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SSSSS [86.1K]
2 years ago
8

sound sensor display reads 40 dB inside and outside it reads 80dB. how many times louder is it outside than inside?

Physics
1 answer:
LuckyWell [14K]2 years ago
6 0

The difference is (80 dB - 40 dB) = 40 dB.

The sound is 40 dB louder outside.

Each 10 dB means 10 times more sound power.

40 dB louder means 10x10x10x10 times more sound power.

That's <em>10,000 times</em> more sound power outside than inside.

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Answer:

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initial velocity (U) = 0 mph = 0 m/s

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(B) average velocity = \frac{V+U}{2}

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(C) distance travelled (S) = ut + 0.5at^{2}

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