Calculate Ambient Noise Intensity Increase: 86.0 dB - 85 dB

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To calculate the increase in sound level from an ambient noise level of 85 dB to a Boom Box playing at 86.0 dB, the correct approach is to convert the decibel levels to intensity. The intensity levels are then added together, as the music does not block the ambient noise but rather adds to it. The difference in decibels alone does not accurately represent the total sound intensity. The discussion emphasizes the importance of understanding that sound levels in decibels cannot simply be added arithmetically. Accurate calculations require converting dB to intensity, combining them, and then converting back to dB for the final result.
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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.

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

I'm not sure really what to do... do I just subtract the two, because that seems to easy I changed the dB to intensity and stuff but not sure what to do with them either or if that is even right... Any help would be nice, or even a hint! Thanks!
 
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Keep in mind that, unless the worker has a Bose noise-cancellation headphone, his music is not going to block the ambient noise, but add to it. Therefore the intensities of the two sources are going to be added together. This does not correspond to the dB's being added though.
 
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Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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