The intensity level in decibels is defined as 10 log10(I/I0), where I0 is a reference intensity equal to the human threshold of hearing, 10–12 W/m2. What is the intensity of the…

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asked Feb 7, 2020 93.6k views

2 Answers

Codtex · Answer 1 · 2020-02-10T07:06:29+0000

Answer:

The intensity of the threshold of pain is 1 W/m²

Step-by-step explanation:

Given that,

Decibels = 120

$I_(0)=10^(-12)\ W/m^2$

The intensity level in decibels is defined as

dB=10 log_(10)((I)/(I_(0)))

Where, I = intensity

I_(0) =Original intensity

Put the value into the formula

120=10log_(10)((I)/(10^(-12)))

(120)/(10)=log_(10)((I)/(10^(-12)))

I=10^(12)*10^(-12)

$I=1\ W/m^2$

Hence, The intensity of the threshold of pain is 1 W/m²

Karuto · Answer 2 · 2020-02-12T00:29:45+0000

Step-by-step explanation:

The intensity level is given by :

$dB=10\ log_(10)(I)/(I_o)$

$I_o=10^(-12)\ W/m^2$

The intensity of the threshold of pain is 120 decibels

$120=10\ log_(10)(I)/(10^(-12))$

$12=\ log_(10)(I)/(10^(-12))$

10^(12)* 10^(-12)=I

$I=1\ W/m^2$

So, the intensity of the threshold pain is 1 W/m². Hence, this is the required solution.