The Calvin cycle<span> refers to the light-independent reactions in photosynthesis that take place in three key steps. Although the </span>Calvin Cycle<span> is not directly dependent on light, it is indirectly dependent on light since the necessary energy carriers (ATP and NADPH) are products of light-dependent reactions.
So basically it indirectly needs the light, even it's called light-independant reaction.
So the answer is the last one.</span>
Geosphere, hydrosphere, atmosphere, and biosphere.
Therefore, the correct answer is D.
The amount of work done by two boys who apply 200 N of force in an unsuccessful attempt to move a stalled car is 0.
Answer: Option B
<u>Explanation:
</u>
Work done is the measure of work done by someone to push an object from its present position. We can also define work done as the amount of forces needed to move an object from its present position to another position. So the amount of work done is directly proportionate to the product of forces acting on the object and the displacement of the object.

So in this present case, as the two boys have done an unsuccessful attempts to push a stalled car so that means the displacement of the car is zero as there is no change in the position of the car. But they have applied a force of 200 N each. So the amount of work done will be

Thus, the amount of work done by two boys will be zero due to their unsuccessful attempt to move a stalled car.
Answer:
C , E , A , D , B
Explanation:
We evaluate the accelerations for each case, using the formula: a = (vf - vi) / t
A) a = (10.3 - 0.5 ) / 1 = 9.8 m/s^2 --> magnitude: 9.8 m/s^2
B) a = (0 - 20) / 1 = - 20 m/s^2 --> magnitude : 20 m/s^2
C) a = (0.02 - 0.004) / 1 = 0.016 m/s^2 --> magnitude : 0.016 m/s^2
D) a = (4.3 - 0) / 0.4 = 10.75 m/s^2 --> magnitude : 10.75 m/s^2
E) a = (1 - 2) / 8.3 = - 0.12 m/s^2 --> magnitude: 0.12 m/s^2
Then, comparing magnitudes from least to greatest:
C , E , A , D , B
It forces the mercury to rise, being pushed up the tube by pressing down on the dish.
I hope I could help :)