Lighting a match changes to :

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

WINSTONCH [101]

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:

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7 0

6 months ago

How will you demonstrate Newton's 1st Law of Motion at home? Brainstorm 2 or 3 possible demonstrations

Archy [21]

1. Roll a ball across a table into an object
2. Drop something

5 0

2 years ago

As a woman walks, her entire weight is momentarily placed on one heel of her high-heeled shoes. Calculate the pressure exerted o

astraxan [27]

Answer:

3.6 x 10⁶ Pa

Explanation:

A = Area of the heel = 1.50 cm² = 1.50 x 10⁻⁴ m²

m = mass of the woman = 55.0 kg

g = acceleration due to gravity = 9.8 m/s²

Force of gravity on the heel is given as

F = mg

Inserting the values

F = (55) (9.8)

F = 539 N

Pressure exerted on the floor is given as

$P = \frac{F}{A}$

$P = \frac{539}{1.5\times 10^{-4}}$

P = 3.6 x 10⁶ Pa

6 0

3 years ago

A 4 kg textbook sits on a desk. It is pushed horizontally with a 50 N applied force against a 15 N frictional force.

GarryVolchara [31]

a) See free-body diagram in attachment

b) The book is stationary in the vertical direction

c) The net horizontal force is 35 N in the forward direction

d) The net force on the book is 35 N in the forward horizontal direction

e) The acceleration is $8.75 m/s^2$ in the forward direction

Explanation:

a)

The free-body diagram of a body represents all the forces acting on the body using arrows, where the length of each arrow is proportional to the magnitude of the force and points in the same direction.

From the diagram of this book, we see there are 4 forces acting on the book:

- The applied force, F = 50 N, pushing forward in the horizontal direction

- The frictional force, $F_f = 15 N$ , pulling backward in the horizontal direction (the frictional force always acts in the direction opposite to the motion)

- The weight of the book, $W=mg$ , where m is the mass of the book and $g=9.8 m/s^2$ is the acceleration of gravity, acting downward. We can calculate its magnitude using the mass of the book, m = 4 kg:

$W=(4)(9.8)=39.2 N$

- The normal reaction exerted by the desk on the book, N, acting upward, and balancing the weight of the book

b)

The book is in equilibrium in the vertical direction, therefore there is no motion.

In fact, the magnitude of the normal reaction (N) exerted by the desk on the book is exactly equal to the weight of the book (W), so the equation of motion along the vertical direction is

$N-W=ma$