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andreyandreev [35.5K]
3 years ago
9

An 8-hour exposure to a sound intensity level of 90.0 dB may cause hearing damage. What energy in joules falls on a 0.800-cm-dia

meter eardrum so exposed?
Physics
2 answers:
Dafna11 [192]3 years ago
7 0

Answer:

1.44x10⁻³J

Explanation:

Given :

Time = 8hrs *(3600secs/1hr)= 28.8*10³seconds

Intensity= 90dB

D= 0.008m

Radius=0.008m/2

=0.004m

the sound level in decibel can be expressed below as

I=10dBlog[I/I₀]

where I₀=10⁻¹²/m² which is the refrence intensity

90=10log[I/10⁻¹²]

I=0.001W/m²

we know that intensity of the wave which id the average rate per unit area which energy is transfered can be calculated using below formular

I = P/A

where P= powerr which is renergy transfer at a time

A= area= πr²

making P subject of formular we have

P= I πr²

= 5.02 x10⁻⁸W

Energy =power x time

E=28800*5*10⁻⁸

=0.001443J

therefore,the energy in joules is 1.44x10⁻³J

11111nata11111 [884]3 years ago
5 0

Answer:

1.4E-3J

Explanation:

Given that

Time = 8hrs = 28.8E3 seconds

Intensity= 90dB

D= 0.008m

Radius= 0.004m

So intensity is sound level Bis

10dBlog(I/Io)

I= 10 (B/10dB)Io

= 10( 90/10) x 10^-12

=0.001W/m²

But we know that

I = P/A

P= I πr²

= 5.02 x10^-8W

But energy is power x time

So E= 5.02E-8 x 28.8E3

= 1.4E-3J

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A very humble bumble bee is flying horizontally due North at a constant speed of 3.11 m/s. At the current location of the bumble
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To solve this problem we will apply the concepts of the Magnetic Force. This expression will be expressed in both the vector and the scalar ways. Through this second we can directly use the presented values and replace them to obtain the value of the magnitude. Mathematically this can be described as,

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You are trying to overhear a most interesting conversation, but from your distance of 10.0 m , it sounds like only an average wh
Alexandra [31]

Answer:

r₂=0.1 m

Explanation:

Given that

r₁= 10 m  , β₁ = 20 dB

At r₂ ,β₂= 60 dB

As we know that intensity level of sound given as

\beta =10\ log\dfrac{I}{10^{-12}}

\beta _1=10\ log\dfrac{I_1}{10^{-12}}

20=10\ log\dfrac{I_1}{10^{-12}}

10² x 10⁻¹² = I₁

I₁=10⁻¹⁰ W/m²

\beta _2=10\ log\dfrac{I_2}{10^{-12}}

60=10\ log\dfrac{I_1}{10^{-12}}

10⁶ x  10⁻¹² = I₂

I₂ = 10⁻⁶ W/m²

I₁=10⁻¹⁰ W/m²

P = I A

P=Power ,I =Intensity  ,A=Area

\dfrac{I_1}{I_2}=\dfrac{r^2_2}{r^2_1}

\dfrac{10^{-10}}{10^{-6}}=\dfrac{r^2_2}{10^2}

r₂=0.1 m

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