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crimeas [40]
3 years ago
12

Pins accompained by a ball bowling dates back to what era

Physics
2 answers:
Afina-wow [57]3 years ago
6 0
That is one of the weirdest questions
jekas [21]3 years ago
5 0

Answer:

Ancient Egypt

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007 (part 1 of 2) 1.0 points
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Answer:

a) The angle of refraction is approximately 34.7

b) The angle the light have to be incident to give an angle of refraction of 90° is approximately 53.42°

Explanation:

According to Snell's law, we have;

\dfrac{n_1}{n_2} = \dfrac{sin (\theta_2)}{sin (\theta_1)}

The refractive index of the glass, n₁ = 1.66

The angle of incident of the light as it moves into water, θ₁ = 27.2°

a) The refractive index of water, n₂ = 1.333

Let θ₂ represent the angle of refraction of the light in water

By plugging in the values of the variables in Snell's Law equation gives;

\dfrac{1.66}{1.333} = \dfrac{sin (\theta_2)}{sin (27.2^{\circ})}

sin (\theta_2) = sin (27.2^{\circ}) \times \dfrac{1.66}{1.333} \approx 0.5692292265

θ₂ = arcsin(0.5692292265) ≈ 34.7°

The angle of refraction of the light in water, θ₂ ≈ 34.7°

b) When the angle of refraction, θ₂ = 90°, we have;

\dfrac{1.66}{1.333} = \dfrac{sin (90^{\circ})}{sin (\theta_1)}

sin (\theta_1) = \dfrac{sin (90^{\circ})}{\left( \dfrac{1.66}{1.333}\right)} = sin (90^{\circ}) \times \dfrac{1.333}{1.66} \approx 0.803

θ₁ ≈ arcsin(0.803) ≈ 53.42°

The angle of incident, θ₁, that would give an angle of refraction of 90° is θ₁ ≈ 53.42°

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