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antiderivativ
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I'm here to check my work again. I hope you don't mind. I'm going to try to check as many as I can today. :)
The sound-level 2.0 m from a pneumatic chipper is 120 dB. Assuming it radiates uniformly in all directions, how far from it must you be in order for the level to drop 40 dB down to something more comfortable?
Given:
D1 = 2.0 m
B1 = 120 dB
D2 = x
B2 = 80 dB
Find x. I decided I could use the inverse square law. As the distance increases, the energy decreases in proportion to the distance squared.
[tex]\frac{B1}{B2}[/tex] = [tex]\frac{D2^2}{D1^2}[/tex]
[tex]\frac{120}{80}[/tex] = [tex]\frac{x^2}{4}[/tex]
x= sqrt((120*4)/80) = 2.4495 m.
In order to hear 40 dB less, I need to walk 2.4495 m from my current distance. The total distance I stand from the chipper is 2 +2.4495 = 4.4495 m. Am I correct?
The sound-level 2.0 m from a pneumatic chipper is 120 dB. Assuming it radiates uniformly in all directions, how far from it must you be in order for the level to drop 40 dB down to something more comfortable?
Given:
D1 = 2.0 m
B1 = 120 dB
D2 = x
B2 = 80 dB
Find x. I decided I could use the inverse square law. As the distance increases, the energy decreases in proportion to the distance squared.
[tex]\frac{B1}{B2}[/tex] = [tex]\frac{D2^2}{D1^2}[/tex]
[tex]\frac{120}{80}[/tex] = [tex]\frac{x^2}{4}[/tex]
x= sqrt((120*4)/80) = 2.4495 m.
In order to hear 40 dB less, I need to walk 2.4495 m from my current distance. The total distance I stand from the chipper is 2 +2.4495 = 4.4495 m. Am I correct?