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lisov135 [29]
3 years ago
12

Describe what happens to radiant energy once it enters the atmosphere? How does Radiant energy cause weather?

Physics
1 answer:
Triss [41]3 years ago
6 0
Once energy from the Sun gets to Earth, several things can happen to it:

Energy can be scattered or absorbed by aerosols in the atmosphere. Aerosols are dust, soot, sulfates and nitric oxides. When aerosols absorb energy, the atmosphere becomes warmer. When aerosols scatter energy, the atmosphere is cooled.
Short wavelengths are absorbed by ozone in the stratosphere.
Clouds may act to either reflect energy out to space or absorb energy, trapping it in the atmosphere.
The land and water at Earth's surface may act to either reflect energy or absorb it. Light colored surfaces are more likely to reflect sunlight, while dark surfaces typically absorb the energy, warming the planet.
Albedo is the percentage of the Sun's energy that is reflected back by a surface. Light colored surfaces like ice have a high albedo, while dark colored surfaces tend to have a lower albedo. The buildings and pavement in cities have such a low albedo that cities have been called "heat islands" because they absorb so much energy that they warm up.
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Infrared waves from the sun are what make our skin feel warm on a sunny day. If an infrared wave has a frequency of 3.0 x 1012 H
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Answer:

The wavelength of the infrared wave is <u>0.0001 m</u>.

Explanation:

Given:

Frequency of an infrared wave is, f=3.0\times 10^{12}\ Hz

We know that, infrared waves are electromagnetic waves. All electromagnetic waves travel with the same speed and their magnitude is equal to the speed of light in air.

So, speed of infrared waves coming from the Sun travels with the speed of light and thus its magnitude is given as:

v=c=3.0\times 10^8\ m/s

Where, 'v' is the speed of infrared waves and 'c' is the speed of light.

Now, we have a formula for the speed of any wave and is given as:

v=f\lambda

Where, \lambda \to \textrm{Wavelength of infrared wave}

Now, rewriting the above formula in terms of wavelength, \lambda, we get:

\lambda=\dfrac{v}{f}

Now, plug in 3.0\times 10^8 for 'v', 3.0\times 10^{12} for 'f' and solve for  \lambda. This gives,

\lambda=\frac{3.0\times 10^8}{3.0\times 10^{12}}\\\\\lambda=0.0001\ m

Therefore, the wavelength of the infrared wave is 0.0001 m.

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