Answer:
at r= 0.56 m far away must you be to be exposed to an average intensity considered to be safe
Explanation:
The maximum safe average intensity of microwaves for human exposure is taken to be 1.00 W/m^2
heat leaked by radar P= 10.0 W
uniformly in the all directions.
Intensity at a distance a r is given by
[tex]I= \frac{P}{4\pi r^2}[/tex]
for safe exposure I= 2.5 W/m^2
and P= 10 W
[tex]r=\sqrt{\frac{P}{4\pi I} }[/tex]
[tex]r=\sqrt{\frac{10}{4\pi 2.5} }[/tex]
on calculating we get
r= 0.56 m
at r= 0.56 m far away must you be to be exposed to an average intensity considered to be safe
The distance you must be exposed average intensity to be considered safe is 0.892 m.
The intensity of the sound wave is given by the following formula as shown below;
I = P/A
where;
I = P/4πr²
r² = P/4πI
r² = (10) / (4π x 1)
r² = 0.796
r = √0.796
r = 0.892 m
Thus, the distance you must be exposed average intensity to be considered safe is 0.892 m.
Learn more about intensity of sound here: https://brainly.com/question/17062836
