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dangina [55]
3 years ago
13

The critical angle for a substance is measured at 53.7 degrees. Light enters from air at 45.0 degrees. At what angle it will con

tinue to travel?
Physics
1 answer:
lys-0071 [83]3 years ago
8 0

Answer:

34.74

Explanation:

Critical angle is represented by the formula

θ(c) = sin^-1 (n2/n1)

And we have been given that the critical angle is 53.7, so we substitute

53.7 = sin^-1 (n2/n1), rearranging, we have

Sin 53.7 = n2/n1

0.8059 = n2/n1

Again, we know that the second medium is air, and the refractive index of air is 1, so

0.8059 = 1/n1, making n1 subject of formula

n1 = 1/0.8059

n1 = 1.24

Now, we have gotten both indices, using Snell's law of refraction, we then find the needed refraction

Sin θ(r) = n(i)/n(r) Sin θ(i)

Now, solving, we have

Sin θ(r) = 1/1.24 * sin 45

Sin θ(r) = 0.8059 * 0.7071

Sin θ(r) = 0.5699

θ(r) = Sin^-1 0.5699

θ(r) = 34.74

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5.0 kg, with a bullet of mass 0.1 kg. The target was mounted on
kari74 [83]

Answer:

306 m/s

Explanation:

Law of conservation of momentum

m1v1 + m2v2 = (m1+m2)vf

m1 is the bullet's mass so it is 0.1 kg

v1 is what we're trying to solve

m2 is the target's mass so it is 5.0 kg

v2 is the targets velocity, and since it was stationary, its velocity is zero

vf is the velocity after the target is struck by the bullet, so it is 6.0 m/s

plugging in, we get

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