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fomenos
3 years ago
8

An oil layer that is 5.0 cm thick is spread smoothly and evenly over the surface of water on a windless day. A light ray from th

e air above enters the oil at 45° with the normal and then goes into the water. What is the angle of refraction of this ray in the water? The index of refraction for the oil is 1.15, and for water it is 1.33.
Physics
1 answer:
marshall27 [118]3 years ago
7 0

Answer:

The angle of refraction in water 32.12°.

Explanation:

Given that,

Thickness = 5.0 cm

Index of refraction for oil = 1.15

Index of refraction for water = 1.33

Angle = 45°

We need to calculate the angle of refraction

When the ray of light enters from air to oil

Using formula of refraction

n_{a}\sin\theta_{a}=n_{o}\sin\theta_{o}

Where, n_{a} = refractive index of air

n_{0} = refractive index of oil

Put the value into the formula

1\times\sin45=1.15\sin\theta_{o}

\sin\theta_{o}=\dfrac{1}{\sqrt{2}\times1.15}

\sin\theta_{o}=0.6148

\theta_{o}=sin^{-1}0.6149

\theta_{o}=37.94^{\circ}

When the ray of light enters from oil to water

Using formula of refraction

n_{0}\sin\theta_{0}=n_{w}\sin\theta_{w}

Where,  n_{w} = refractive index of water

1.15\times\sin37.94^{\circ}=1.33\sin\theta_{w}

\sin\theta_{w}=\dfrac{1.15\sin37.94^{\circ}}{1.33}

\theta_{w}=\sin^{-1}0.53163

\theta_{w}=32.12^{\circ}

Hence, The angle of refraction in water 32.12°.

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The mass of fuel the engine burn each second to produce a thrust of 7.66×10⁵ N is 2.5×10² kg/s.

<h3 /><h3>What is mass?</h3>

Mass can be defined as the quantity of matter contained in a body. The S.I unit of mass is kilogram(kg)

To calculate the mass the engine burns each seconds, we use the formula below.

Formual:

  • M = T/v............. Equation

Where:

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  • T = Thrust
  • v = Velocity

From the question,

Given:

  • T = 7.66×10⁵ N
  • v = 3.05×10³ m/s

Substitute these values into equation 1

  • M = (7.66×10⁵)/(3.05×10³)
  • M = 2.5×10² kg/s

Hence, the mass of fuel burned in each second is 2.5×10² kg/s.

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Suppose that a person gets hit by a bus moving at 30 mi/h with a 58,000 lbs of force in the direction of motion. If the mass of
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The impulse of a force is due to the change in the motion of an object

A. The persons speed after impact is approximately 59.38 mi/h

B. The expected speed is <u>29.89 mi/h</u> which is less than the findings

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Known parameters are;

The speed of the bus, v = 30 mi/h

The force with which the person was hit, F = 58,000 lbs

Mass of the bus, M = 40,000 lbs

Mass of the person, m = 150 lbs

Duration of the impact, Δt = 0.007 seconds

A. The speed of the person at the end of the impact, <em>v</em>, is given as follows;

The impulse of the force = F × Δt = m × Δv

For the person, we get;

58,000 lbf ≈ 1866094.816 lb·ft./s²

58,000 lbf × 0.007 s = 150 lbs × Δv

1,866,094.816 lb·ft./s²

\Delta v = \dfrac{1,866,094.816\ lbs \times 0.007 \, s}{150 \, lbs} \approx  87.084  \ ft./s

Δv = v₂ - v₁

The initial speed of the person at the instant, can be as v₁ = 0

The final speed, v₂ = Δv - v₁

∴ v₂ ≈  87.084 ft./s - 0 = 87.084 ft./s

≈ <u>87.084 ft./s</u>

<u />v_2 \approx \dfrac{87.084 \ ft./s}{y} \times\dfrac{1 \ mi}{5280 \ ft.} \times \dfrac{3,600 \ s}{1 \, hour} \approx 59.38 \ mi/h<u />

The speed of the person at the end of the impact, v₂ ≈ <u>59.38 mi/h</u>

B. Where the momentum is conserved, we have;

m₁·v₁ + m₂v₂ = (m₁ + m₂)·v

v = \dfrac{m_1 \cdot v_1 + m_2 \cdot v_2}{m_2 + m_1}

v = \dfrac{40,000 \times 30  + 150 \times 0}{40,000 + 150} \approx 29.89

The expected speed of the person at the end of the impact is 29.89 mi/h, and therefore, <u>the findings does not agree with the expectation</u>

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