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Tpy6a [65]
3 years ago
15

A Ray of light in air is incident on an air to glass boundary at an angle of 30. degrees with the normal. If the index of refrac

tion of the glass is 1.52, what is the angle of the refracted ray within the glass with respect to the normal? A. 19 degrees B. 22 degrees C. 30 degrees D. 1.52
Physics
1 answer:
elena55 [62]3 years ago
5 0

Answer:

A

Explanation:

Snell's law states:

n₁ sin θ₁ = n₂ sin θ₂

where n is the index of refraction and θ is the angle of incidence (relative to the normal).

The index of refraction of air is approximately 1.  So:

1 sin 30° = 1.52 sin θ

θ ≈ 19°

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A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, and
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Answer:

(a) t = 1.14 s

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(b)

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(2)(9.8\ m/s^2)h = (0\ m/s)^2-(4\ m/s)^2\\

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Now, for the time in air during upward motion we use first equation of motion:

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(c)

Now we will consider the downward motion and use the third equation of motion:

2gh = v_f^2-v_i^2

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vf = final speed = ?

Therefore,

2(9.8\ m/s^2)(2.62\ m) = v_f^2 - (0\ m/s)^2\\

<u>vf = 7.17 m/s</u>

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v_f = v_i + gt_1\\7.17\ m/s = 0\ m/s + (9.8\ m/s^2)t_2\\t_2 = 0.73\ s

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t = 0.41 s + 0.73 s

<u>t = 1.14 s</u>

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