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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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A particle moving along the x-axis has its velocity described by the function vx =2t2m/s, where t is in s. Its initial position
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Answer:

Follows are the solution to this question:

Explanation:

In point a:

Place of particles

X(t)=\int V_{x}(t)dt

       =\int 2t^{2}dt\\\\=\frac{2}{3}t^{3}+C

\to t=0\\\\ \to X(0)=2.3 \ m

\to X(0)=0+C\\\\ \to C=2.3\  m

\to X(t)=( \frac{2}{3})t^3 + 2.3\\\\ \to t=2.2\\\\\to X=( \frac{2}{3})\times 2.2^3 +2.3 \\\\

        = \frac{2}{3}\times 10.648 +2.3\\\\= \frac{21.296}{3}+2.3\\\\  = 7.09+2.3\\\\ =9.39\\\\ =9.4\ m

In point b:

when t=2.2 \ s

the Particle velocity  (V)=2 \times 2.22 =9.68\  \frac{m}{s}

In point c:

Calculating the Particle acceleration:

\to a=\frac{dV}{dt} =4\ t\\\\\to t=2.2 \ s\\\\\to a=4\times 2.2  =8.8 \ \frac{m}{s^2}

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3 years ago
If a planet has an eccentricity of 0.92 and a distance of the major axis is 5,000,000 km, then what is the distance between foci
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Answer:

it’s the third option: 4,600,000 km

Explanation:

just took the test and that was correct

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Hailey is preparing a debate on the benefits of using synthetic polymers over natural polymers, and she wants to create a list t
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Answer:

D. Synthetic polymers are inexpensive to produce.

Explanation:

Hailey should include that synthetic polymers are inexpensive to produce as a benefit of synthetic polymers.

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Most of these polymers are very cheap to produce and does not cost too much. This is why they almost found everywhere.

Explanation:

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In which point the electric intensity of a sphere is maximum​
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Here, as the charge is uniformly distributed in the sphere, we will consider s as an area of the sphere which is, s=4πr2 and r is radius of the gaussian surface shown in the figure above. From this, it can be seen that
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A ball of mass m is thrown into the air in a 45° direction of the horizon, after 3 seconds the ball is seen in a direction 30° f
Rzqust [24]

Answer:

Velocity (magnitude) is 98.37 m/s

Explanation:

We use the vertical component of the initial velocity, which is:

v_{0y}=v_0*sin(45)=\frac{\sqrt{2} }{2}v_0

Using kinematics expression of vertical velocity (in y direction) for an accelerated motion (constant acceleration, which is gravity):

v_{y}=v_{0y}+a*t=\frac{\sqrt{2} }{2}v_0-9.8t

Now we need to find v_y as a function of v_0. We use the horizontal velocity, which is always the same as follow:

v_x=v_0cos(45\º)=\frac{\sqrt{2} }{2}v_0=v_{t=3}*cos(30\º) \\

We know the angle at 3 seconds:

v_y(t=3)=v_{t=3}*sin(30\º)\\v_{t=3}=\frac{v_y}{sin(30\º)}

Substitute  v_{t=3} in  v_x and then solve for  v_y

\frac{\sqrt{2} }{2}v_0=\frac{v_y*cos(30\º) }{sin(30\º)} \\v_y=\frac{\sqrt{6} }{6}v_0

With this expression we go back to the kinematic equation and solve it for initial speed

\frac{\sqrt{6} }{6} v_0 =\frac{\sqrt{2} }{2}v_0-29.4\\v_0(\frac{\sqrt{6}-3\sqrt{2}}{6} )=-29.4\\v_0=98.37 m/s

3 0
4 years ago
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