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inn [45]
3 years ago
9

A rescue team is searching for Andrew, a geologist who was stranded while conducting research in the mountains of Colorado. The

team uses electronic listening devices in order to detect any shouts for help. The sound waves from his shouts reaching the base camp can be approximated by a sinusoidal wave with a frequency f = 530 Hz and displacement amplitude A = 2.00×10−8 m , where the sound wave properties are valid at the base camp where the measurements are being made. What sound intensity level will the rescue team measure from the frightened researcher? Assume the speed of sound is v=328m/s and the density of air rho = 1.34 kg/m3
Physics
1 answer:
Sindrei [870]3 years ago
4 0

Answer:

I = 9.82 10⁻⁷ W / m²

Explanation:

The intensity of the sound wave is the energy of the wave between the order per unit area of ​​the same

      I = P / A = E / T A

the energy is calculated by integrating the mechanical energy in a period, where the mass is changed by the density and ‘s’ is the amplitude of the sound wave

     I = ½ ρ v (w s)²

   

     I = ½ 1.35 328 (2π 530 2.00 10⁻⁸)²

     I = 221.4 (4.435 10⁻⁹)

     I = 9.82 10⁻⁷ W / m²

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soldi70 [24.7K]

Answer:

F_m= 8.28 \times 10^{-16} N

Explanation:

The magnitude of the force is F_m=evB\sin\theta\\\implies F_m=1.602 \times 10^{-19}\times 300\times 19\times\sin 65^0= 8.28 \times 10^{-16} N

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2 years ago
In the diagram, q1= -2.60*10^-9 C and
Alekssandra [29.7K]

Answer:

The magnitude of the net electric field is:

E_{net}=90.37\: N/c

Explanation:

The electric field due to q1 is a vertical positive vector toward q1 (we will call it E1).

On the other hand, the electric field due to q2 is a horizontal positive vector toward q2(We will call it E2).

Knowing this, the <u>magnitude of the net electric</u> field will be the<u> E1 + E2. </u>

Let's find first E1 and E2.

The electric field equation is given by:

|E_{1}|=k\frac{|q_{1}|}{d_{1}^{2}}

Where:

  • k is the Coulomb constant (k = 9*10^{9} Nm²/C²)
  • q1 is the first charge
  • d1 is the distance from q1 to P

|E_{1}|=(9*10^{9})\frac{|-2.60*10^{-9}|}{0.538^{2}}

|E_{1}|=80.84\: N/C

And E2 will be:

|E_{2}|=k\frac{|q_{2}|}{d_{2}{2}}

|E_{2}|=(9*10^{9})\frac{|-8.30*10^{-9}|}{1.36^{2}}

|E_{2}|=40.39\: N/C

Finally, we need to use the  Pythagoras theorem to find the magnitude of the net electric field.

E_{net}=\sqrt{E_{1}^{2}+E_{2}^{2}}

E_{net}=\sqrt{80.84^{2}+40.39^{2}}

E_{net}=90.37\: N/c

I hope it helps you!

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2 years ago
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Answer:

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Explanation: you're welcome

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A microphone is located on the line connecting two speakers that are 0.513 m apart and oscillating in phase. The microphone is 1
pantera1 [17]

Answer:

frequency 1 = 334.30 Hz

frequency 2 = 1002.92 Hz

Explanation:

Given data

speaker distance y = 0.513 m

microphone distance D = 1.80 m

to find out

lowest two frequencies

solution

we know velocity of sound is 343 m/s

so we consider point x

so at 1st speaker distance from x   = D + (y/2)

1st speaker distance from x   = 1.80 + (0.513/2) = 2.0565 m   .....1

and

at 2nd speaker distance from x   = D - (y/2)

2nd speaker distance from x   = 1.80 - (0.513/2) = 1.5435 m     .........2

so destructive interference from 1 and 2  we know

1st - 2nd = ( m + 0.5 ) wavelength

2.0565 m - 1.5435 m = (  0+ 0.5) wavelength

wavelength  = 1.026 m

so here 1st min frequency will be

frequency 1 = velocity of sound / wavelength

frequency 1 = 343 / 1.026 =334.30 Hz

and

2nd  min frequency will be

frequency 2 =

2.0565 m - 1.5435 m = (  1 + 0.5) wavelength

wavelength  = 0.342 m

frequency 2 =  velocity of sound / wavelength

frequency 2 = 343 / 0.342 = 1002.92 Hz

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