Answer:
The specific question is not stated, however the general idea is given in the attached picture. The electric field in each region can be found by Gauss’ Law.
at r < R:
Since the solid sphere is conducting, the total charge Q is distributed over the surface, and the electric field inside the sphere is zero.
E = 0.
at R < r < 2R:
The electric field can be found by Gauss’ Law as in the attachment. The green pencil shows this exact region.
at 2R < r:
The electric field can again be found by Gauss’ Law, the blue pencil shows the calculations for this region.
Explanation:
Gauss’ Law is straightforward when applied to spheres. The area of the sphere is 
, and the enclosed charge is given in the question as Q for the inner sphere, and 2Q for the whole system.
<span>The "scientific method" consists of forming a hypothesis (a statement about what the researchers think is true), collecting data that might confirm or refute the hypothesis, and then examining the data to see how well or poorly they support the original hypothesis.</span>
Answer:
The amount of heat transfer is 21,000J .
Explanation:
The equation form of thermodynamics is,
ΔQ=ΔU+W
Here, ΔQ is the heat transferred, ΔU is the change in internal energy, and W is the work done.
Substitute 0 J for W and 0 J for ΔU
ΔQ = 0J+0J
ΔQ = 0J
The change in internal energy is equal to zero because the temperature changes of the house didn’t change. The work done is zero because the volume did not change
The heat transfer is,
ΔQ=Q (in
) −Q (out
)
Substitute 19000 J + 2000 J for Q(in) and 0 J for Q(out)
ΔQ=(19000J+2000J)−(0J)
=21,000J
Thus, the amount of heat transfer is 21,000J .
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