All categories

2 years ago

How does the Earth release energy back into the atmosphere?

1 answer:

8 0

In general, the Earth releases energy back to the atmosphere through reflection, evaporation, and radiation. The Earth gets energy from the sunlight, part of which it absorbs, while part it reflects backwards, thus giving energy to the atmosphere. Also, the heating up of the Earth by the absorbed sunlight, radiates back in the lower layers of the atmosphere, again giving back energy to it. The water vapor is another way in which the Earth gives back energy tot he atmosphere as through the evaporation, the water vapor gets into the lower parts of the atmosphere and gives energy to it.

You might be interested in

Suppose that a comet has a very eccentric orbit that brings it quite close to the Sun at closest approach (perihelion) and beyon

Nutka1998 [239]

Answer:

16.63min

Explanation:

The question is about the period of the comet in its orbit.

To find the period you can use one of the Kepler's law:

$T^2=\frac{4\pi}{GM}r^3$

T: period

G: Cavendish constant = 6.67*10^-11 Nm^2 kg^2

r: average distance = 1UA = 1.5*10^11m

M: mass of the sun = 1.99*10^30 kg

By replacing you obtain:

$T=\sqrt{\frac{4\pi}{GM}r^3}=\sqrt{\frac{4\pi^2}{(6.67*10^{-11}Nm^2/kg^2)(1.99*10^{30}kg)}(1.496*10^8m)^3}\\\\T=997.9s\approx16.63min$

the comet takes around 16.63min

8 0

2 years ago

What is the magnitude of the gravitational force acting on the sun due to the earth?The earth does not exert any gravitational f

Sholpan [36]

I haven't worked on Part-A, and I don't happen to know the magnitude of the gravitational force that the Sun exerts on the Earth.

But whatever it is, it's exactly, precisely, identical, the same, and equal to the magnitude of the gravitational force that the Earth exerts on the Sun.

I think that's the THIRD choice here, but I'm not sure of that either.

4 0

2 years ago

Suppose that you observe light emitted from a distant star to be at a wavelength of 525 nm. The wavelength of light to an observ

Tomtit [17]

Answer:

The velocity of the star is 0.532 c.

Explanation:

Given that,

Wavelength of observer = 525 nm

Wave length of source = 950 nm

We need to calculate the velocity

If the direction is from observer to star.

From Doppler effect

$\lambda_{0}=\sqrt{\dfrac{c+v}{c-v}}\times\lambda_{s}$

Put the value into the formula

$525=\sqrt{\dfrac{c+v}{c-v}}\times950$

$\dfrac{c+v}{c-v}=(\dfrac{525}{950})^2$

$\dfrac{c+v}{c-v}=0.305$

$c+v=0.305\times(c-v)$

$v(1+0.305)=c(0.305-1)$

$v=\dfrac{0.305-1}{1+0.305}c$

$v=−0.532c$

Negative sign shows the star is moving toward the observer.

Hence, The velocity of the star is 0.532 c.

7 0

2 years ago

The measure of the average kinetic energy of the particles of a substance is the ______.

denis23 [38]

The kelvin temperature/scale i think

7 0

2 years ago

a large heavy truck and a baby carriage roll down a hill. Neglecting friction, at the bottom of the hill, the baby carriage will

Lisa [10]

Answer:

please give me brainlist and follow

Explanation:

At the bottom of the hill, the baby carriage will likely have less momentum Therefore, option D is correct. Solution: ... Therefore, at the bottom of the hill, the heavy truck will have more momentum and baby carriage will have less momentum.

5 0

2 years ago