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2 years ago

Through what process is carbon pulled from the atmosphere in the carbon cycle

1 answer:

7 0

Carbon is pulled from the atmosphere in the carbon cycle through the process of photosynthesis. Details about photosynthesis can be found below.

<h3>What is photosynthesis?</h3>

Photosynthesis is the process whereby green plants obtain their nutrition by utilizing energy from sunlight.

Green plants absorb carbon in the form of carbon dioxide from the atmosphere and use it in the photosynthetic process.

This means that one way that carbon is removed from the atmosphere during the carbon cycle is through photosynthesis.

Learn more about photosynthesis at: brainly.com/question/1388366

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Do quasars reside within or without side of galaxies?

sveticcg [70]

They almost entirely reside within galaxies because quasars are a subset of blackholes with a large and fast enough accretion disk to generate a beam of interstellar material perpendicular to itself. This typically only occurs in the largest black holes at the center of galaxies (supermassive blackholes) or at least stellar black holes---which still occur within galaxies because the material is necessary to form them.

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3 years ago

How did people studying lunar eclipses learn that Earth is round?

qwelly [4]

Answer:

Scientists have studied eclipses since ancient times. Aristotle observed that the Earth's shadow has a circular shape as it moves across the moon. He posited that this must mean the Earth was round. Another Greek astronomer named Aristarchus used a lunar eclipse to estimate the distance of the Moon and Sun from Earth

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3 years ago

Read 2 more answers

Which statements are true concerning Newton's law of gravitation? The gravitational force is related to the mass of each object.

OLEGan [10]

Answer:

The gravitational force is related to the mass of each object.

The gravitational force is an attractive force.

Explanation:

Gravitational force is a long range force of attraction between any two masses.

Mathematically given as :

$F=G.\frac{m_1.m_2}{r^2}$

where:

$m_1 & m_2$ are the masses

r= distance between the center of mass of the two objects.

G= gravitational constant = $6.67\times 10^{-11} m^3.kg^{-1}.s^{-2}$

From the above relation of eq. (1) it is clear that,

Gravitational force is inversely proportional to the square of the distance and directly proportional to the masses.

The mass of an object is independent of its size due to the fact that density may vary for different objects.

The force of gravity varies with height as:

$\frac{g}{g_x} =(\frac{r_x}{r} )^2$

where:

$g=9.8\,m.s^{-2}$

$g_x=$ gravity at height $r_x$ of the center of mass of the object from the center of mass of the earth.

and we know that force:

$F=m\times g$

where: m= mass of the object.

5 0

3 years ago

A 0.5 kg turntable with a radius of 15 cm is rotating 78 times per minute.

alexandr1967 [171]

Answer:

Explanation:

8 0

2 years ago

A 2.45-kg frictionless block is attached to an ideal spring with force constant 355 N/m. Initially the spring is neither stretch

ANTONII [103]

Answer:

A. A = 0.913 m

B. amax = 132.24m/s^2

C. Fmax = 324.01N

Explanation:

When the block is moving at the equilibrium point , its velocity is maximum.

A. To find the amplitude of the motion you use the following formula for the maximum velocity:

$v_{max}=A\omega$ (1)

vmax = maximum velocity = 11.0 m/s

A: amplitude of the motion = ?

w: angular frequency = ?

Then, you have to calculate the angular frequency of the motion, by using the following formula:

$\omega=\sqrt{\frac{k}{m}}$ (2)

k: spring constant = 355 N/m

m: mass of the object = 2.54 kg

$\omega = \sqrt{\frac{355N/m}{2.45kg}}=12.03\frac{rad}{s}$

Next, you solve the equation (1) for A and replace the values of vmax and w:

$A=\frac{v}{\omega}=\frac{11.0m/s}{12.03rad/s}=0.913m$

The amplitude of the motion is 0.913m

B. The maximum acceleration of the block is given by:

$a_{max}=A\omega^2 = (0.913m)(12.03rad/s)^2=132.24\frac{m}{s^2}$

The maximum acceleration is 132.24 m/s^2

C. The maximum force is calculated by using the second Newton law and the maximum acceleration:

$F_{max}=ma_{max}=(2.45kg)(132.24m/s^2)=324.01N$

It is also possible to calculate the maximum force by using:

Fmax = k*A = (355N/m)(0.913m) = 324.01N

The maximum force exertedbu the spring on the object is 324.01 N

4 0

3 years ago