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Goryan [66]
3 years ago
13

A spotlight on a boat is y = 2.2 m above the water, and the light strikes the water at a point that is x = 8.5 m horizontally di

splaced from the spotlight (see the drawing). The depth of the water is 4.0 m. Determine the distance d, which locates the point where the light strikes the bottom.
Physics
1 answer:
slava [35]3 years ago
5 0

Answer:

The answer to the question is

The distance d, which locates the point where the light strikes the bottom is   29.345 m from the spotlight.

Explanation:

To solve the question we note that Snell's law states that

The product of the incident index and the sine of the angle of incident is equal to the product of the refractive index and the sine of the angle of refraction

n₁sinθ₁ = n₂sinθ₂

y = 2.2 m and strikes at x = 8.5 m, therefore tanθ₁ = 2.2/8.5 = 0.259 and

θ₁ =  14.511 °

n₁ = 1.0003 = refractive index of air

n₂ = 1.33 = refractive index of water

Therefore sinθ₂ =  \frac{n_1sin\theta_1}{n_2}  = \frac{1.003*0.251}{1.33} = 0.1885 and θ₂ = 10.86 °

Since the water depth is 4.0 m we have tanθ₂ = \frac{4}{x_2} or x₂ = \frac{4}{tan\theta_2 } =\frac{4}{tan(10.86)} = 20.845 m

d = x₂ + 8.5 = 20.845 m + 8.5 m = 29.345 m.

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24.8m/s

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Given data

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Read 2 more answers
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irina [24]

Answer:

Both bullets will hit the ground at the same time.

Explanation:

Let's only analyze the vertical problem.

Any object that is not in the floor or resting in some site is being affected by the gravitational force (remember that we are ignoring air resistance)

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The object will hit the ground when p(t) = 0

Then we need to solve for t the next equation:

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Notice that the only things we need to know are:

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This means that both bullets will hit the ground at the same time.

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