Answer:
(a) The magnitude of the electric dipole moment is 1.68 x 10⁻¹⁴ C.m
(b) The difference between the potential energies ΔU, is 4.6704 x 10⁻¹¹ J
Explanation:
Given;
magnitude of charge, q = 2 nC = 2 x 10⁻⁹ C
distance of separation, d = 8.4 μm = 8.4 x 10⁻⁶ m
strength of electric field, E = 1390 N/C
(a) the magnitude of the electric dipole moment
p = qd
p = (2 x 10⁻⁹ C)(8.4 x 10⁻⁶ m)
p = 1.68 x 10⁻¹⁴ C.m
(b) the difference between the potential energies for dipole orientations parallel and anti-parallel to E
ΔU = U(180) - U(0)
ΔU = 2pE
ΔU = 2(1.68 x 10⁻¹⁴ )(1390)
ΔU = 4.6704 x 10⁻¹¹ J
Not sure, nobody knows hopefully soon we will know
Increasing the pitch makes a note higher
The answer to your question is B. When light hits something relatively flat, smooth and shiny, it is reflected by the surface, allowing nearby objects to be seen on the surface.
Answer:
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