The loudness level of a sound, D, in decibels, is given by the formula Upper D equals 10 log (10 Superscript 12 Baseline Upper I )D=10log1012I, where I is the intensity of the sound, in watts per metersquared2. Decibel levels range from 0, a barely audible sound, to 160, a sound resulting in a ruptured eardrum. The sound of a certain animal can be heard 500 miles away, reaching an intensity of 2.62.6times×10 Superscript 6106 watts per meter squared .meter2. Determine the decibel level of this sound. At close range, can the sound of this animal rupture the human eardrum?

complete question

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188.0 decibels

Explanation:

Substitute I = 6.3 x 10^6 into the given formula  

D(6.3 x 10^6) = 10log(10^12 x 6.3 x 10^6) = 188.0 to one decimal place.  

This is greater than 160 decibels, so the sound could rupture the human eardrum.  

Substitute I = 6.3 x 10^6 into the given formula  

at right angles to each other

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j