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Leno4ka [110]
3 years ago
11

Although the reactions of the calvin cycle do not depend directly on light. True or False

Physics
1 answer:
Harlamova29_29 [7]3 years ago
4 0

Answer:

True

Explanation:

The complete question is:

<em>"Although the reactions of the Calvin cycle do not depend directly on light, they do not usually occur at night. True o False"</em>

<em> </em>The Calvin cycle is also known as the Calvin-Benson cycle or as the CO₂ fixation phase in the photosynthesis process.

The Calvin cycle generates the reactions necessary to fix the carbon in a solid structure for the formation of glucose and, in turn, regenerates the molecules for the continuation of the cycle.

The Calvin cycle is known as the dark phase of photosynthesis, or the carbon fixation phase. It is called the dark phase because this cycle is not dependent on light like other parts that make up the photosynthesis process. But it uses the energy that is produced in the light phase of photosynthesis to fix carbon.

It can be said that it consists of or forms the second stage of photosynthesis, in which the carbon of the carbon dioxide that is absorbed is fixed.

So, the statement is true because the Calvin cycle uses the energy that is produced in the light phase of photosynthesis to fix carbon.

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2 years ago
A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.15
Oksanka [162]

1) 5.5 N

When the ball is at the bottom of the circle, the equation of the forces is the following:

T-mg = m\frac{v^2}{R}

where

T is the tension in the string, which points upward

mg is the weight of the string, which points downward, with

m = 0.158 kg being the mass of the ball

g = 9.8 m/s^2 being the acceleration due to gravity

m \frac{v^2}{R} is the centripetal force, which points upward, with

v = 5.22 m/s being the speed of the ball

R = 1.1 m being the radius of the circular trajectory

Substituting numbers and re-arranging the formula, we find T:

T=mg+m\frac{v^2}{R}=(0.158 kg)(9.8 m/s^2)+(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=5.5 N

2) 3.9 N

When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:

T=m\frac{v^2}{R}

And by substituting the numerical values, we find

T=(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=3.9 N

3) 2.3 N

When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is

T+mg=m\frac{v^2}{R}

And substituting the numerical values and re-arranging it, we find

T=m\frac{v^2}{R}-mg=(0.158 kg)\frac{5.22 m/s)^2}{1.1 m}-(0.158 kg)(9.8 m/s^2)=2.3 N

4) 3.3 m/s

The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:

T=0\\mg = m\frac{v^2}{R}

and re-arranging the equation, we find

v=\sqrt{gR}=\sqrt{(9.8 m/s^2)(1.1 m)}=3.3 m/s

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2 years ago
wo slightly out of tune instruments will create beats when the same note is played. This is a pattern of louder and softer sound
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Answer: All apply

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