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klio [65]
3 years ago
8

The law of reflection states that the angle of incidence is equal to the angle of reflection.

Physics
2 answers:
djyliett [7]3 years ago
5 0
Yes ... the law of reflection states that the angle of incidence is equal to the angle of reflection <em /><em>
it also states that the incident ray reflected ray and the normal at the point of incidence lie on the same plane....;

</em>
Minchanka [31]3 years ago
3 0

Answer:

yes, it is true.

Explanation:

When a light ray falls on a highly smooth and polished surface, it bounces back into the same medium. This phenomenon is called reflection of light.

There are two laws of reflection.

1. The angle of incidence is equal to the angle of reflection.

2. Incident ray, reflected ray and the normal all lie in the same plane.

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Permeability is a measure of how fast a liquid can pass through a layer of solid. In this case, the lesser the time, the more permeable the solid is. Or the other way around, the bigger the time, the less permeable the solid is. Therefore the answer in this problem is:

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4 years ago
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a toy dart gun generates a dart with .140kg.m/s momentum and a velocity of 4m/s. what is the mass of the dart in grams?
sleet_krkn [62]

Answer: m = 0.035kg = 35g

Explanation: Momentum p=0.140kgm/s

Velocity v=4m/s

Mass m=?

Formula-

Momentum depends on the mass of the object in motion and its velocity.

The equation for momentum is

p = mv

m = p/v

m = 0.140/4

m = 0.035kg

m = 35g

Hence, in the toy dart gun mass of the dart is 0.035kg.

4 0
2 years ago
Two astronauts of mass 100 kg are 2 m apart in outer space. What is the
fredd [130]

The force of gravity between the astronauts is 1.67\cdot 10^{-7}N

Explanation:

The magnitude of the gravitational force between two objects is given by:

F=G\frac{m_1 m_2}{r^2}

where :

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2} is the gravitational constant

m_1, m_2 are the masses of the two objects

r is the separation between them

In this problem, we have two astronauts, whose masses are:

m_1 = 100 kg\\m_2 = 100 kg

While the separation between the astronauts is

r = 2 m

Substituting into the equation, we can find the gravitational force between the two astronauts:

F=\frac{(6.67\cdot 10^{-11})(100)(100)}{2^2}=1.67\cdot 10^{-7}N

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4 0
3 years ago
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Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 6.1 m/s. Ignore f
tatiyna

Answer:

a) h=3.16 m, b)  v_{cm }^ = 6.43 m / s

Explanation:

a) For this exercise we can use the conservation of mechanical energy

Starting point. Highest on the hill

           Em₀ = U = mg h

final point. Lowest point

           Em_{f} = K

Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere

           K = ½ m v_{cm }^{2} + ½ I_{cm} w²

angular and linear speed are related

           v = w r

           w = v / r

            K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²

            Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)

as there are no friction losses, mechanical energy is conserved

             Em₀ = Em_{f}

             mg h = ½ v_{cm }^{2} (m + I_{cm} / r²)         (1)

             h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)

for the moment of inertia of a basketball we can approximate it to a spherical shell

             I_{cm} = ⅔ m r²

we substitute

            h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)

            h = ½ v_{cm }^{2}/g    5/3

             h = 5/6 v_{cm }^{2} / g

           

let's calculate

           h = 5/6 6.1 2 / 9.8

           h = 3.16 m

b) this part of the exercise we solve the speed of equation 1

          v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)

in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia

              I_{cm} = ½ m r²

we substitute

             v_{cm } = √ [2gh / (1 + ½)]

             v_{cm } = √(4/3 gh)

let's calculate

             v_{cm } = √ (4/3 9.8 3.16)

             v_{cm }^ = 6.43 m / s

4 0
3 years ago
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