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The low-frequency speaker of a stereo set has a surface area of 0.05 $\mathrm{m}^{2}$ and produces 1 $\mathrm{W}$ of acoustical power. What is the intensity at the speaker? If the speaker projects sound uniformly in all directions, at what distance from the speaker is the intensity 0.1 $\mathrm{W} / \mathrm{m}^{2} ?$

$\begin{aligned} I_{a} &=20 \mathrm{W} / \mathrm{m}^{2} \\ r &=3.99 \mathrm{m} \end{aligned}$

Physics 101 Mechanics

Physics 103

Chapter 16

Oscillatory Motion and Waves

Periodic Motion

Wave Optics

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so here for party. We know that the intensity is equaling the power output divided by the area. In this case, this would be the power output divided by the length times the width. This would be equaling the power output of 1.0 watts, divided by 0.500 meters squared, and this is equaling 20.0 watts per square meter. This would be the intensity of the speaker. For part a four part be solving for the area. This would be equaling the power output divided by the intensity. This is equaling the power output of 1.0 watts. Divided by now, the intensity is 0.100 square meters. This is equaling 10 0.0 square. Rather, my apologies, this intensity unit would be watts per square meter the magnitudes, of course, the same. And so the area would be 10.0 square meters. And we know that here the area is four pies squared and so we can then say that here the distance are at which the intensity reduces 2.100 watts per square meter would be equal in the square root of 10.0 square meters, divided by four pi. And so essentially we can say that are again. The distance at which the intensity decreases 2.100 watts per square meter is equal in 0.89 two meters. This would be our final answer for part B. That is the end of the solution. Thank you for once.

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