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Nadusha1986 [10]
3 years ago
14

The electrons of photosystem ii are excited and transferred to electron carriers. From which molecule or structure do the photos

ystem ii replacement electrons come?
Physics
1 answer:
AysviL [449]3 years ago
7 0

Answer:

water

Explanation:

The light energy trapped in the chlorophyll a reactive molecule of Photosystem II releases electrons at a higher energy level. These electrons are replaced in the chlorophyll a molecule by electrons that come indirectly from water molecules that are cleaved and also release protons (H +) and oxygen gas. The electrons pass from the primary electron acceptor, along an electron transport chain, at a lower energy level, the reaction center of Photosystem I. As they pass along this electron transport chain , a proton gradient is formed from which ATP is synthesized. The light energy absorbed by Photosystem I releases electrons to another primary acceptor. From this acceptor, electrons are transferred by other transporters to NADP + and NADPH is formed. The electrons removed from Photosystem I are replaced by those from Photosystem II. ATP and NADPH represent the net gain of the reactions that capture energy.

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A parallel plate capacitor fully charged to voltage V is connected to the battery (the voltage on the plates remains fixed). If
Papessa [141]

Charge will decreases.

A parallel plate capacitor when it is fully charged to voltage V is given as:

                    C = Q/V

The capacitance of parallel plate capacitor with two plates of Area A separated by distance d and no dielectric material between plates is

                    C = ε₀ A /d

since from above equation it shows C is proportional to Q and also C is inversely proportional to distance d.

So, ATQ when d increases C will decrease which in result decreases charge on the capacitor.

Thus,  Charge will decrease.

Learn more about capacitance here:

     brainly.com/question/17115454

          #SPJ4

7 0
2 years ago
A 4kg brick is dropped from the top of a building whose hight is 30m.what is the velocity with which it reaches the ground​
miss Akunina [59]

Answer:

=24.25 ^−1

Explanation:

Let   and   be initial and final velocity of the body respectively,  

be acceleration due to gravity ( 9.8^−2 ),  ℎ be the height of the body.

=0 ^ −1

ℎ=30

we know that, ^2−^ 2=2ℎ

^2=2∗9.8∗30

^2=588

=24.25 ^−1

4 0
3 years ago
Light incident on a glass sheet is partly reflected and partly refracted. How is the reflected ray different from the refracted
kkurt [141]

Answer:

E. The refracted ray is vertically polarized whereas the reflected ray is horizontally polarized.

Explanation:

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7 0
3 years ago
A point charge q1 is held stationary at the origin. A second charge q2 is placed at point a, and the electric potential energy o
likoan [24]

Answer:

U_{b}=+7.3*10^{-8}J

Explanation:

Given data

Electric potential at point a is Ua=5.4×10⁻⁸J

q₂ moves to point b where a negative work done on it  W_{a-b}=-1.9*10^{-8}J

Required

Electric potential energy Ub

Solution

When a particle moves from a point where the potential is Ua to a point where it is Ub the change in potential energy is equal to work done where the force exerted on the charge is conservative and work done is given by:

W_{a-b}=U_{a}-U_{b}\\U_{b}=U_{a}-W_{a-b}

Now  substitute the given values

So

U_{b}=5.4*10^{-8}J-(-1.9*10^{-8}J)\\U_{b}=+7.3*10^{-8}J

3 0
3 years ago
A 5000 kg African elephant has a resting metabolic rate of 2500 W. On a hot day, the elephant's environment is likely to be near
alina1380 [7]

Answer:

36 kg

Explanation:

To answer this question, a few assumptions have to be made:

  • That the temperature on the day is 35 °C
  • That all the heat from the elephant is goes to warming/evaporating the water on the surface of the elephant

Energy released per hour = 2500 J/s * 3600 s = 9 000 000 J

Q = mcΔT

9 000 000 J= m *4.186 J/g-K * (373K - 308K) + m*2260 J/g

m =  36 000 g = 36 kg

3 0
3 years ago
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