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2 years ago

9

You are going to an outdoor concert, and you'll be standing near a speaker that emits 60 W of acoustic power as a spherical wave

. What minimum distance should you be from the speaker to keep the sound intensity level below 94 dB?

1 answer:

Bas_tet [7]2 years ago

3 0

Answer:

$r=44m$

Explanation:

β is calculated as:

$\beta =(10dB)log_{10}(I/I_{o} )\\ I=I_{o}10^{\frac{\beta }{10dB} }\\ I=(1.0*10^{-12}W/m^{2} )10^{\frac{\(94dB }{10dB} }\\I=2.51mW/m^{2}$

The distance r is defined as the radius of spherical wave.solve for r

We have

$I=\frac{P_{source} }{4\pi r^{2} }\\ r=\sqrt{\frac{P_{source}}{4\pi I} }\\ r=\sqrt{\frac{60W}{4\pi (2.51mW/m^{2} )} }\\r=44m$

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They do the method 3 times to be sure. Because if you do it once, that could mean anything. If you do it twice, it may or may not have the same result. If you do it 3 times and it matches one of the previous answers, then it's likely that it's correct.

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3 years ago

What is the volume of a bar of soap that is 9 cm long, 5 cm wide, and 2 cm high?

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Answer:

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2x5=10

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3 years ago

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A box of mass 26 kg is initially at rest on a flat floor. The coefficient of kinetic friction between the box and the floor is 0

Kazeer [188]

Answer:

$\Delta K = 52J$

Explanation:

The change in kinetic energy will be simply the difference between the final and initial kinetic energies: $\Delta K=K_f-K_i$

We know that the formula for the kinetic energy for an object is:

$K=\frac{mv^2}{2}$

where <em>m </em>is the mass of the object and <em>v</em> its velocity.

For our case then we have:

$\Delta K = K_f-K_i=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}=\frac{m(v_f^2-v_i^2)}{2}$

Which for our values is:

$\Delta K = \frac{m(v_f^2-v_i^2)}{2} = \frac{(26Kg)((2m/s)^2-(0m/s)^2)}{2} = 52J$

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3 years ago

A damped mass/spring system takes 14.0 s for its amplitude of the oscillator to decrease by a factor of 9. By what factor does t

fiasKO [112]

Answer:

The correct answer is "0.246".

Explanation:

Given that the amplitude is decreased by a factor of 9, then

$A \rightarrow (A-\frac{A}{9} )$

$A \rightarrow \frac{8A}{9}$

As we know,

Energy will be:

⇒ $E_{1}=\frac{1}{2}KA^2$

and,

⇒ $E_{2}=\frac{1}{2}K(\frac{8A}{9} )^2$

$=\frac{64KA^2}{162}$

⇒ $\Delta E=E_1-E_2$

On putting the estimated values, we get

$=\frac{1}{2}KA^2-\frac{64KA^2}{162}$

⇒ $\frac{\Delta E}{E}=\frac{\frac{20}{162}KA^2}{\frac{1}{2}KA^2}$

$=\frac{40}{162}$

$=0.246$

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2 years ago

An electron is placed 5.40 cm due north of a charged sphere and experienced a force of 8.30 x 10^-13 N due north. What is the el

ira [324]

Answer:

The electric field is $E = 5.1*10^6N/C.$

Explanation:

The force $F$ on a charge $q$ in an electric field $E$ is given by

$F = qE$ ,

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$E = \dfrac{F}{q }$

Now, the force on the electron is $F = 8.30*10^{-13}N$ , and its charge is

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3 years ago