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Musya8 [376]
3 years ago
8

Which of the following is mainly responsible for the removal of carbon dioxide from the atmosphere?

Physics
2 answers:
olga55 [171]3 years ago
7 0

Answer:

plants

Explanation:

:)

allochka39001 [22]3 years ago
3 0

4.plants

Explanation:

One of the main processes that is responsible for the removal of carbon dioxide in the atmosphere is photosynthesis.

  • Plants are autotrophs that produce their own food in the ecosystem.
  • Other form of life depends on them for nourishment.
  • A certain process known as photosynthesis is used by plants to manufacture their own food.
  • In this process, green plants use carbon dioxide and water in the presence of sunlight to produce their own food.
  • Through this process oxygen is released as a by product.

Learn more ;

Photosynthesis brainly.com/question/4324141

#learnwithBrainly

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An electromagnetic wave in a vacuum has a wavelength of 0.032 m. What is its frequency?
Sergio [31]
L=wavelength
L=0.032
c=Lf
f=c/L
f=3e8/0.032
f=9.375GHz
7 0
3 years ago
A(n) ____________ stretch is one done where antagonist muscles are used to stretch the muscles. But the _____________ stretch is
aliina [53]

Answer: Dynamic - Static Flexibility

Explanation:

3 0
3 years ago
(ik it says physics but astronomy is a field of physics sooo) A recently discovered planet in a different solar system is locate
marusya05 [52]

The distance, to the nearest tenth, is 314.9 light-years.

<em>The given data is:</em>

A recently discovered planet is located 1.85*10^5 miles from Earth.

Now we want to transform this distance to light-years.

Remember that a light-year is defined as "the distance that the light would travel in one year".

using the relation:

distance = speed*time

The speed of light is:

speed = 6.706*10^8 mi/h

And in one year has 8760 hours, then we have:

time = 8760 h

replacing these in the equation we get:

distance = speed*time

distance = (6.706*10^8 mi/h)*(8760 h)  = 5,874,456,000,000 miles

Son one light-year is equivalent to 5,874,456,000,000 miles

1 light-year = 5,874,456,000,000 miles

So to transform a distance in miles to light-years, we just need to divide that distance by 5,874,456,000,000 miles:

The distance between the new planet and Earth was:

D = 1.85*10^15 mi = ( 1.85*10^15)/(5,874,456,000,000) = 314.9 light-years.

if you want to learn more about this, you can read:

brainly.com/question/1302132

3 0
3 years ago
As a cold air mass advances on a warm air mass, what usually comes before it?​
Marina CMI [18]

Answer: A cold front occurs when a cold air mass advances into a region occupied by a warm air mass. If the boundary between the cold and warm air masses doesn't move, it is called a stationary front.

Explanation: Two types of occluded front exist: the warm-type and the cold-type. They’re distinguished by the relative temperatures of the air mass ahead of the occlusion – in other words, the air mass ahead of the original warm front – and the air mass behind the cold front. If the air behind the cold front is colder than the air ahead of the occlusion, it shoves beneath that air (because it’s denser) to form a cold-type occluded front. If the air behind the cold front is warmer than the air ahead, it rides over it to form a warm-type occluded front – which appears to be the more common case. In either situation, the lighter warm air representing the air mass originally between the warm and cold fronts sits above the boundary between the two cooler air masses.

Hope this helps!!

8 0
3 years ago
Learning Goal: To be able to calculate work done by a constant force directed at different angles relative to displacement
svlad2 [7]

Answer:

Woke done, W = 4156.92 Joules

Explanation:

The work done by the force can be calculated as :

W=F\times s

W=Fs\ cos\theta

\theta is the angle between force and the displacement

It is assumed to find the work done for the given parameters i.e.

Force, F = 30 N

Distance travelled, s = 160 m

Angle between force and displacement, \theta=30

Work done is given by :

W=Fs\ cos\theta

W=30\times 160\ cos(30)

W = 4156.92 Joules

So, the work done by the object is 4156.92 Joules. Hence, this is the required solution.

5 0
3 years ago
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