The answer is given so just please explain how to get there and write clearly 3.

Question

The answer is given so just please explain how to get there and write clearly

3. Once he sneaks inside the un genius of Doctor Octopus a machinery at the far end of a The machinery e f 392 Hz. At a distance of d speed of sound is 343 m/s. (a) How many wavelengths has the sound traveled in those 6.30 m (b) What is the actual intensity of the waves at that distance, measured in W/m2? (c) At what distance will the sound intensity level be 78 dB? derground lair, Spider-Man quickly recognizes the technological t work. Scouting around the area, he hears a steady hum of n Dery large cavern. The machinery emits a pure tone at frequency 6.30 m, the sound intensity level is 98 dB. Assume that the

Explanation / Answer

a) we know, v = lamda*f

v ---> wave speed
lamda --> wavelength
f --> frequency

lamda = v/f

= 343/392

= 0.875 m

d/lamda = 6.3/0.875

= 7.2

d = 7.2*lamda

d = 7.2 wavelengths <<<<<<<--------------------Answer

b) use the formula for sound intensity level,

beta = 10*log(I/Io)

98 = 10*log(I/10^-12)

9.8 = log(I/10^-12)

10^9.8 = I/10^-12

==> I = 10^(9.8 - 12)

= 6.31*10^-3 W/m^2

= 6.31 mW/m^2 <<<<<<<--------------------Answer

c) let d' is the distance and let I' is the correspoding intensity of sound.

beta' = 10*log(I'/Io)

78 = 10*log(I'/10^-12)

7.8 = log(I'/10^-12)

10^7.8 = I'/10^-12

==> I' = 10^(7.8 - 12)

= 6.31*10^-5 W/m^2

= 6.31 mW/m^2

we know, I is onversly proportional to d^2

so,

I'/I = (d/d')^2

sqrt(I'/I) = d/d'

==> d' = d*sqrt(I/I')

= 6.3*sqrt( (6.31*10^-3)/(6.31*10^-5))

= 63.0 m <<<<<<<--------------------Answer

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