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kupik [55]
4 years ago
14

Light travels faster in warmer air. On a sunny day, the sun can heat a road and create a layer of hot air above it. Let's model

this layer as a uniform one with a sharp boundary separating it from the cooler air above. Use this model to explain the formation of a mirage appearing like the shiny surface of a pool of water.
Physics
1 answer:
AysviL [449]4 years ago
6 0

Answer:

Explanation:

If we assume there is a sharp boundary between the two masses of air, there will be a refraction. The refractive index of each medium will depend on the relative speeds of light.

n = c / v

If light travels faster in warmer air, it will have a lower refractive index

nh < nc

Snell's law of refraction relates angles of incidence and refracted with the indexes of refraction:

n1 * sin(θ1) = n2 * sin(θ2)

sin(θ2) = sin(θ1) * n1/n2

If blue light from the sky passing through the hot air will cross to the cold air, then

n1 = nh

n2 = nc

Then:

n1 < n2

So:

n1/n2 < 1

The refracted light will come into the cold air at angle θ2 wich will be smaller than θ1, so the light is bent upwards, creating the appearance of water in the distance, which is actually a mirror image of the sky.

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Both Newton's Law of Universal Gravitation and Coulomb's Law follow the ...
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Answer:

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Both the Newton's law of universal gravitation and coulomb's law have their force inversely proportion to the square of the distance between the bodies.

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Gravitational potential energy is related to–A. what an object is made of. B.an object’s motion.C. an object’s position.D. an ob
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3 years ago
A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3.0 m is initially at rest. A 20 kg boy approaches the m
nekit [7.7K]

Answer:

The velocity of the merry-go-round after the boy hops on the merry-go-round is 1.5 m/s

Explanation:

The rotational inertia of the merry-go-round = 600 kg·m²

The radius of the merry-go-round = 3.0 m

The mass of the boy = 20 kg

The speed with which the boy approaches the merry-go-round = 5.0 m/s

F_T \cdot r = I \cdot \alpha  = m \cdot r^2  \cdot \alpha

Where;

F_T = The tangential force

I =  The rotational inertia

m = The mass

α = The angular acceleration

r = The radius of the merry-go-round

For the merry go round, we have;

I_m \cdot \alpha_m  = I_m \cdot \dfrac{v_m}{r \cdot t}

I_m = The rotational inertia of the merry-go-round

\alpha _m = The angular acceleration of the merry-go-round

v _m = The linear velocity of the merry-go-round

t = The time of motion

For the boy, we have;

I_b \cdot \alpha_b  = m_b \cdot r^2  \cdot \dfrac{v_b}{r \cdot t}

Where;

I_b = The rotational inertia of the boy

\alpha _b = The angular acceleration of the boy

v _b = The linear velocity of the boy

t = The time of motion

When the boy jumps on the merry-go-round, we have;

I_m \cdot \dfrac{v_m}{r \cdot t} = m_b \cdot r^2  \cdot \dfrac{v_b}{r \cdot t}

Which gives;

v_m = \dfrac{m_b \cdot r^2  \cdot \dfrac{v_b}{r \cdot t} \cdot r \cdot t}{I_m} = \dfrac{m_b \cdot r^2  \cdot v_b}{I_m}

From which we have;

v_m =  \dfrac{20 \times 3^2  \times 5}{600} =  1.5

The velocity of the merry-go-round, v_m, after the boy hops on the merry-go-round = 1.5 m/s.

5 0
3 years ago
ASAP pls answer right and I will mark brainiest
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1. D:Asteroids have geological activity
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3.D:All of the above
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I hope this helps
6 0
3 years ago
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