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SSSSS [86.1K]
1 year ago
9

how much power is radiated as a sound from a band whose intensity is 1.6x10-3 w/m2 at a distance of 12m

Physics
1 answer:
Vesna [10]1 year ago
8 0

2.89watts.

<h3>What is meant by sound intensity?</h3>
  • The average rate at which sound energy moves across a unit area normal to a given direction is used to determine a sound's intensity. This rate is generally stated in ergs per second per square centimeter.
  • Decibels are the units used to measure sound intensity, often known as sound power or sound pressure. The decibel (dB) unit is named after Alexander Graham Bell, who also created the audiometer and the telephone. An audiometer is a tool to gauge a person's hearing capacity for various noises.
  • Our ability to measure the flow of sound energy as a time-averaged vector quantity makes sound intensity measuring an effective method. We can identify sound sources and tell direct sound from reverberant sound in a room using the characteristics of sound intensity.

How much power is radiated as a sound from a band whose intensity is 1.6x10-3 w/m2 at a distance of 12m:

Formula: I\frac{P}{4\pi r^{2} }

I=1.6x10-3 w/m2

r=12m

I\frac{P}{4\pi r^{2} }

P=I4\pi r^{2}

   =(1.6x10-3\frac{W}{m^{2} } ) (4\pi (12m)^{2} )

   =2.89watts.

To learn more about sound intensity, refer to:

brainly.com/question/17062836

#SPJ9

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Find the density of seawater at a depth where the pressure is 500 atm if the density at the surface is 1100 kg/m^3 . Seawater ha
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The density of seawater at a depth where the pressure is 500 atm is 1124kg/m^3

Explanation:

The relationship between bulk modulus and pressure is the following:

B=\rho_0 \frac{\Delta p}{\Delta \rho}

where

B is the bulk modulus

\rho_0 is the density at surface

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\Delta \rho is the variation of density

In this problem, we have:

B=2.3\cdot 10^9 N/m^2 is the bulk modulus

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Therefore, we can find the density of the water where the pressure is 500 atm as follows:

\rho = \rho_0 + \Delta \rho = \rho_0+\frac{\rho_0 \Delta p}{B}=\rho_0 (1+\frac{\Delta p}{B})=(1100)(1+\frac{5.05\cdot 10^7}{2.3\cdot 10^9})=1124kg/m^3

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