Respuesta :
Explanation:
The intensity level is given by :
[tex]dB=10\ log_{10}\dfrac{I}{I_o}[/tex]
[tex]I_o=10^{-12}\ W/m^2[/tex]
The intensity of the threshold of pain is 120 decibels
[tex]120=10\ log_{10}\dfrac{I}{10^{-12}}[/tex]
[tex]12=\ log_{10}\dfrac{I}{10^{-12}}[/tex]
[tex]10^{12}\times 10^{-12}=I[/tex]
[tex]I=1\ W/m^2[/tex]
So, the intensity of the threshold pain is 1 W/m². Hence, this is the required solution.
Answer:
The intensity of the threshold of pain is 1 W/m²
Explanation:
Given that,
Decibels = 120
[tex]I_{0}=10^{-12}\ W/m^2[/tex]
The intensity level in decibels is defined as
[tex]dB=10 log_{10}(\dfrac{I}{I_{0}})[/tex]
Where, I = intensity
[tex]I_{0}[/tex]=Original intensity
Put the value into the formula
[tex]120=10log_{10}(\dfrac{I}{10^{-12}})[/tex]
[tex]\dfrac{120}{10}=log_{10}(\dfrac{I}{10^{-12}})[/tex]
[tex]I=10^{12}\times10^{-12}[/tex]
[tex]I=1\ W/m^2[/tex]
Hence, The intensity of the threshold of pain is 1 W/m²
