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3 years ago

A brand of earplugs reduces the sound intensity level by 27 dB.By what factor do these earplugs reduce the acoustic intensity?

Physics

1 answer:

Fofino [41]3 years ago

3 0

Answer:

Explanation:

27dB = 2.7 B

So I / I₀ = 10⁻²°⁷ , I₀ is intensity of main sound and I is intensity after reduction.

= 1.99 X 10⁻³

So intensity will reduce by 1.99 X 10⁻³ .

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coldgirl [10]

Answer:

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Explanation:

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3 years ago

The filament of a certain lamp has a resistance that increases linearly with temperature. When a constant voltage is switched on

alekssr [168]

Answer:

The change in temperature is $\Delta T = 1795 K$

Explanation:

From the question we are told that

The temperature coefficient is $\alpha = 4 * 10^{-3 }\ k^{-1 }$

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$R = R_o [1 + \alpha \Delta T]$

Where $R_o$ is the initial resistance

Making the change in temperature the subject of the formula

$\Delta T = \frac{1}{\alpha } [\frac{R}{R_o} - 1 ]$

Now from ohm law

$I = \frac{V}{R}$

This implies that current varies inversely with current so

$\frac{R}{R_o} = \frac{I_o}{I}$

Substituting this we have

$\Delta T = \frac{1}{\alpha } [\frac{I_o}{I} - 1 ]$

From the question we are told that

$I = \frac{I_o}{8}$

Substituting this we have

$\Delta T = \frac{1}{\alpha } [\frac{I_o}{\frac{I_o}{8} } - 1 ]$

=> $\Delta T = \frac{1}{3.9 * 10^{-3}} (8 -1 )$

$\Delta T = 1795 K$

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3 years ago

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3 years ago