What is normal for a spring that obeys hook's law ?
2 answers:
You might be interested in
Answer:
Explanation:
Energy of signal being radiated per second on all sides = 71 x 10³ J .
At a distance of 220 m it is spread over an area of 4 π x (220)² because it is spreading uniformly on all sides.
So energy crossing per unit area
=
= 11.67 x 10⁻² Wm⁻²s⁻¹.
This is the intensity of the signal.
At 2200 m this intensity will further reduce by 100 times
So there it becomes equal to
11.67 x 10⁻⁴ Wm⁻² s⁻¹.
Answer:
emf induced in the loop, at the instant when 9.0s have passed = 1.576 * 10 ⁻² V.
Direction is counter clockwise.
Explanation:
See attached pictures.
