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Mila [183]
1 year ago
8

In addition to ozone, what four other greenhouse gases or groups of greenhouse gases are included in nearly all the climate mode

ls?
Physics
1 answer:
Musya8 [376]1 year ago
5 0

In addition to ozone, the four other greenhouse gases or groups of greenhouse gases are included in nearly all the climate models are Carbon dioxide, water vapor, and methane, nitrous oxide.

Carbon dioxide, absorbs energy at a variety of wavelengths, a range that overlaps with that of infrared energy. As CO2 soaks up this infrared energy, it vibrates and re-emits the infrared energy back in all directions. About half of that energy goes out into space, and about half of it returns to Earth as heat, contributing to the ‘greenhouse effect.’

As methane is emitted into the air, it reacts in several ways. It reacts and  after oxidization methane forms water vapor and carbon dioxide. So, not only does methane contribute to global warming directly but also, indirectly through the release of carbon dioxide.

Nitrous oxide enhances the greenhouse effect by capturing reradiated infrared radiation from the Earth’s surface and subsequently warming the troposphere . It is chemically inert in the troposphere and stays in the troposphere for about 120 years before moving into the stratosphere where it ultimately leads to destruction of stratospheric ozone.

Water vapor does absorb longwave radiation and radiates it back to the surface, thus contributing to warming.

To learn more about greenhouse gases here

brainly.com/question/14131369?referrer=searchResults

#SPJ4

You might be interested in
Which is the correct order from least amount of atomic (particle) movement to the most amount of atomic particle movement?
babymother [125]

Answer: particles movement in solid< particles movement in liquids< particles movement in gases.

Explanation:

Atoms are very small, it is not easily seen even with the help of light microscopes. However, We use multiple models of atoms toexplain describe particles of an atom behaviour.

In solids, the particles are packed together tightly in an ordered arrangement. The particles only vibrate about their position in the structure because the particles are held together too strongly to allow movement. Thereby,making the particles MOVE THE LEAST

In liquids, the particles are close together and they move with random motion in the container. The particles move rapidly in all directions but there is more colision between itself even more than particles in gases. This means that the particles here are MORE FASTER THAN THAT OF THE SOLID.

Particles in gases move the FASTEST, more than the particles in solids and liquids. Although, the average speed of the particles depends on their mass and the temperature.

8 0
3 years ago
I NEED HELP WITH THE LAST QUESTION PLS HELP!! (The one below 8)
maksim [4K]

Answer:

I. 6 cells .

II. Series connection.

Explanation:

I. Determination of the number of cells needed.

From the question given above,

Total voltage (V) = 9 V

1.5 V = 1 cell

Number of cells needed =?

The number of cells needed to make the 9V battery can be obtained as follow:

1.5 V = 1 cell

Therefore,

9 V = 9 V × 1 cell / 1.5 V

9 V = 6 cells

Thus, 6 cells of 1.5 V each is needed

II. Determination of the connection line

Total voltage (Vₜ) = 9 V

Cell 1 (V₁) = 1.5 V

Cell 2 (V₂) = 1.5 V

Cell 3 (V₃) = 1.5 V

Cell 4 (V₄) = 1.5 V

Cell 5 (V₅) = 1.5 V

Cell 6(V₆ ) = 1.5 V

For parrall connection:

Vₜ = V₁ = V₂ = V₃ = V₄ = V₅ = V₆

9 V = 1.5 V =... = 1.5 V

For series connection:

Vₜ = V₁ + V₂ + V₃ + V₄ + V₅ + V₆

9 = 1.5 + 1.5 + 1.5 + 1.5 + 1.5 + 1.5

9 V = 9 V

From the illustration above, we can see that series connection of each cells will give a total volt of 9 V unlike the parallel connection which resulted to 1.5 V.

Therfore, the cells should be arranged in series connection

8 0
2 years ago
What are three inventions used to increase friction?<br>
Sophie [7]
The answer I believe is shoes, tires and gloves.
4 0
3 years ago
A point charge is used to determine the electric field around a charged particle. Why is it necessary that the point charge does
Mnenie [13.5K]
The strength, and possibly the shape and direction, of the electric field
around a charged particle depends on the location of the particle. 

If the process of measuring the field causes the particle to move, then
the measurement you get wouldn't mean anything. 

Your measurements wouldn't show the ACTUAL field around the particle.
They would show what the field is like AFTER something comes along
and distorts it, and that's not what you're trying to measure.

It would be like carrying a flame thrower into a freezer when you go in
to measure the temperature in there.

Or if you had to measure how much light is leaking into a dark room,
and you carried a flashlight with you to see your way around in there.
7 0
3 years ago
Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of -0.9537 N when separated by
scoray [572]

Answer:

<em>The initial charges on the spheres are  </em>6.796\ 10^{-6}\ c and -3.898\ 10^{-6}\ c

Explanation:

<u>Electrostatic Force </u>

Two charges q1 and q2 separated a distance d exert a force on each other which magnitude is computed by the known Coulomb's formula

\displaystyle F=\frac{K\ q_1\ q_2}{d^2}

We are given the distance between two unknown charges d=50 cm = 0.5 m and the attractive force of -0.9537 N. This means both charges are opposite signs.

With these conditions we set the equation

\displaystyle F_1=\frac{K\ q_1\ q_2}{0.5^2}=-0.9537

Rearranging

\displaystyle q_1\ q_2=\frac{-0.9537(0.5)^2}{k}

Solving for q1.q2

\displaystyle q_1\ q_2=-2.6492.10^{-11}\ c^2\ \ ......[1]

The second part of the problem states the spheres are later connected by a conducting wire which is removed, and then, the spheres repel each other with an electrostatic force of 0.0756 N.

The conducting wire makes the charges on both spheres to balance, i.e. free electrons of the negative charge pass to the positive charge and they finally have the same charge:

\displaystyle q=\frac{q_1+q_2}{2}

Using this second condition:

\displaystyle F_2=\frac{K\ q^2}{0.5^2}=\frac{K(q_1+q_2)^2}{(4)0.5^2}=0.0756

\displaystyle q_1+q_2=2.8983\ 10^{-6}\ C

Solving for q2

\displaystyle q_2=2.8983\ 10^{-6}\ C-q_1

Replacing in [1]

\displaystyle q_1(2.8983\ 10^{-6}-q_1)=-2.64917.10^{-11}

Rearranging, we have a second-degree equation for q1.  

\displaystyle q_1^2-2.8983.10^{-6}q_1-2.64917.10^{-11}=0

Solving, we have two possible solutions

\displaystyle q_1=6,796.10^{-6}\ c

\displaystyle q_1=-3.898.10^{-6}\ c

Which yields to two solutions for q2

\displaystyle q_2=-3.898.10^{-6}\ c

\displaystyle q_2=6.796.10^{-6}\ c

Regardless of their order, the initial charges on the spheres are 6.796\ 10^{-6}\ c and -3.898\ 10^{-6}\ c

8 0
3 years ago
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