The molarity remains the same so the ratio does not change
Answer:
180 W
Explanation:
The work done by the man against gravity is equal to its gain in gravitational potential energy:

where
(mg) = 720 N is the weight of the man
is the change in height
Substituting,

The power he must deliver is given by

where
W = 3600 J
t = 20 s is the time taken
Substituting,

Answer:
The answer is: True.
Explanation:
If free electrons or other substances could travel through the electrolyte, they would disrupt the chemical reaction. Whether they combine at anode or cathode, together hydrogen and oxygen form water, which drains from the cell. As long as a fuel cell is supplied with hydrogen and oxygen, it will generate electricity.
(Credit: Google)
We can use the law of conservation of energy to solve the problem.
The total mechanical energy of the system at any moment of the motion is:

where U is the potential energy and K the kinetic energy.
At the beginning of the motion, the ball starts from the ground so its altitude is h=0 and therefore its potential energy U is zero. So, the mechanical energy is just kinetic energy:

When the ball reaches the maximum altitude of its flight, it starts to go down again, so its speed at that moment is zero: v=0. So, its kinetic energy at the top is zero. So the total mechanical energy is just potential energy:

But the mechanical energy must be conserved, Ef=Ei, so we have

and so, the potential energy at the top of the flight is
Angular acceleration = (change in angular speed) / (time for the change)
change in angular speed = (zero - 2,600 RPM) = -2,600 RPM
time for the change = 10 sec
Angular acceleration = -2600 RPM / 10 sec = -260 rev / min-sec
(-260 rev/min-sec) x (1 min / 60 sec) = <em>-(4 1/3) rev / sec²</em>
Since the acceleration is negative, the motor is slowing down.
You might call that a 'deceleration' of (4 1/3) rev/sec² .
The average speed is 1/2(2,600 + 0) = 1,300 rev/min = (21 2/3) rev/sec.
Number of revs = (average speed) x (time) = (21 2/3) x (10sec) = <em>(216 2/3) revs</em>