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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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Name 7 characteristics that are unique to mammals.
SSSSS [86.1K]
Mammals are endothermic vertebrates
Have hair and fur on the body
Have mammary glands
Four chambered hearts
Have sebaceous (fat secreting glands), sudoriferus (sweat), and scent glands.
Have heterodont dentation (different types of teeth)
Posses diaphragm
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3 0
3 years ago
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Order the sequence of ideas that led to Marie Curie’s discovery of radioactive elements. Number the events in chronological orde
Marysya12 [62]

Answer:

2 1 3

Explanation:

6 0
3 years ago
15) On a cold day, you take in 4.2 L (i.e., 4.2 x 10-3 m3) of air into your lungs at a temperature of 0°C. If you hold your brea
Rudiy27

Answer:

4.8L ( i.e 4.8 x 10^-3 m3)

Explanation:

Step 1:

Data obtained from the question.

Initial volume (V1) = 4.2L

Initial temperature (T1) = 0°C

Final temperature (T2) = 37°C

Final volume (V2) =?

Step 2:

Conversion of celsius temperature to Kelvin temperature. This is illustrated below

K = °C + 273

T1 = 0°C = 0°C + 273 = 273K

T2 = 37°C = 37°C + 273 = 310K

Step 3:

Determination of the final volume.

Since the pressure is constant,

Charles' Law equation will be applied as shown below:

V1 /T1 = V2/T2

4.2/273 = V2 /310

Cross multiply to express in linear form

273 x V2 = 4.2 x 310

Divide both side by 273

V2 = (4.2 x 310)/273

V2 = 4.8L ( i.e 4.8 x 10^-3 m3)

Therefore, the volume of the air in the lungs at that point is 4.8L ( i.e 4.8 x 10^-3 m3)

3 0
3 years ago
A girl pushes a 1.04 kg book across a table with a horizontal applied force 10 points
mr Goodwill [35]

Answer:

Approximately 11.0\; \rm m \cdot s^{-1}. (Assuming that g = 9.81 \; \rm N \cdot kg^{-1}, and that the tabletop is level.)

Explanation:

Weight of the book:

W = m \cdot g = 1.04 \; \rm kg \times 9.81\; \rm N \cdot kg^{-1} \approx 10.202\; \rm N.

If the tabletop is level, the normal force on the book will be equal (in magnitude) to weight of the book. Hence, F(\text{normal force}) \approx 10.202\; \rm N.

As a side note, the F_N and W on this book are not equal- these two forces are equal in size but point in the opposite directions.

When the book is moving, the friction F(\text{kinetic friction}) on it will be equal to

  • \mu_{\rm k}, the coefficient of kinetic friction, times
  • F(\text{normal force}), the normal force that's acting on it.

That is:

\begin{aligned}& F(\text{kinetic friction}) \\ &= \mu_{\rm k}\cdot F(\text{normal force})\\ &\approx 0.35 \times 10.202\; \rm N \approx 3.5708\; \rm N\end{aligned}.

Friction acts in the opposite direction of the object's motion. The friction here should act in the opposite direction of that 15.0\; \rm N applied force. The net force on the book shall be:

\begin{aligned}& F(\text{net force})  \\ &= 15.0 \; \rm N - F(\text{kinetic friction}) \\& \approx 15.0 - 3.5708\; \rm N \approx 11.429\; \rm N\end{aligned}.

Apply Newton's Second Law to find the acceleration of this book:

\displaystyle a = \frac{F(\text{net force})}{m} \approx \frac{11.429\; \rm N}{1.04\; \rm kg} \approx 11.0\; \rm m \cdot s^{-2}.

6 0
3 years ago
A spaceship accelerates from 0m/s to 40m/s in 5 seconds. What is the acceleration of the spaceship
nadya68 [22]
Acceleration = change in velocity/time
= 40/5
=8m/s^2
6 0
3 years ago
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