Question

The intensity, or loudness, of a sound can be measured in decibels (dB), according to the equation I(dB)=10log[1/10], where I is the intensity of a given sound and I0 is the threshold of a hearing intensity. What is the intensity, in decibles, [I(dB)], when I=10^32(I0)? Round to the nearest whole number.

Answer 1

Answer:

C. 320

Step-by-step explanation:

Answer 2

According to it's formula, the intensity when $l = 10^{32}$ is of 440 db.

What is the formula for the intensity of a sound?

The intensity of a sound l is given by:

$L(l) = 10\log{\left(\frac{l}{l_0}\right)}$

In which $l_0 = 10^{-12}$ is the threshold of a hearing intensity.

In this problem, we have that the sound is of $l = 10^{32}$, hence:

$L(l) = 10\log{\left(\frac{10^{32}}{10^{-12}}\right)} = 10\log{10^{44}} = 10 \times 44 = 440$

Hence, the intensity of the sound is of 440 db.

More can be learned about the intensity of a sound at https://brainly.com/question/17062836