pushed 5.0N at 45°. What is the magn
Firlakuza [10]

Answer:

3.5\:\mathrm{m/s^2}

Explanation:

Newton's 2nd law is given as \Sigma F = ma.

To find the acceleration in the horizontal direction, you need the horizontal component of the force being applied.

Using trigonometry to find the horizontal component of the force:

\cos 45^{\circ}=\frac{x}{5},\\\frac{\sqrt{2}}{{2}}=\frac{x}{5},\\x=\frac{5\sqrt{2}}{2}

Use this horizontal component of the force to solve for for the acceleration of the object:

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Answer:

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\Phi_B=BScos\theta

Therefore, there will be a variable magnetic flux, when the magnitude of the magnetic field (B) changes over time, when the area of the loop (S) changes over time and / or when the angle (\theta) between the field and the surface vector changes over time.

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