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melomori [17]
3 years ago
5

A concert loudspeaker suspended high off the ground emits 28.0 W of sound power. A small microphone with a 0.700 cm2 area is 55.

0 m from the speaker. What is the sound intensity at the position of the microphone?
Physics
1 answer:
grandymaker [24]3 years ago
7 0

Explanation:

It is known that wave intensity is the power to area ratio.

Mathematically,    I = \frac{P}{A}

As it is given that power is 28.0 W and area is 7 \times 10^{-5} m^{2}.

Therefore, sound intensity will be calculated as follows.

             I = \frac{P}{A}

               = \frac{28.0 W}{4 \times 3.14 \times 7 \times 10^{-5} m^{2}}

                = 0.318 \times 10^{5} W/m^{2}

or,             = 3.18 \times 10^{4} W/m^{2}

Thus, we can conclude that sound intensity at the position of the microphone is 3.18 \times 10^{4} W/m^{2}.

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