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3 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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Answer:

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Explanation:

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3 years ago

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What happens when the temperature of a substance decreases significantly?

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A 54.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 126 m/s from the top of a cliff

Nikolay [14]

Answer:

a) The initial total mechanical energy of the projectile is 498556.296 joules.

b) The work done on the projectile by air friction is 125960.4 joules.

c) The speed of the projectile immediately before it hits the ground is approximately 82.475 meters per second.

Explanation:

a) The system Earth-projectile is represented by the Principle of Energy Conservation, the initial total mechanical energy ( $E$ ) of the project is equal to the sum of gravitational potential energy ( $U_{g}$ ) and translational kinetic energy ( $K$ ), all measured in joules:

$E = U_{g} + K$ (Eq. 1)

We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:

$E = m\cdot g\cdot y + \frac{1}{2}\cdot m\cdot v^{2}$ (Eq. 1b)

Where:

$m$ - Mass of the projectile, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$y$ - Initial height of the projectile above ground, measured in meters.

$v$ - Initial speed of the projectile, measured in meters per second.

If we know that $m = 54\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $y = 132\,m$ and $v = 126\,\frac{m}{s}$ , the initial mechanical energy of the earth-projectile system is:

$E = (54\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (132\,m)+\frac{1}{2}\cdot (54\,kg)\cdot \left(126\,\frac{m}{s} \right)^{2}$

$E = 498556.296\,J$

The initial total mechanical energy of the projectile is 498556.296 joules.

b) According to this statement, air friction diminishes the total mechanical energy of the projectile by the Work-Energy Theorem. That is:

$W_{loss} = E_{o}-E_{1}$ (Eq. 2)

Where:

$E_{o}$ - Initial total mechanical energy, measured in joules.

$E_{1}$ - FInal total mechanical energy, measured in joules.

$W_{loss}$ - Work losses due to air friction, measured in joules.

We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:

$W_{loss} = E_{o}-K_{1}-U_{g,1}$

$W_{loss} = E_{o} -\frac{1}{2}\cdot m\cdot v_{1}^{2}-m\cdot g\cdot y_{1}$ (Eq. 2b)

Where:

$m$ - Mass of the projectile, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$y_{1}$ - Maximum height of the projectile above ground, measured in meters.

$v_{1}$ - Current speed of the projectile, measured in meters per second.

If we know that $E_{o} = 498556.296\,J$ , $m = 54\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $y_{1} = 297\,m$ and $v_{1} = 89.3\,\frac{m}{s}$ , the work losses due to air friction are:

$W_{loss} = 498556.296\,J -\frac{1}{2}\cdot (54\,kg)\cdot \left(89.3\,\frac{m}{s} \right)^{2} -(54\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (297\,m)$

$W_{loss} = 125960.4\,J$

The work done on the projectile by air friction is 125960.4 joules.

c) From the Principle of Energy Conservation and Work-Energy Theorem, we construct the following model to calculate speed of the projectile before it hits the ground:

$E_{1} = U_{g,2}+K_{2}+1.5\cdot W_{loss}$ (Eq. 3)

$K_{2} = E_{1}-U_{g,2}-1.5\cdot W_{loss}$

Where:

$E_{1}$ - Total mechanical energy of the projectile at maximum height, measured in joules.

$U_{g,2}$ - Potential gravitational energy of the projectile, measured in joules.

$K_{2}$ - Kinetic energy of the projectile, measured in joules.

$W_{loss}$ - Work losses due to air friction during the upward movement, measured in joules.

We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:

$\frac{1}{2}\cdot m \cdot v_{2}^{2} = E_{1}-m\cdot g\cdot y_{2}-1.5\cdot W_{loss}$ (Eq. 3b)

$m\cdot v_{2}^{2} = 2\cdot E_{1}-2\cdot m \cdot g \cdot y_{2}-3\cdot W_{loss}$

$v_{2}^{2} = 2\cdot \frac{E_{1}}{m}-2\cdot g\cdot y_{2}-3\cdot \frac{W_{loss}}{m}$

$v_{2} = \sqrt{2\cdot \frac{E_{1}}{m}-2\cdot g\cdot y_{2}-3\cdot \frac{W_{loss}}{m} }$

If we know that $E_{1} = 372595.896\,J$ , $m = 54\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $y_{2} =0\,m$ and $W_{loss} = 125960.4\,J$ , the final speed of the projectile is:

$v_{2} =\sqrt{2\cdot \left(\frac{372595.896\,J}{54\,kg}\right)-2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0\,m)-3\cdot \left(\frac{125960.4\,J}{54\,kg}\right) }$

$v_{2} \approx 82.475\,\frac{m}{s}$

The speed of the projectile immediately before it hits the ground is approximately 82.475 meters per second.

7 0

3 years ago

how long would it take for a radio wave sent from a space satellite circling mars to reach Earth? Assume that radio waves (a for

kati45 [8]

Answer:

3 Minutes 2 seconds to 22 minutes 16 seconds

Explanation:

Lets assume the Mars to be at the closest distance to Earth. This distance (D) = 54.6 Million km

The signal travels at the speed of light, so speed of the signal (V) = 300000 km/s

So, the time (T) taken by the radio wave to reach Earth from Mars will be,

$T = \frac{D}{V}$

$T = \frac{54600000}{300000}$

Thus, T = 182 Sec = 3 Minutes 2 seconds.

The radio wave will take minimum 03 Minutes 02 seconds to reach Earth. Here is should be noted that the distance between the two planets keep on changing as they revolve around the Sun. There will come a point when Mars is farthest from Earth and the distance (D) will be 401 Million km. Then, the time will change to ,

$T = \frac{401000000}{300000}$

T = 1336.67 sec = 22.27 Minutes.

So the maximum time will be 22 minutes 16 seconds.

3 0

3 years ago

A box is at rest at the top of a frictionless inclined plane. As the box slides down the ramp where along the ramp would the box

Yuliya22 [10]

Answer:

I would say near the end of the slide.

3 0

3 years ago