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:
Option D. The average speed is 2.5 meters/second, and the average velocity is 0 meters/second.
Explanation:
we know that
To find out the average speed divide the total distance by the total time
Let
d -----> the total distance in meters
t -----> the time in seconds
s ----> the speed in meters per second
Remember that
we have
substitute
<u><em>Find out the average velocity</em></u>
To find out the average velocity divide the displacement) by the time
The displacement is the distance from the start point to the end point regardless of the route
In this problem
The start point is A and the end point is A
so
The displacement is equal to zero
therefore
The average velocity is 0 m/sec