What is the field outside the capacitor plates in a parallel capacitor?
2 answers:
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Answer:
The answer cannot be determined.
Explanation:
The energy of the diver when he hits the pool will be equal to its potential energy , and for the temperature of the pool to rise up, this energy has to be converted into the heat energy of the pool.
The change in temperature then will be
Where m is the mass of water in the pool, c is the specific heat capacity of water, and is the added heat which in this case is the energy of the diver.
Since we do not know the mass of the water in the pool, we cannot make this calculation.
Answer:
6.0 × W/
Explanation:
From Wien's displacement formula;
Q = e A
Where: Q is the quantity of heat transferred, e is the emissivity of the surface, A is the area, and T is the temperature.
The emissive intensity = = e
Given from the question that: e = 0.6 and T = 1000K, thus;
emissive intensity = 0.6 ×
= 0.6 × 1.0 ×
= 6.0 ×
Therefore, the emissive intensity coming out of the surface is 6.0 × W/
.