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4 years ago

If a sound is 30 dB and its absolute pressure was 66 x 10-9 Pa, what must have been the reference pressure?

Physics

1 answer:

Ksenya-84 [330]4 years ago

4 0

Given:

I = 30dB

P = 66 × $10^{-9}$ Pa

Solution:

Formula used:

I = $20\log_{10}(\frac{P}{P_{o}})$ (1)

where,

I = intensity of sound

P = absolute pressure

$P_{o}$ = reference pressure

Using Eqn (1), we get:

$30 = 20\log _{10}\frac{66\times 10^{-9}}{P^{o}}$

$P_{o}$ = $\frac{66\times 10^{-9}}{10^{1.5}}$

$P_{o}$ = 2.08 × $10^{-9}$ Pa

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OSCilloscope shows wave starts cycle minimum 20 units reaches max of 100 units before completing 5 milliseconds what is relation

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Answer:

$T=0.02\ s$

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Explanation:

Given:

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Time taken to reach from the first minima to the first maxima, $t=5\times 10^{-3}\ s$

As we know that an oscilloscope executes a wave cycle represented by a sine wave. So we can deduce that it has executed one-fourth of the cycle in going from the amplitude of 20 units to 100 units in 0.005 seconds.

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4 years ago

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3 years ago

A cyclist is riding his bike up a mountain trail. When he starts up the trail , he is going 8 m/s. As the trail gets steeper, he

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4 years ago

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3 years ago

a cement block accidentally falls from rest from the ledge of a 80.6-m-high building. When the block is 10.8 m above the ground,

fomenos

Answer:

0.229 seconds

Explanation:

Given:

y₀ = 80.6 m

v₀ = 0 m/s

a = -9.8 m/s²

We need to find the difference in times when y = 10.8 m and y = 2.10 m.

When y = 10.8 m:

y = y₀ + v₀ t + ½ at²

10.8 = 80.6 + (0) t + ½ (-9.8) t²

10.8 = 80.6 − 4.9 t²

4.9 t² = 69.8

t = 3.774

When y = 2.10 m:

y = y₀ + v₀ t + ½ at²

2.10 = 80.6 + (0) t + ½ (-9.8) t²

2.10 = 80.6 − 4.9 t²

4.9 t² = 78.5

t = 4.003

The difference is:

4.003 − 3.774 = 0.229

The man has 0.229 seconds to get out of the way.

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4 years ago