The highest trophic level has the least available energy in kilojoules.
Even though the food web is not shown in the question, but we know that energy decreases steadily as it is passed on from one trophic level to the next according to the second law of thermodynamics.
Energy enters into the system from the sun. The primary producers utilize this energy to produce food. As plants are eaten by animals, this energy is transferred along the food web an diminishes at each higher trophic level.
At the highest trophic level, the the least available energy in kilojoules in this food web is found.
Learn more: brainly.com/question/2233704
Water enters the atmosphere through evaporation, transpiration, excretion and sublimation: Transpiration is the loss of water from plant
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Answer:
Magnitude of net force will be 432.758 N
Explanation:
We have given x component of acceleration 
And vertical component of acceleration 
Mass of the ball m = 0.40 kg
So net acceleration 
Now according to second law of motion
Force = mass × acceleration
So F = 0.40×1081.896 = 432.758 N
As we move above up from one trophic level to another in
an energy pyramid, what happens to the energy?
a. It decreases from one trophic level to another.
b. It remains the same for each trophic level.
c. It increases from one trophic level to another.
As we move above up from one trophic level to another in
an energy pyramid, the energy level decreases from one trophic level to
another. The answer is letter A.
Answer:
See the answers below.
Explanation:
to solve this problem we must make a free body diagram, with the forces acting on the metal rod.
i)
The center of gravity of the rod is concentrated in half the distance, that is, from the end of the bar to the center there is 40 [cm]. This can be seen in the attached free body diagram.
We have only two equilibrium equations, a summation of forces on the Y-axis equal to zero, and a summation of moments on any point equal to zero.
For the summation of forces we will take the forces upwards as positive and the negative forces downwards.
ΣF = 0

Now we perform a sum of moments equal to zero around the point of attachment of the string with the metal bar. Let's take as a positive the moment of the force that rotates the metal bar counterclockwise.
ii) In the free body diagram we can see that the force acts at 18 [cm] of the string.
ΣM = 0
![(15*9) - (18*W) = 0\\135 = 18*W\\W = 7.5 [N]](https://tex.z-dn.net/?f=%2815%2A9%29%20-%20%2818%2AW%29%20%3D%200%5C%5C135%20%3D%2018%2AW%5C%5CW%20%3D%207.5%20%5BN%5D)