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Butoxors [25]
3 years ago
9

A box of weight 50 N is at rest on a floor where the coefficient of static friction is 0.20. A rope is attached to the box and p

ulled horizontally to the right. What is the friction force that exerts by the floor on the box and which direction
Physics
1 answer:
Bas_tet [7]3 years ago
5 0

Answer:

The friction force that exerts by the floor on the box is 10 N.        

Explanation:

Given that,

Weight of the box, W = 50 N

The coefficient of static friction is 0.2

A rope is attached to the box and pulled horizontally to the right. We need to find the friction force that exerts by the floor on the box. The direction of frictional force is in left side as we know that the frictional force acts in the opposite direction of motion. The force of friction is given by :

F=\mu N

N is normal reaction, N = W = 50 N

F=0.2\times 50\\\\F=10\ N

So, the friction force that exerts by the floor on the box is 10 N.

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When a 0.622 kg basketball hits the floor, its velocity changes from 4.23 m/s down to 3.85 m/s up. If the ball was in contact wi
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when the ball hits the floor and bounces back the momentum of the ball changes.

the rate of change of momentum is the force exerted by the floor on it.

the equation for the force exerted is

f = rate of change of momentum

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v is the final velocity which is - 3.85 m/s

u is initial velocity - 4.23 m/s

m = 0.622 kg

time is the impact time of the ball in contact with the floor - 0.0266 s

substituting the values

f = \frac{0.622 kg (3.85 m/s - (-)4.23 m/s)}{0.0266}

since the ball is going down, we take that as negative and ball going upwards as positive.

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A 8.8 cm diameter circular loop of wire is in a 1.04 T magnetic field. The loop is removed from the field in 0.30 s . Assume tha
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Answer:

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Explanation:

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<u>Where:</u>

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<em>ΔΦ = ΔB*A                                            </em>

<em>Δt = is the time = 0.3 s   </em>

<em>A = is the loop of wire area = πr² = πd²/4 </em>

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Hence the average induced emf is:

E = - \frac{\Delta B*A}{\Delta t} = - \frac{(0- 1.04 T) \pi (0.088 m)^{2}}{4*0.3 s} = 0.021 V                      

Therefore, the average induced emf is 0.021 V.

I hope it helps you!

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