Intensely Loud Question~H/w check

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The intensity of a wave at twice the distance is calculated to be 1.25 mW/cm, as it decreases by a factor of 1/4 according to the inverse square law. The decibel level change was determined using the formula 10log(I2/I1), resulting in a value of -6.02 dB, indicating a decrease in sound intensity. For absolute decibel levels, the reference intensity I_0 of 10^{-12} W/m^2 must be considered. The calculations provided are confirmed to be correct. Overall, the approach to both intensity and decibel calculations is accurate.
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at certain point the intensity of the wave is 5mw/cm. How intense would it be at a distance twice as far away. and find the dB level of the waves loudness according to the first.

For the intensity since it is twice as far away i found out that it is 1/4 as intense so i got an intensity of 1.25mw/cm.
For dB I used the 10log*I2/I1, and got -6.02, or 6.02 depending on how you look at it. Do these answers sound right?? thanx
 
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Intensity goes as 1/r2, so the first part is right. If you're looking for the difference in decibel level, the second part looks right as well. If you're looking for absolute decibel levels, you need to consider it relative to the standard:

I_0=10^{-12}~W/m^2
 


Yes, your calculations and answers seem to be correct. When calculating intensity, the inverse square law states that the intensity decreases by a factor of 1/4 when the distance is doubled. And for dB, the formula you used is correct as well. The negative sign indicates that the sound intensity has decreased while the positive value indicates the increase in the loudness level. Keep up the good work!
 

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