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3 years ago

7

If two firecrackers produce a sound level of 81 dBdB when fired simultaneously at a certain place, what will be the sound level

if only one is exploded?

1 answer:

Mrac [35]3 years ago

5 0

Answer:

77.96dB

Explanation:

Recall that decibels are a unit of measuring intensity of sound, and depend on the logarithm of the intensity

the intensity, measured in decibels is given by:

I(db)=10log(I/I0)

I is the intensity in MKS units; I0 is the threshold intensity for human hearing (10^-12 W/m^2)

Thus, if the two sounds together have a dB of 81, we know:

81=10log(I/I0)

using the data above, we can find the intensity of the two sounds to be

0.000125 W/m^2

therefore, one firecracker has an intensity half of that, or 0.0000625W/m^2

now use this value to find the dB of one firecracker:

I(dB0=10log(0.0000625/10^-12)=77.96dB

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3 years ago

Two loudspeakers in a 20c room emit 686 hz sound waves along the x-axis.a. if the speakers are in phase, what is the smallest di

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<u>Answer :</u>

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<u>Explanation :</u>

It is given that,

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$d=\dfrac{\lambda}{2}$ ........(1)

Velocity of sound wave is given by :

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$d=\dfrac{v}{2f}$

$d=\dfrac{343\ m/s}{2\times 686\ Hz}$

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(2) For constructive interference, the path difference is integral multiple of wavelengths i.e.

$d=n\lambda$ ( n = integers )

Let n = 1

So, $d=\dfrac{v}{f}$

$d=\dfrac{343\ m/s}{686\ Hz}$

$d=0.5\ m$

Hence, the smallest distance between the speakers for which the interference of the sound waves is maximum constructive is 0.5 m.

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3 years ago

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