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4 years ago

What is the speed between reflected ray and the incident ray

Physics

1 answer:

Korvikt [17]4 years ago

7 0

Answer:

The speed is the same as long as the reflection is regular.

Explanation:

This is because in regular reflection, the angle of incidence is equal to the angle of reflection in accordance with the second law of reflection.

Since speed of light depends on the angle of the light ray it makes with the reflecting surface, the speed is the same

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vivado [14]

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3 years ago

A small box is held in place against a rough vertical wall by someone pushing on it with a force directed upward at 27 ∘ above t

ycow [4]

Answer:

Explanation:

Applied force, F = 18 N

Coefficient of static friction, μs = 0.4

Coefficient of kinetic friction, μs = 0.3

θ = 27°

Let N be the normal reaction of the wall acting on the block and m be the mass of block.

Resolve the components of force F.

As the block is in the horizontal equilibrium, so

F Cos 27° = N

N = 18 Cos 27° = 16.04 N

As the block does not slide so it means that the syatic friction force acting on the block balances the downwards forces acting on the block .

The force of static friction is μs x N = 0.4 x 16.04 = 6.42 N .... (1)

The vertically downward force acting on the block is mg - F Sin 27°

= mg - 18 Sin 27° = mg - 8.172 ... (2)

Now by equating the forces from equation (1) and (2), we get

mg - 8.172 = 6.42

mg = 14.592

m x 9.8 = 14.592

m = 1.49 kg

Thus, the mass of block is 1.5 kg.

6 0

3 years ago

You push your friend, whose mass is 54kg, down a hill so she can go sledding. Her acceleration is 3m/s2. Calculate the amount of

Alecsey [184]

Answer:

18 newtons

Explanation:

Divide weight by speed

3 0

3 years ago

1. Does the validity of the principle of conservation of momentum depend on the validity of Newton's 3rd law of motion?

OverLord2011 [107]

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4 years ago

A box rests on top of a flat bed truck. The box has a mass of m = 16.0 kg. The coefficient of static friction between the box an

3241004551 [841]

Answer:

1) 1.31 m/s2

2) 20.92 N

3) 8.53 m/s2

4) 1.76 m/s2

5) -8.53 m/s2

Explanation:

1) As the box does not slide, the acceleration of the box (relative to ground) is the same as acceleration of the truck, which goes from 0 to 17m/s in 13 s

$a = \frac{\Delta v}{\Delta t} = \frac{17 - 0}{13} = 1.31 m/s2$

2)According to Newton 2nd law, the static frictional force that acting on the box (so it goes along with the truck), is the product of its mass and acceleration

$F_s = am = 1.31*16 = 20.92 N$

3) Let g = 9.81 m/s2. The maximum static friction that can hold the box is the product of its static coefficient and the normal force.

$F_{\mu_s} = \mu_sN = mg\mu_s = 16*9.81*0.87 = 136.6N$

So the maximum acceleration on the block is

$a_{max} = F_{\mu_s} / m = 136.6 / 16 = 8.53 m/s^2$

4)As the box slides, it is now subjected to kinetic friction, which is

$F_{\mu_s} = mg\mu_k = 16*9.81*0.69 = 108.3 N$

So if the acceleration of the truck it at the point where the box starts to slide, the force that acting on it must be at 136.6 N too. So the horizontal net force would be 136.6 - 108.3 = 28.25N. And the acceleration is

28.25 / 16 = 1.76 m/s2

5) Same as number 3), the maximum deceleration the truck can have without the box sliding is -8.53 m/s2

3 0

3 years ago