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Scrat [10]
2 years ago
12

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 86.0 dB.
1. Calculate the INCREASE in the sound level from the ambient work environment level (in dB).
2. 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 116 dB. By what factor does that sound intensity exceed the 1.5- Hours/day intensity limits from the graph?
Physics
1 answer:
andrew-mc [135]2 years ago
7 0

(1) The INCREASE in the sound level from the ambient work environment level (in dB) is 1 dB.

(2) The factor that sound intensity exceed the 1.5- Hours/day intensity limits from the graph is 19.6 %.

<h3> Increase in the sound level </h3>

The INCREASE in the sound level from the ambient work environment level (in dB) is calculated as follows;

Increase in sound level = final sound level - original sound level

Increase in sound level = 86 dB - 85 dB = 1 dB

<h3>Factor of sound level increase</h3>

from the graph at 1.5 hours/day, sound level = 97 dB

Increase in sound intensity = final sound level - original sound level

Increase in sound intensity = 116 dB - 97 dB = 19 dB

Factor increase = 19/97 = 0.196 = 19.6%

Thus, the factor that sound intensity exceed the 1.5- Hours/day intensity limits from the graph is 19.6 %.

Learn more about sound intensity here: brainly.com/question/14261338

#SPJ1

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