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3 years ago

A sound is recorded at 19 decibels. What is the intensity of the sound?

Physics

1 answer:

aalyn [17]3 years ago

4 0

Answer:

79·433 × $10^{-12}$ W/m²

Explanation:

Given intensity of sound in decibels is 19

Formula for intensity of sound in decibels is

where

$I_{1}$ is the intensity of sound measured in decibels

I is the intensity of sound measured in W/m²

$I_{0}$ is the intensity of the sound that is taken as standard and it's value is equal to $10^{-12}$ W/m²

By applying the formula

19 = 10 × ㏒(I ÷ $I_{0}$ )

⇒ 1·9 = ㏒(I ÷ $I_{0}$ )

⇒ antilog(1·9) = I ÷ $I_{0}$

⇒ 79·433 = I ÷ $10^{-12}$

⇒ I = 79·433 × $10^{-12}$ W/m²

∴ Intensity of sound = 79·433 × $10^{-12}$ W/m²

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Fynjy0 [20]

Answer:

The fraction of the object that is below the surface of the water is ¹⁷/₂₀

Explanation:

Given;

specific gravity of the object, γ = 0.850

Specific gravity is given as;

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Fraction of the object's weight below the surface of water is calculated as;

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3 years ago

Q.Solve the following circuit find total resistance RT. Also find value of voltage across resister RC.

Drupady [299]

Answer:

14.57 ohms

Explanation:

Here in the figure ,Rb & R₄are in series & also Rc & R₅ are in series. As they are in series , ( Rb + R₄ ) & (Rc & R₅) are in parallel . So the equivalent resistance in that branch = ( 2 + 18 ) ║ ( 3 + 12 )

= 20 ║ 15

= (20×15) / (20 + 15)

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Also Ra ( 6 ohm ) is in series with that branch ,. So the equivalent resistance of the whole circuit = 8.57 + 6 = 14.57 ohms.

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2 years ago

A train starts at v=10m/s and accelerates at 1m/s2 for 5s. Its final speed is *

ad-work [718]

Answer:

D) 15 m/s

Explanation:

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2 years ago

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3 years ago

Suppose that the acceleration of a model rocket is proportional to the difference between 160 ft/sec and the rocket's velocity.

Levart [38]

Answer:

Explanation:

Given ,

dv / dt = k ( 160 - v )

dv / ( 160 - v ) = kdt

ln ( 160 - v ) = kt + c , where c is a constant

when t = 0 , v = 0

Putting the values , we have

c = ln 160

ln ( 160 - v ) = kt + ln 160

ln ( 160 - v / 160 ) = kt

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1 - v / 160 = $e^{kt }$

v / 160 = 1 - $e^{kt }$

v = 160 ( 1 - $e^{kt }$ )

differentiating ,

dv / dt = - 160k $e^{kt }$

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given when t = 0 , a = 280

280 = - 160 k

k = - 175

a = - 160 x - 175 $e^{kt}$

a = 28000 $e^{kt}$

when a = 128 t = ?

128 = 28000 $e^{kt}$

$e^{kt }$ = .00457

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2 years ago