The photosystem channels the excitation energy gathered by absorption of light by any one of the pigment molecules to a specific "reaction center chlorophyll," which in turn passes the energy to
-photosystem I.
-photosystem II.
-the primary electron acceptor.
-the secondary electron center.
-cytochrome.
Answer: -the primary electron acceptor.
Explanation:
The photosystem II has a reaction center, in the reaction center the energy from sunlight is converted into high energy electrons. At the center of the reaction center, chlorophyll molecule is present which absorbs the light and one of its electron is promoted to the higher energy.
The high energy electron is hop downward and it is transferred to the plastoquinone A, which is a primary electron acceptor. Then the electron is transferred to the plastoquinone B. The plastoquinone B will receives enough electrons it delivers its electron to the electron transfer chain.
Answer: B
Explanation:
A machine multiplies the work output of a something by some amount. In this question, the ax is the machine. Thus, the ax is multiplying the work outputted by the lumberjack, so the answer is B. The reason it is not C is because there could be other forces interacting on the wood which the ax can affect.
Answer:
Higher, Windward side, Condenses
Explanation:
The Windward side refers to that side of a mountain that faces the direction from which the wind is blowing. In this direction, the moisture containing hot air blowing from a distant place moves upward and strikes the mountain at a greater height, where the air mass is thin and the temperature is relatively cold. As the temperature and pressure decrease with altitude, the hot uprising air cools and gradually condenses. This results in the occurrence of high precipitation over this region i.e. the windward side of the mountain.
Therefore, the precipitation is always higher on the windward side of a mountain as the hot air undergoes condensation at greater height as it rises upward.
Answer:
330.5 m
Explanation:
In this case, the object is launched horizontally at 30° with an initial velocity of 40 m/s .
The maximum height will be calculated as;
where ∝ is the angle of launch = 30°
vi= initial launch velocity = 40 m/s
g= 10 m/s²
h= 40²*sin²40° / 2*10
h={1600*0.4132 }/ 20
h= 661.1/2 = 330.5 m
