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Nana76 [90]
1 year ago
12

The electrons lost from chlorophyll photooxidation are replaced by the oxidation of water. how many electrons are generated from

the oxidation of one water molecule?
Physics
1 answer:
White raven [17]1 year ago
3 0

There are 2 electrons generated from the oxidation of one water molecule.

<h3>Describe photooxidation.</h3>

The process of a substance interacting with oxygen or losing electrons from chemical species under the influence of light is known as photooxidation. Photooxidation happens in plants when there is environmental stress. It is called photooxidative stress as a result. Reactive oxygen species are produced by the absorption of excess excitation energy in plant tissues. Chloroplasts are harmed by the accumulation of these reactive oxygen species, which is a damaging process in plants. High-intensity light and little CO_{2} are the two conditions that cause this photooxidative stress to occur most frequently. It is a procedure that requires light. Photorespiration in C_{3} plants guards against photooxidation.

To know more about Photooxidation visit:

brainly.com/question/14788790

#SPJ4

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A dolphin in an aquatic show jumps straight up out of the water at a velocity of 13.0 m/s.
SVEN [57.7K]

Answer:

a) Knowns, initial speed v_{i}=13.0 m/s, final speed v_{f}=0 m/s and gravity due it is a constant g=9.8m/s^{2}

b) The maximum high reached by the dolphin is y_{max}=8.62 m

c) Total time is t=2.65s

Explanation:

a) First of all the initial speed is given at the start of the problem, gravity is constant and final speed is known any object thrown straight up reaches its max high at 0m/s speed.

b) Second, now that we know final speed we use v_{f} =v_{i}-gt, as we clear for t=\frac{v_{i}-v_{f} }{g}=\frac{20.0m/s}{9.8m/s^{2} }=1.32 s.

Then we use y=v_{i}t-\frac{1}{2} gt^{2}=(20.0m/s)(1.32s)-\frac{1}{2} (9.8m/s^{2} ) (1.32s)^{2}  =8.62m

c)Third, finally we can use y=v_{i}t-\frac{1}{2} gt^{2}, as we know y=0m when the dolphin fall into the water again and v_{i} =13.0m/s, then we have 0=(13m/s)t-\frac{1}{2} (9.8m/s^{2}  )t^{2} is a quadratic form 0=t(13.0-4.9t) so we have t_{i}=0s and t_{f}=\frac{13}{4.90}  =2.65s

6 0
3 years ago
A rod of length Lo moves iwth a speed v along the horizontal direction. The rod makes an angle of (θ)0 with respect to the x' ax
Colt1911 [192]

Answer:

From the question we are told that

  The length of the rod is  L_o

    The  speed is  v  

     The angle made by the rod is  \theta

     

Generally the x-component of the rod's length is  

     L_x =  L_o cos (\theta )

Generally the length of the rod along the x-axis  as seen by the observer, is mathematically defined by the theory of  relativity as

       L_xo  =  L_x  \sqrt{1  - \frac{v^2}{c^2} }

=>     L_xo  =  [L_o cos (\theta )]  \sqrt{1  - \frac{v^2}{c^2} }

Generally the y-component of the rods length  is mathematically represented as

      L_y  =  L_o  sin (\theta)

Generally the length of the rod along the y-axis  as seen by the observer, is   also equivalent to the actual  length of the rod along the y-axis i.e L_y

    Generally the resultant length of the rod as seen by the observer is mathematically represented as

     L_r  =  \sqrt{ L_{xo} ^2 + L_y^2}

=>  L_r  = \sqrt{[ (L_o cos(\theta) [\sqrt{1 - \frac{v^2}{c^2} }\ \ ]^2+ L_o sin(\theta )^2)}

=>  L_r= \sqrt{ (L_o cos(\theta)^2 * [ \sqrt{1 - \frac{v^2}{c^2} } ]^2 + (L_o sin(\theta))^2}

=>   L_r  = \sqrt{(L_o cos(\theta) ^2 [1 - \frac{v^2}{c^2} ] +(L_o sin(\theta))^2}

=> L_r =  \sqrt{L_o^2 * cos^2(\theta)  [1 - \frac{v^2 }{c^2} ]+ L_o^2 * sin(\theta)^2}

=> L_r  =  \sqrt{ [cos^2\theta +sin^2\theta ]- \frac{v^2 }{c^2}cos^2 \theta }

=> L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }

Hence the length of the rod as measured by a stationary observer is

       L_r = L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }

   Generally the angle made is mathematically represented

tan(\theta) =  \frac{L_y}{L_x}

=>  tan {\theta } =  \frac{L_o sin(\theta )}{ (L_o cos(\theta ))\sqrt{ 1 -\frac{v^2}{c^2} } }

=> tan(\theta ) =  \frac{tan\theta}{\sqrt{1 - \frac{v^2}{c^2} } }

Explanation:

     

     

       

7 0
3 years ago
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