Answer:
2.83m
Step-by-step explanation:
The information that we have is
Intensity at 2.0 m:
and
![r_(1)=2m](https://img.qammunity.org/2021/formulas/physics/middle-school/5i7phnqiyrbjnznx2zqvxbod5qq5lidbrf.png)
we need an intensity level of:
![I_(2)=40dB](https://img.qammunity.org/2021/formulas/physics/middle-school/2nhqbt1my1tyrcl01h329d05ueekt2nbrs.png)
thus, we are looking for the distance
.
which we can find with the law for intensity and distance:
![((r_(2))/(r_(1)) )^2=(I_(1))/(I_(2))](https://img.qammunity.org/2021/formulas/physics/middle-school/jdeitjsgh0i1ys3fo792pcloyrvmkz55iy.png)
we solve for
:
![(r_(2))/(r_(1))=\sqrt{(I_(1))/(I_(2)) }\\\\r_(2)=r_(1)\sqrt{(I_(1))/(I_(2)) }](https://img.qammunity.org/2021/formulas/physics/middle-school/rgquke38366oc6a33ij09idaxv8y3gv77u.png)
and we substitute the known values:
![r_(2)=(2m)\sqrt{(80dB)/(40dB) }\\\\r_(2)=(2m)√(2)\\ r_(2)=2.83m](https://img.qammunity.org/2021/formulas/physics/middle-school/hykvw72v7p3gxe9yyq3sljthe6pa39p47u.png)
at a distance of 2.83m the intensity level is 40dB