The graph shows the US Department of Labor noise regulation for working without

Question

The graph shows the US Department of Labor noise regulation for working without ear protection. A machinist is in an environment where the ambient sound level is of 85 dB, i.e., corresponding to the 8 Hours/day noise level. The machinist likes to listen to music, and plays a Boom Box at an average level of 88.0 dB.

Calculate the INCREASE in the sound level from the ambient work environment level (in dB).

A student at a rock concert has a seat close to the speaker system, where the average sound intensity corresponds to a sound level of 114 dB. By what factor does that sound intensity exceed the 1.5- Hours/day intensity limits from the graph?

Explanation / Answer

I cant see the graph to answer part b. Heres part a): turn the dB levels in powers of 10. To do this simply divide the dB level by 10. 85 = 10^(8.5) 86 = 10^(8.6) 10log(10^(8.5) + 10^(8.6)) = 88.539 dB 88.539 - 85 = 3.539 dB 3.54 rounded (2dp)

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