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3 years ago

What are the products of linear electron flow during the light reactions of photosynthesis?

Physics

1 answer:

Katena32 [7]3 years ago

7 0

Answer:

NADPH and ATP

Explanation:

In the clear stage the light that "hits" chlorophyll excites an electron to a higher energy level. In a series of reactions, energy is converted (throughout an electron transport process) into ATP and NADPH. Water breaks down in the process releasing oxygen as a secondary product of the reaction. ATP and NADPH are used to make the C-C bonds in the dark stage.

Photophosphorylation is the process of converting the energy of the electron excited by light into a pyrophosphate bond of an ADP molecule. This occurs when water electrons are excited by light in the presence of P680. The energy transfer is similar to the chemosmotic electron transport that occurs in the mitochondria.

Light energy causes the removal of an electron from a P680 molecule that is part of Photosystem II, the electron is transferred to an acceptor molecule (primary acceptor), and then passes downhill to Photosystem I through a conveyor chain of electrons The P680 requires an electron that is taken from the water by breaking it into H + ions and O-2 ions. These O-2 ions combine to form O2 that is released into the atmosphere.

The light acts on the P700 molecule of Photosystem I, causing an electron to be raised to a higher potential. This electron is accepted by a primary acceptor (different from the one associated with Photosystem II).

The electron goes through a series of redox reactions again, and finally combines with NADP + and H + to form NADPH, a carrier of H needed in the independent phase of light.

Electron of photosystem II replaces the excited electron of the P700 molecule.

There is therefore a continuous flow of electrons (non-cyclic) from water to NADPH, which is used for carbon fixation.

Cyclic electron flow occurs in some eukaryotes and in photosynthetic bacteria. NADPH does not occur, only ATP. This also occurs when the cell requires additional ATP, or when there is no NADP + to reduce it to NADPH.

In Photosystem II, the "pumping" of H ions into the thylakoids (from the stroma of the chloroplast) and the conversion of ADP + P to ATP is motorized by an electron gradient established in the thylakoid membrane.

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1. A 930-kg car traveling 56 km/h comes to a complete stop in 2.0 s. What is the

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The force exerted on the car during this stop is 6975N

<u>Explanation:</u>

Given-

Mass, m = 930kg

Speed, s = 56km/hr = 56 X 5/18 m/s = 15m/s

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Force, F = ?

F = m X a

F = m X s/t

F = 930 X 15/2

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3 years ago

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3 years ago

Two 60.o-g arrows are fired in quick succession with an initial speed of 82.0 m/s. The first arrow makes an initial angle of 24.

olganol [36]

Answer:

a) The first arrow reaches a maximum height of 56.712 meters, whereas second arrow reaches a maximum height of 342.816 meters, b) Both arrows have a total mechanical energy at their maximum height of 201.720 joules.

Explanation:

a) The first arrow is launch in a parabolic way, that is, horizontal speed remains constant and vertical speed changes due to the effects of gravity. On the other hand, the second is launched vertically, which means that velocity is totally influenced by gravity. Let choose the ground as the reference height for each arrow. Each arrow can be modelled as particles and by means of the Principle of Energy Conservation:

First arrow

$U_{g,1} + K_{x,1} + K_{y,1} = U_{g,2} + K_{x,2} + K_{y,2}$

Where:

$U_{g,1}$ , $U_{g,2}$ - Initial and final gravitational potential energy, measured in joules.

$K_{x,1}$ , $K_{x,2}$ - Initial and final horizontal translational kinetic energy, measured in joules.

$K_{y,1}$ , $K_{y,2}$ - Initial and final vertical translational kinetic energy, measured in joules.

Now, the system is expanded and simplified:

$m \cdot g \cdot (y_{2} - y_{1}) + \frac{1}{2}\cdot m \cdot (v_{y, 2}^{2} -v_{y, 1}^{2}) = 0$

$g \cdot (y_{2}-y_{1}) = \frac{1}{2}\cdot (v_{y,1}^{2}-v_{y,2}^{2})$

$y_{2}-y_{1} = \frac{1}{2}\cdot \frac{v_{y,1}^{2}-v_{y,2}^{2}}{g}$

Where:

$y_{1}$ . $y_{2}$ - Initial and final height of the arrow, measured in meters.

$v_{y,1}$ , $v_{y,2}$ - Initial and final vertical speed of the arrow, measured in meters.

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

The initial vertical speed of the arrow is:

$v_{y,1} = v_{1}\cdot \sin \theta$

Where:

$v_{1}$ - Magnitude of the initial velocity, measured in meters per second.

$\theta$ - Initial angle, measured in sexagesimal degrees.

If $v_{1} = 82\,\frac{m}{s}$ and $\theta = 24^{\circ}$ , the initial vertical speed is:

$v_{y,1} = \left(82\,\frac{m}{s} \right)\cdot \sin 24^{\circ}$

$v_{y,1} \approx 33.352\,\frac{m}{s}$

If $g = 9.807\,\frac{m}{s^{2}}$ , $v_{y,1} \approx 33.352\,\frac{m}{s}$ and $v_{y,2} = 0\,\frac{m}{s}$ , the maximum height of the first arrow is:

$y_{2} - y_{1} = \frac{1}{2}\cdot \frac{\left(33.352\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{9.807\,\frac{m}{s^{2}} }$

$y_{2} - y_{1} = 56.712\,m$

Second arrow

$U_{g,1} + K_{y,1} = U_{g,3} + K_{y,3}$

Where:

$U_{g,1}$ , $U_{g,3}$ - Initial and final gravitational potential energy, measured in joules.

$K_{y,1}$ , $K_{y,3}$ - Initial and final vertical translational kinetic energy, measured in joules.

$m \cdot g \cdot (y_{3} - y_{1}) + \frac{1}{2}\cdot m \cdot (v_{y, 3}^{2} -v_{y, 1}^{2}) = 0$

$g \cdot (y_{3}-y_{1}) = \frac{1}{2}\cdot (v_{y,1}^{2}-v_{y,3}^{2})$

$y_{3}-y_{1} = \frac{1}{2}\cdot \frac{v_{y,1}^{2}-v_{y,3}^{2}}{g}$

If $g = 9.807\,\frac{m}{s^{2}}$ , $v_{y,1} = 82\,\frac{m}{s}$ and $v_{y,3} = 0\,\frac{m}{s}$ , the maximum height of the first arrow is:

$y_{3} - y_{1} = \frac{1}{2}\cdot \frac{\left(82\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{9.807\,\frac{m}{s^{2}} }$

$y_{3} - y_{1} = 342.816\,m$

The first arrow reaches a maximum height of 56.712 meters, whereas second arrow reaches a maximum height of 342.816 meters.

b) The total energy of each system is determined hereafter:

First arrow

The total mechanical energy at maximum height is equal to the sum of the potential gravitational energy and horizontal translational kinetic energy. That is to say:

$E = U + K_{x}$