Respuesta :
Answer:
[tex]d_2=19.8m[/tex]
Explanation:
The problem can be solved easily, because the relationship between intensities [tex]I_1[/tex] and [tex]I_2[/tex] at distances [tex]d_1[/tex] and [tex]d_2[/tex] respectively can be written as:
[tex]\frac{I_2}{I_1} =\frac{d_1^2}{d_2^2}[/tex]
Where:
[tex]I_1=Intensity\hspace{3}1\\I_2=Intensity\hspace{3}2\\d_1=Distance\hspace{3}1\\d_2=Distance\hspace{3}2[/tex]
In this case:
[tex]I_1=80dB\\I_2=40dB\\d_1=14m\\d_2=?[/tex]
Solving for [tex]d_2[/tex]
[tex]d_2=\sqrt{\frac{d_1^2 *I_1}{I_2} } =\sqrt{\frac{14^2*80}{40} } =\sqrt{392} =19.79898987m\approx19.8m[/tex]
Answer:
1400m
Explanation:
For spherical waves we can use the following relationship between distance and intensity:
[tex]\frac{I_{1}}{I_{2}}=\frac{r_{2}^2}{r_{1}^2}[/tex]
Where [tex]I_{1}[/tex] and [tex]I_{2}[/tex] are the first and second intensity, in [tex]W/m^2[/tex]. And [tex]r_{1}[/tex] is the first distance: [tex]r_{1}=14m[/tex] , and [tex]r_{2}[/tex] the distace we want to find.
Clearing the previous equation or [tex]r_{2}[/tex]
[tex]r_{2}^2=\frac{I_{1}r_{1}^2}{I_{2}} \\r_{2}=\sqrt{\frac{I_{1}r_{1}^2}{I_{2}}} \\r_{2}=r_{1}\sqrt{\frac{I_{1}}{I_{2}}}[/tex].
This is what we will be using to find the answer, but fist we must convert the quantities 80dB and 40dB to [tex]W/m^2[/tex]
We will call the quantities in dB [tex]\beta_{1}[/tex] and [tex]\beta _{2}[/tex]:
[tex]\beta_{1}=80dB[/tex]
[tex]\beta_{2}=40dB[/tex]
We will use the following to find the corresponding intensities [tex]I_{1}[/tex] and [tex]I_{2}[/tex]:
[tex]\beta_{1}=10log\frac{I_{1}}{I_{o}}[/tex]
[tex]\beta_{2}=10log\frac{I_{2}}{I_{o}}[/tex]
for both of these: [tex]I_{0}=10^{-12}W/m^2[/tex], and it is the minimum intensity that a human being perceives.
Thus, for [tex]\beta_{1}=80dB[/tex] we have:
[tex]80dB=10log\frac{I_{1}}{I_{o}}[/tex]
[tex]8dB=log\frac{I_{1}}{I_{o}}[/tex]
And to eliminate the logarithm, we use its inverse operation, exponentiation.
[tex]10^8=10^{log\frac{I_{1}}{I_{0}} }[/tex]
[tex]10^8=\frac{I_{1}}{I_{0}}[/tex]
[tex]I_{0}(10^8)=I_{1}[/tex]
replacing [tex]I_{0}=10^{-12}W/m^2[/tex]:
[tex]10^{-12}W/m^2(10^8)=I_{1}[/tex]
[tex]1x10^{-4}W/m^2=I_{1}[/tex]
and similarly for [tex]I_{2}[/tex]
[tex]40dB=10log\frac{I_{2}}{I_{o}}[/tex]
[tex]4dB=log\frac{I_{2}}{I_{o}}[/tex]
[tex]10^4=10^{log\frac{I_{2}}{I_{0}} }[/tex]
[tex]10^4=\frac{I_{2}}{I_{0}}[/tex]
[tex]I_{0}(10^4)=I_{2}[/tex]
replacing [tex]I_{0}=10^{-12}W/m^2[/tex]:
[tex]10^{-12}W/m^2(10^4)=I_{2}[/tex]
[tex]1x10^{-8}W/m^2=I_{2}[/tex]
Now we substitute in the equation we had found for [tex]r_{2}[/tex]
[tex]r_{2}=r_{1}\sqrt{\frac{I_{1}}{I_{2}}}\\r_{2}=(14m)\sqrt{\frac{1x10^{-4}W/m^2}{1x10^{-8}W/m^2}}\\r_{2}=(14m)\sqrt{10000}\\ r_{2}=(14m)(100)\\r_{2}=1400m[/tex]
