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Gemiola [76]
1 year ago
5

The net reaction of the calvin cycle is the conversion of co2 into the three-carbon sugar g3p. Along the way, reactions rearrang

e carbon atoms among intermediate compounds and use the atp and nadph produced by the light reactions. In this exercise, you will track carbon atoms through the calvin cycle as required for the net production of one molecule of g3p.
Physics
1 answer:
WINSTONCH [101]1 year ago
7 0

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.

Know more about the Calvin cycle here:

brainly.com/question/15205817

#SPJ4

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If 1495 j of heat is needed to raise the temperature of a 351 g sample of a metal from 55.0°c to 66.0°c, what is the specific he
forsale [732]
The amount of heat needed to increase the temperature of a substance by \Delta T is given by
Q= mC_s \Delta T
where m is the mass of the substance, Cs is its specific heat capacity and \Delta T is the increase of temperature.

If we re-arrange the formula, we get
C_s =  \frac{Q}{m \Delta T}
And if we plug the data of the problem into the equation, we can find the specific heat capacity of the substance:
C_s =  \frac{1495 J}{(351 g)(66.0^{\circ}C-55.0^{\circ}C)}=0.39 J/g^{\circ}C
6 0
3 years ago
Read 2 more answers
A 0.20 kg baseball is traveling at 40 m/s toward the batter. The ball is hit by the bat with a force of 200N, and is
Paraphin [41]

Answer:

Time, t = 0.015 seconds.

Explanation:

Given the following data;

Mass, m = 0.2kg

Force, F = 200N

Initial velocity, u = 40m/s

Final velocity, v = 25m/s

To find the time;

Ft = m(v - u)

Time, t = m(v - u)/f

Substituting into the equation, we have;

Time, t = 0.2(25 - 40)/200

Time, t = 0.2(-15)/200

Time, t = 3/200

Time, t = 0.015 seconds.

Note: We ignored the negative sign because time can't be negative.

8 0
3 years ago
To avoid an accident, a driver steps on the brakes to stop a 1000kg car traveling at 65km/h. If the braking distance is 35m, how
Alenkinab [10]
To stop the car it would be 100m because if the car is going to 65km/h then it would still be 100km/h
8 0
3 years ago
The initial kinetic energy imparted to a 0.25 kg bullet is 1066 J. The acceleration of gravity is 9.81 m/s 2 . Neglecting air re
lubasha [3.4K]

Answer:

The range of the bullet is 0.435 kilometers.

Explanation:

According to the problem, maximum height is equal to the range of the bullet. That is:

\Delta x = \Delta y

Where:

\Delta x - Range of the bullet, measured in meters.

\Delta y - Maximum height of the bullet, measured in meters.

By the Principle of Energy Conservation, gravitational potential energy reaches its maximum at the expense of the initial kinetic energy. That is to say:

K_{1} = U_{2}

Where:

K_{1} - Kinetic energy at point 1, measured in joules.

U_{1} - Gravitational potential energy at point 2, measured in joules, and:

U_{2} = m\cdot g \cdot \Delta y

Where:

m - Mass of the bullet, measured in kilograms.

g - Gravitational constant, measured in meters per square second.

The maximum height is now cleared:

K_{1} = m\cdot g \cdot \Delta y

\Delta y = \frac{K_{1}}{m\cdot g}

If K_{1} = 1066\,J, m = 0.25\,kg and g = 9.81\,\frac{m}{s^{2}}, the maximum height is now computed:

\Delta y = \frac{1066\,J}{(0.25\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}

\Delta y = 434.791\,m

\Delta y = 0.435\,km

Lastly, the range of the bullet is 0.435 kilometers.

3 0
3 years ago
How much is the moon in the first quarter actually lit
grandymaker [24]
Half of the moon is illuminated.
8 0
4 years ago
Read 2 more answers
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