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3 years ago

Identify the ways in which carbon is added to the atmosphere?

Physics

1 answer:

marissa [1.9K]3 years ago

7 0

Each time you exhale, you are releasing carbon dioxide gas (CO2) into the atmosphere. Animals and plants get rid ofcarbon dioxide gas through a process called respiration. Carbon moves from fossil fuels to the atmosphere when fuels are burned.

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It takes you 10 seconds and 100 n of force to move a shopping cart 10 meters to the checkstand. what is your power?

Stells [14]

Power = Work/Time =(Force* Displacement)/Time

Power = (100N* 10m)/10s = 100 Watts

6 0

3 years ago

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You are out for a bicycle ride on a calm, windless day. You are heading northward on a level road and are experiencing a pressur

goldenfox [79]

The answer is (C)= zero but that it is swirling rapidly in all directions.

8 0

3 years ago

If the ball leaves the projectile launcher at a speed of 2.2 m/s at an angle of 30ᴼ, and the projectile launcher is on a table a

IgorC [24]

Answer: 1.12 m

Explanation:

This situation is related to parabolic motion, hence we can use the following equations:

$y=y_{o}+V_{o}sin \theta t-\frac{g}{2}t^{2}$ (1)

$x=V_{o} cos \theta t$ (2)

Where:

$y=0 m$ is the ball final height (when it hits the ground)

$y_{o}=1.1 m$ is the ball initial height

$V_{o}=2.2 m/s$ is the initial velocity

$\theta=30\°$ is the angle at which the ball was launched

$t$ is the time

$g=9.8 m/s^{2}$ is the acceleration due gravity

$x$ is the horizontal distance the ball travels

Rewriting (1) with the given values:

$0 m=1.1 m+(2.2 m/s)(cos 30\°)t-\frac{9.8 m/s^{2}}{2}t^{2}$ (3)

Multiplying all the eqquation by -1 and rearranging:

$4.9 m/s^{2} t^{2}-1.1 m/s t-1.1 m=0$ (4)

So, since we have a quadratic equation here (in the form of $0=at^{2}+bt+c$ , we will use the quadratic formula to find $t$ :

$t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ (5)

Where $a=4.9$ , $b=-1.1$ , $c=-1.1$

Substituting the known values and choosing the positive result of the equation, we have:

$t=\frac{-(-1.1)\pm\sqrt{(-1.1)^{2}-4(4.9)(-1.1)}}{2(4.9)}$

$t=0.59 s$ (6)

Now, substituting (6) in (2):

$x=(2.2 m/s)(cos 30\°)(0.59 s)$ (7)

$x=1.12 m$ (8) This is the horizontal distance at which the ball hits the ground.

3 0

4 years ago

Calculate the density of a solid of mass 20kg if it occupies a volume of 4m³

choli [55]

Answer:

Density=Mass/Volume

=20/4 kg/m^3

=5 kg/m^3

4 0

3 years ago

If we put negative charge between two similar positive charges then what is it's equilibrium? And how?

Gnesinka [82]

Your question has been heard loud and clear.

Well it depends on the magnitude of charges. Generally , when both positive charges have the same magnitude , their equilibrium point is towards the centre joining the two charges. But if magnitude of one positive charge is higher than the other , then the equilibrium point will be towards the charge having lesser magnitude.

Now , a negative charge is placed in between the two positive charges. So , if both positive charges have same magnitude , they both pull the negative charge towards each other with an equal force. Thus the equilibrium point will be where the negative charge is placed because , both forces are equal , and opposite , so they cancel out each other at the point where the negative charge is placed. However if they are of different magnitudes , then the equilibrium point will be shifted towards the positive charge having less magnitude.

Thank you

5 0

4 years ago

Read 2 more answers