The conversion of CO2 into the three-carbon sugar G3p is the Calvin cycle's net reaction.
<h3 /><h3>What is Calvin Cycle?</h3>
Calvin cycle uses ATP and NADPH, which are created during the light reaction, to decrease atmospheric carbon dioxide in 3C sugar; Three steps can be distinguished in the Calvin Cycle reaction:
Carboxylation: A reaction that is catalyzed by RUBISCO, where RUBIP is the CO2 acceptor, results in the formation of 6 molecules of PGA (phosphoglycerate).
Here, there are 6 molecules with 18 carbons in PGA and 3 molecules with 3 carbons in CO2.
Reduction: One molecule of the output, glyceraldehyde 3 phosphate, is used to create sugar while the remaining five molecules go through regeneration.
Here, one glyceraldehyde-3-phosphate molecule with three carbon atoms is present.
Regeneration: From glyceraldehyde 3 phosphate, CO2 acceptor (RUBIP) is generated.
Here, 3 molecules of RUBIP and 5 molecules of dihydroxy acetone phosphate each have 15 carbons.
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1) The law of motion of the projectile is

To find the velocity, we should compute the derivative of h(t):

So now we can calculate the speed at t=2 s and t=4 s:


The negative sign in the second speed means the projectile has already reached its maximum height and it is now going downward.
2) The projectile reaches its maximum height when the speed is equal to zero:

So we have

And solving this we find

3) To find the maximum height, we take h(t) and we just replace t with the time at which the projectile reaches the maximum height, i.e. t=2.30 s:

4) The time at which the projectile hits the ground is the time at which the height is zero: h(t)=0. So, this translates into

This is a second-order equation, and if we solve it we get two solutions: the first solution is negative, so we can ignore it since it's physically meaningless; the second solution is

And this is the time at which the projectile hits the ground.
5) The velocity of the projectile when it hits the ground is the velocity at time t=4.68 s:

with negative sign, because it is directed downward.
Those two units can be compared to a 'mile per hour' and a 'mile per hour - hour'.
One is a rate. The other is a quantity, after maintaining a rate for some time.
-- 'Joule' is a unit of energy. It's the amount of work (energy) you do
when you push with a force of 1 newton though a distance of 1 meter.
Lifting 10 pound of beans 3 feet off the floor takes about 40.7 joules of energy.
-- 'Watt' is a <u><em>rate</em></u> of using energy . . . 1 joule per second.
If you lift 10 pounds 3 feet off the floor in 1 second, your <em>power</em> is 40.7 watts.
-- 'Watt-second' is the amount of energy used in one second,
at the rate of 1 joule per second . . . 1 joule.
-- 'Watt-hour' is the amount of energy used in one hour,
at the rate of 1 joule per second . . . 3,600 joules.
-- 'Kilowatt' is a bigger <em>rate</em> of using energy . . . 1,000 joules per second.
-- 'Kilowatt - second' is the amount of energy used in one second,
at the rate of 1,000 joules per second . . . 1,000 joules .
-- 'Kilowatt - hour' is the amount of energy used in one hour,
at the rate of 1,000 joules per second . . . 3,600,000 joules .
Depending on where you live, 3,600,000 joules of energy bought
from the electric company costs something between 5¢ and 25¢.
Work = Force x Distance
Assuming that this work is being done parallel to the displacement that is, but under that assumption:
W = (50)(10)
W = 500 J
Reflect parallel of the principal axis