The amount of energy lost at the transition between each trophic level of the pyramid of energy is about 90%.
There are a total of 4 trophic levels.
The producers which bare plants represent the first trophic level. Herbivores represent the second trophic level. Carnivores represent the third trophic level. Top carnivores represent the fourth trophic level.
The 10 % rule is followed by the energy flow in a food chain. It means that moving from one trophic level to another, only 10% of energy is transferred and the rest is lost in the atmosphere.
So the 90% percent energy is lost at the transition between each trophic level of the pyramid of energy.
If you need to learn more about the trophic levels, click here
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A) <u>Weight = mass × acceleration (due to gravity) </u>
= 60×9.8
= 588 N
<u>B) Potential energy = mass x gravity x change in height
</u>
1,000 = 60.0 x 9.8 x h
h = 1.7 m
<u>C) Kinetic energyF = potential energyI
</u>
KEF = 1/2mv2
PEI = mgh = 1,000 J
1/2mv2 = 1,000
1/2(60.0)v2 = 1,000
v2 = 33.33
v = 5.77 m/s
The best choice would be letter C hope this helps
Answer:
The answer is 13 however make sure if they ask for a certain measurement like meter answer it by saying 13 meters.
Explanation:
This basically turns into basic algebra if you know the formula for work. The formula for work is W=F*d
Here are the variables that you know 650J=50N*d so you need d.
All you do is divide 650J by 50N and you get a total of 13 (meters since I don't know what they want you to put it in).
(a) 1200 rad/s
The angular acceleration of the rotor is given by:

where we have
is the angular acceleration (negative since the rotor is slowing down)
is the final angular speed
is the initial angular speed
t = 10.0 s is the time interval
Solving for
, we find the final angular speed after 10.0 s:

(b) 25 s
We can calculate the time needed for the rotor to come to rest, by using again the same formula:

If we re-arrange it for t, we get:

where here we have
is the initial angular speed
is the final angular speed
is the angular acceleration
Solving the equation,
