The answer to the question above is c hope this helps !!!!
I think the correct answer would be one half the wavelength. Waves would "feel bottom" when the water is at the depth of 0.5 of the wavelength. "Feel bottom" is a term used to describe that the depth of water affects the wave properties. Hope this answers the question.
Explanation:
It is given that, a long, straight wire is surrounded by a hollow metal cylinder whose axis coincides with that of the wire.
The charge per unit length of the wire is
and the net charge per unit length is
.
We know that there exist zero electric field inside the metal cylinder.
(a) Using Gauss's law to find the charge per unit length on the inner and outer surfaces of the cylinder. Let
are the charge per unit length on the inner and outer surfaces of the cylinder.
For inner surface,



For outer surface,



(b) Let E is the electric field outside the cylinder, a distance r from the axis. It is given by :


Hence, this is the required solution.
The car has an initial velocity
of 23 m/s and a final velocity
of 0 m/s. Recall that for constant acceleration,

The car stops in 7 s, so the acceleration is

