A Carnot engine operates with a cold reservoir at a temperature of 474 K and a hot reservoir at a temperature of 628 K. What is
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Answer:
The overall velocity of the water when it hits the bottom is:
Explanation:
Use the law of conservation of energy.
Call it instant [1] to the moment when the water is just before reaching the falls.
At this moment its height h is 206 meters and its velocity horizontally is
m/s.
At the instant [1] the water has gravitational power energy
The water also has kinetic energy Ek.
Then the Total E1 energy is:
In the instant [2] the water is within an instant of touching the ground. At this point it only has kinetic energy, since the height h = 0. However at time [2] the water has maximum final velocity
So:
As the energy is conserved then
Now we solve for .
The potential energy will be 1.46*10^-4J.
To find the answer, we have to know about the torque acting on a current loop in a uniform magnetic field.
<h3>How to find the potential energy of the loop?</h3>- We have the expression for torque acting on a current loop in a uniform magnetic field as,
where; M is the magnetic dipole moment, B is the magnetic field , and theta is the angle between M and B.
- As we know that, the torque is equal to force times the perpendicular distance. Thus, it is equivalent to the work done. This work is stored as the potential energy in the loop.
- Thus, the potential energy will be,
Thus, we can conclude that, the potential energy will be 1.46*10^-4J.
Learn more about the torque here:
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