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3 years ago

Light travels from crown glass (n=1:52) into air (n=1.00). The angle of refraction in

Physics

1 answer:

RUDIKE [14]3 years ago

8 0

The angle of incidence in glass is $34.7^{\circ}$

Explanation:

We can solve this problem by applying Snell's law of refraction, which states that:

$n_1 sin \theta_1 = n_2 sin \theta_2$

where

$n_1, n_2$ are the index of refraction of the first and second medium, respectively

$\theta_1, \theta_2$ are the angle of incidence and refraction, respectively

In this problem we have:

$n_1 = 1.52$ is the index of refraction of the first medium (glass)

$n_2 = 1.00$ is the index of refraction of the second medium (air)

$\theta_2 =60^{\circ}$ is the angle of refraction in glass

Solving for $\theta_i$ , we find the angle of incidence:

$\theta_1 = sin^{-1} (\frac{n_2 sin \theta_2}{n_1})=\sin^{-1}(\frac{(1.00)(sin 60^{\circ})}{1.52})=34.7^{\circ}$

Learn more about refraction:

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For a freely falling object weighing 3 kg : A. what is the object's velocity 2 s after it's release. B. What is the kinetic ener

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A) 19.6 m/s (downward)

B) 576 J

C) 19.6 m

D) Velocity: not affected, kinetic energy: doubles, distance: not affected

Explanation:

An object in free fall is acted upon one force only, which is the force of gravity.

Therefore, the motion of an object in free fall is a uniformly accelerated motion (constant acceleration). Therefore, we can find its velocity by applying the following suvat equation:

$v=u+at$

where:

v is the velocity at time t

u is the initial velocity

$a=g=9.8 m/s^2$ is the acceleration due to gravity

For the object in this problem, taking downward as positive direction, we have:

$u=0$ (the object starts from rest)

$a=9.8 m/s^2$

Therefore, the velocity after

t = 2 s

is:

$v=0+(9.8)(2)=19.6 m/s$ (downward)

The kinetic energy of an object is the energy possessed by the object due to its motion.

It can be calculated using the equation:

$KE=\frac{1}{2}mv^2$

where

m is the mass of the object

v is the speed of the object

For the object in the problem, at t = 2 s, we have:

m = 3 kg (mass of the object)

v = 19.6 m/s (speed of the object)

Therefore, its kinetic energy is:

$KE=\frac{1}{2}(3)(19.6)^2=576 J$

In order to find how far the object has fallen, we can use another suvat equation for uniformly accelerated motion:

$s=ut+\frac{1}{2}at^2$

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

For the object in free fall in this problem, we have:

u = 0 (it starts from rest)

$a=g=9.8 m/s^2$ (acceleration of gravity)

t = 2 s (time)

Therefore, the distance covered is

$s=0+\frac{1}{2}(9.8)(2)^2=19.6 m$

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M = 6 kg

For part A) (final velocity of the object), we notice that the equation that we use to find the velocity does not depend at all on the mass of the object. This means that the value of the final velocity is not affected.

For part B) (kinetic energy), we notice that the kinetic energy depends on the mass, so in this case this value has changed.

The new kinetic energy is

$KE'=\frac{1}{2}Mv^2$

where

M = 6 kg is the new mass

v = 19.6 m/s is the speed

Substituting,

$KE'=\frac{1}{2}(6)(19.6)^2=1152 J$

And we see that this value is twice the value calculated in part A: so, the kinetic energy has doubled.

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