When a sound's intensity increases by a factor of 10, its decibel reading increases by how many decibels?

If sound intensity increases by a factor of 10, the decibel reading increases by 10.

Explanation:

Decibel scale:

In order to express levels of sound meaningfully in numbers that are more manageable, a logarithmic scale is used, rather than a linear one. This scale is the decibel scale. every increase of 3 dB represents a doubling of sound intensity.

Increase in decibel scale is given by the formula:

n = 10*log(Increased Intensity of sound/Original Intensity of Sound

n = 10*log(I/I₀)

In our case the sound intensity increases by a factor of 10. So I/I₀ increased by 10 times and will be 10 I/I₀

n = 10*log(10 I/I₀)

n = 10*log(10) + 10log(I/I₀)

n = 10 + 10*log(I/I₀)

This shows that the decibel reading is increased by 10 points.

hi there!

your question asks at which segment does freezing occur.

since we're talking about water, the freezing point of water will be 32 degrees fahrenheit or below.

hope this ! ~

wavelength divides in half and speed stays the same.