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goblinko [34]
3 years ago
8

If the speed of light in a vacuum is twice that of the speed of light in a given material, what is the index of infraction for t

hat material?

Physics
2 answers:
Virty [35]3 years ago
6 0
I halved the speed of light in vacuum to get the speed of light in the material, since the speed of light in vacuum is twice the speed of light in material.

The index of refraction is essentially the speed of light through a vacuum divided by the speed of light through given medium.

I attached some notes on refraction too, hopefully this helps :D

Dovator [93]3 years ago
3 0

Answer:

Index of Refraction

Explanation:

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A 300 g wooden block on a smooth, level surface is firmly attached to a very light horizontal spring with a spring constant of 2
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the solution for the oscillatory movement allows to find the result for the amplitude of the initial displacement is:

  • The range of motion is: A = 4.44 cm

<h3>Oscillatory movement.</h3>

The oscillatory periodic motion of a system occurs when there is a recovered force, in the special case that this force is proportional to the displacement is called simple harmonic motion, which is described by the expression.

            x = A cos (wt + Ф)

            w² = k/m

where x is the displacement, A the amplitude, w the angular velocity, t the time, k the spring constant, m the mass, and Ф a phase constant determined by the initial conditions.

Let's find the angular velocity/

            w= \sqrt{ \frac{200}{0.300} }

            w = 25.8 rad/s

Let's look for the constant Ф, as the system is released from rest its initial velocity is zero, for zero time. The definition of speed is:

             v= \frac{dx}{dt}

             v= - A w sin (wt +Ф)

             

             0 = -A w sin Ф

            Ф= 0

They indicate that at a given instant of the time the velocity is v= 50.0 cm/s and it is in a position x= 4.00 cm, let us write the equations for this time

Position.

               4.00 = A cos 25.8t

Speed.

              50.0 = - At 25.8 sin 25.8t

To solve the system, ;et's square and add.

              Cos² 25.8t = \frac{16}{A^2}

              sin² 25.8t = \frac{3.756}{A^2 }

              1 = \frac{1}{A^2} \ (16 + 3.756)

               A = \sqrt{19.756}

               A= 4.44 cm

In conclusion using the solution for the oscillatory movement we can find the result for the amplitude of the initial displacement is:

The range of motion is: A = 444 cm

Learn more about oscillatory motion here:  brainly.com/question/14311816

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