What is the decibel level of a sound at 4% intensity of a 95dB sound?

  • Thread starter Thread starter slaw155
  • Start date Start date
  • Tags Tags
    Intensity
AI Thread Summary
To determine the decibel level of a sound at 4% intensity of a 95dB sound, the formula dB = 10log(I/I0) is used, where I0 is the reference intensity. First, calculate 4% of the intensity corresponding to 95dB, which is I = 0.04 * I95dB. Applying the logarithmic properties, the decibel level can be derived as dB = 10log(0.04 * I95dB/I0). The calculation results in a lower decibel level, reflecting the reduced intensity. This method effectively incorporates the percentage into the decibel calculation.
slaw155
Messages
41
Reaction score
0
What is the decibel level of a sound whose intensity is 4% that of a 95dB sound?

I know dB = 10log(I/Io) but don't know how to incorporate this for this question.
 
Physics news on Phys.org
Look at incorporating the percent to the I in the formula and then use log rules to pull out the percent factor.
 
Thanks solved it
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
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}...

Similar threads

Back
Top