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rewona [7]
1 year ago
15

How does a mirage occur in a deser?explain with a labelled diagram.​

Physics
2 answers:
Ronch [10]1 year ago
8 0

Answer:

Mirages happen when the ground is very hot and the air is cool.

Explanation:

They happen when light passes through two layers of air with different temperatures. The desert sun heats the sand, which in turn heats the air just above it. The hot air bends light rays and reflects the sky.

When you see it from a distance, the different air masses colliding with each other act as a mirror.

azamat1 year ago
6 0

Answer:

When it is very hot and sunny, roads can become very hot. When light rays from the sun reach this air pocket just above the road, the speed of the photon increases slightly, causing its path to alter, or bend from an observer's point of view. This makes something that looks like a puddle of water appear on the road.

People sometimes label a mirage as an illusion or as a hallucination. But, a mirage is neither one of those. Illusions and hallucinations are products of the mind. But the physics of Earth's atmosphere causes a mirage.

Mirages really have nothing to do with water at all. It's really all about how light travels through the air.


Here is a labelled diagram:

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2 years ago
A big wheel at a theme park has a diameter of 14m and people on the ride complete one revolution in 24s. calculate the distance
just olya [345]

Explanation:

We'll call the radius r and the diameter d:

We also assume that the riders are at a distance r = d/2 = 7m from the center of the wheel.

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New 5G networks utilize millimeter-wave radiation. Millimeter-wave radiation refers to electromagnetic waves with frequencies in
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Answer:

It corresponds to 1mm-10 mm range.

Explanation:

  • Electromagnetic waves (such as the millimeter-wave radiation) travel at the speed of light, which is 3*10⁸ m/s in free space.
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        v = \lambda * f  (1)

  • Replacing v= c=3*10⁸ m/s, and the extreme values of f (which are givens), in (1) and solving for λ, we can get the free-space wavelengths that correspond to the 30-300 GHz range, as follows:

       \lambda_{low} = \frac{c}{f_{high}}  = \frac{3e8m/s}{300e9Hz} = 1 mm (2)

      \lambda_{high} = \frac{c}{f_{low}}  = \frac{3e8m/s}{30e9Hz} = 10 mm (3)

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