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Vilka [71]
3 years ago
12

What are the two most important processes in the cycling of oxygen in and out of the atmosphere?

Physics
2 answers:
butalik [34]3 years ago
5 0
C. 
It is C because photosynthesis uses CO2 to form O2 and respiration uses O2 to create CO2, both processes filter oxygen through the atmosphere.
Ainat [17]3 years ago
5 0

Answer: Option C

Explanation:

Photosynthesis is the process by which plants makes food by the help of the carbon dioxide found in the atmosphere. This process take place in the presence of sunlight.

The end product of this process is glucose with oxygen as a by-product. The oxygen is released into the atmosphere.

The oxygen is used by all the living forms of life on earth and the carbon dioxide is released out by the animals in the atmosphere by the help of the process known as cellular respiration.

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Assign a positive velocity to the red box and negative velocity to the blue box. Are they moving in opposite direction?
attashe74 [19]

Answer:

<em>Yes, they are moving in opposite direction one to the other.</em>

Explanation:

Velocity is a vector quantity, which means that it has both magnitude and direction. The magnitude shows the size of the velocity, and the direction shows which way it is moving in reference to a chosen reference direction. If the red box is assigned a positive velocity, and the blue box is assigned a negative velocity, as indicated in the question, then it means that the red box, and the blue box, both move in opposite direction to the other.

4 0
3 years ago
¿ Cuáles son los principales aportes de las ciencias en el desarrollo del mundo? Menciona una aplicación de Física del contenido
maria [59]

Answer:earth, water, air, fire, and (later) aether, which were proposed to explain the nature and complexity of all matter in terms of simpler substances.

Explanation:

3 0
3 years ago
Why aren't there lunar and solar eclipses every month?
Shalnov [3]

The moon's orbital plane is different from that of the earth's orbit around the sun due to which lunar and solar eclipses do not occur every month.

<h3 /><h3>What is a solar eclipse?</h3>

When the moon passes in front of Earth and the sun, it creates a solar eclipse and casts a shadow across the globe.

Only during the new moon phase, when the moon travels squarely between the sun and Earth and casts shadows on its surface, can a solar eclipse occur.

Due to the moon's orbit being five degrees off-center from Earth's orbit around the Sun, lunar eclipses do not happen every month and the Moon often passes above or below the shadow.

Lunar eclipses would happen every month if there were no tilt. Eclipses of the sun and the moon happen equally often.

Hence lunar and solar eclipses aren't coming every month.

To learn more about the solar eclipse refer;

brainly.com/question/17749647

#SPJ1

3 0
2 years ago
Read 2 more answers
A 210 Ohm resistor uses 9.28 W of
stiv31 [10]
  • Resistance=R=210Ohm
  • Power=9.28W=P
  • Current=I

\boxed{\sf P=I^2R}

\\ \sf\longmapsto I^2=\dfrac{P}{R}

\\ \sf\longmapsto I^2=\dfrac{9.28}{210}

\\ \sf\longmapsto I^2\approx0.04

\\ \sf\longmapsto I\approx\sqrt{0.04}

\\ \sf\longmapsto I\approx\sqrt{\dfrac{4}{100}}

\\ \sf\longmapsto I\approx\dfrac{\sqrt{4}}{\sqrt{100}}

\\ \sf\longmapsto I\approx\dfrac{2}{10}

\\ \sf\longmapsto I\approx0.2A

8 0
3 years ago
In 0.601 s, a 13.1-kg block is pulled through a distance of 4.19 m on a frictionless horizontal surface, starting from rest. The
Naya [18.7K]

Answer:

0.615 m

Explanation:

We need to determine the force on the spring first. By Newton's second law of motion, force is the product of the mass and acceleration. The mass is given.

The acceleration is determined using the equation of motion.

Given parameters:

Initial velocity, <em>u</em> = 0.00 m/s

Distance, <em>s</em> = 4.19 m

Time, <em>t</em> = 0.601 s

We use the equation

s = ut+\frac{1}{2}at^2

With <em>u</em> = 0.00 m/s,

s = \frac{1}{2}at^2

a = \dfrac{2s}{t^2}

a = \dfrac{2\times4.19\text{ m}}{(0.601\text{ s})^2} = 23.2\text{ m/s}^2

The force is

F = (13.1\text{ kg})(23.2\text{ m/s}^2) = 303.92 \text{ N}

From Hooke's law, the extension, <em>e</em>, of a string is given by

e = \dfrac{F}{k}

where <em>k</em> is the spring constant.

Hence,

e = \dfrac{303.92\text{ N}}{494\text{ N/m}} = 0.615\text{ m}

4 0
3 years ago
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