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Jim quadruples the distance between himself and a sound source. What is the change in decibels of the sound he hears?

A. -6 dB

B. -12 dB

C. +6 dB

D. +12 dB

Respuesta :

AL2006

Quadrupling the distance reduces the sound to 1/16 of its original intensity.

10 log(1/16) = -12 dB (choice-B)

huntermiller4

Answer:

b

Explanation:

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