Question

The intensity, or loudness, of a sound can be measured in decibels (dB), according to the equation I (d B) = 10 log left-bracket StartFraction I Over I Subscript 0 Baseline EndFraction Right-bracket, where I is the intensity of a given sound and I0 is the threshold of hearing intensity. What is the intensity, in decibels, [I(dB)], when I = 10 Superscript 32 Baseline (I Subscript 0)?

Answer 1

The intensity level of a sound where I is 10 to the power of 32 times the threshold of hearing (I0) can be calculated using the formula I (dB) = 10 log [I / I0]. When I = 10^32 * I0, the intensity level is found to be 320 decibels (dB).

Step-by-step explanation:

The subject of the question is about calculating the sound intensity level in decibels (dB), which is a concept in Physics. According to the given formula I (dB) = 10 log [I / I0], where I is the intensity of the sound in watts per meter squared and I0 is the threshold of hearing intensity, we are asked to find the intensity level when I is 1032 times I0. Given that I0 is 10-12 W/m2, the reference intensity threshold for normal hearing at 1000 Hz, we can substitute these values into the equation to find the decibel level.

Let's calculate the intensity level using the given values:

• I = 1032 * I0
• I0 = 10-12 W/m2
• I (dB) = 10 log [1032 * I0 / I0]
• I (dB) = 10 log [1032]
• I (dB) = 10 * 32
• I (dB) = 320 dB

The intensity of the sound in this case is 320 decibels.

Answer 2

Its D or 80

Step-by-step explanation:

I=10^8 (I subscript 0) can be written as I/I subscript 0, and you can plug that right into the log to get 80.