A charge of 1.5 × 1015 coulombs moves from point A to a lower potential at point B. The electric field is uniform at 2.5 × 10-5
1 answer:
The work done by the electric field is given by:
where q is the charge, and
is the potential difference between point A and B. But the potential difference is equal to
where E is the intensity of the electric field, while d is the distance between the two points A and B.
Therefore, the work done by the field is
where q is the charge, and
where E is the intensity of the electric field, while d is the distance between the two points A and B.
Therefore, the work done by the field is
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Answer:
a. 572Btu/s
b.0.1483Btu/s.R
Explanation:
a.Assume a steady state operation, KE and PE are both neglected and fluids properties are constant.
From table A-3E, the specific heat of water is , and the steam properties as, A-4E:
Using the energy balance for the system:
Hence, the rate of heat transfer in the heat exchanger is 572Btu/s
b. Heat gained by the water is equal to the heat lost by the condensing steam.
-The rate of steam condensation is expressed as:
Entropy generation in the heat exchanger could be defined using the entropy balance on the system:
Hence,the rate of entropy generation in the heat exchanger. is 0.1483Btu/s.R
Answer:
mu=12Tm^2
Explanation:
the magnetic moment mu of a single loop is given by:
where I is the current, B is the magnetic field and A is the area of the loop. By replacing we obtain:
hope this helps!!