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3 years ago

12

Quadrupling the power output from a speaker emitting a single frequency will result in what increase in loudness (in units of dB

)

1 answer:

Leviafan [203]3 years ago

6 0

Answer:

<em>6.02 dB increase </em>

<em></em>

Explanation:

Let us take the initial power from the speaker P' = P Watt

then, the final power P = 4P Watt

for a given unit area, initial intensity (power per unit area) will be

I' = P Watt/m^2

and the final quadrupled sound will produce a sound intensity of

I = 4P Watt/m^2

Increase in loudness is gotten from the relation

ΔL = $10log_{10} \frac{I}{I'}$

where

I = final sound intensity

I' = initial sound intensity

imputing values of the intensity into the equation, we have

==> $10log_{10} \frac{4P}{P}$ = $10log_{10} 4$ = <em>6.02 dB increase </em>

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Which statements are true about galaxies, stars, and the universe? (more than one true answer by the way)

Elina [12.6K]

Answer:

statements <em><u>2, 3, 4, and 7</u></em> are true

Explanation:

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3 years ago

Metals are used in many products because of the characteristic properties that most metals share. Which properties of the metal

Anastaziya [24]

The most important characteristics that are exhibited by metals are-

1- Metals are ductile

2-Most metals are conductive in nature.

3-Most metals are malleable.

4- Metals have strong inter molecular force of attraction between the.

5-Metals have luster.

6-Metals are sonorous.

Here we are given Tungsten filament.

Tungsten is a metal.So it must be conductive and as well as ductile in nature.

The electric filament that we are using in our electric bulb glows due to the heating effect of current.Hence the chosen substances for glowing electric bulb must have high melting point.

The melting point of tungsten is 1650 degree celsius which is very high.That's why it is used in electric bulb.

Hence the correct answer to the question is the third one i.e Tungsten is ductile,has a high melting point, and is electrically conductive.

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3 years ago

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When you measure something in meters cubed, you are measuring ____.

Shalnov [3]

I think your answer is volume

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3 years ago

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In an attempt to reduce the extraordinarily long travel times for voyaging to distant stars, some people have suggested travelin

alexandr402 [8]

Answer:

a) $v=0.999124c$

b) $E=7.566*10^{22}$

c) $E_a=760 times\ larger$

Explanation:

From the question we are told that

Distance to Betelgeuse $d_b=430ly$

Mass of Rocket $M_r=20000$

Total Time in years traveled $T_d=36years$

Total energy used by the United States in the year 2000 $E_{2000}=1.0*10^20$

Generally the equation of speed of rocket v mathematically given by

$v=\frac{2d}{\triangle t}$

$v=860ly/ \triangle t$

where

$\triangle t=\frac{\triangle t'}{(\sqrt{1-860/ \triangle t)^2}}$

$\triangle t=\frac{36}{(\sqrt{1-860/ \triangle t)^2}}$

$\triangle t=\sqrt{(860)^2+(36)^2}$

$\triangle t=860.7532$

Therefore

$v=\frac{860ly}{ 860.7532}$

$v=0.999124c$

b)

Generally the equation of the energy E required to attain prior speed mathematically given by

$E=\frac{1}{\sqrt{1-(v/c)^2} }-1(20000kg)(3*10^8m/s)^2$

$E=7.566*10^{22}$

c)Generally the equation of the energy $E_a$ required to accelerate the rocket mathematically given by

$E_a=\frac{E}{E_{2000}}$

$E_a=\frac{7.566*10^{22}}{1.0*10^{20}}$

$E_a=760 times\ larger$

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3 years ago

If we double the frequency of a system undergoing simple harmonic motion, which of the following statements about that system ar

AnnyKZ [126]

Answer:

a. The angular frequency is doubled.

e. The period is reduced to one-half of what it was.

Explanation:

Angular frequency is given as;

ω = 2πf

$\frac{\omega _1}{f_1} = \frac{\omega _2}{f_2}$

when the frequency is doubled

$\frac{\omega _1}{f_1} = \frac{\omega _2}{(2f_1)} \\\\\omega _1 = \frac{\omega _2}{2}\\\\\omega _2 = 2\omega _1$

Thus, the angular frequency will be doubled.

Amplitude in simple harmonic motion is the maximum displacement.

Frequency is related to period in simple harmonic motion as given in the equation below;

$f = \frac{1}{T} \\\\f_1T_1= f_2T_2\\\\T_2 = \frac{f_1T_1}{f_2}$

when the frequency is doubled;

$T_2 = \frac{f_1T_1}{2f_1} \\\\T_2 = \frac{T_1}{2}$

Thus, the period will be reduced to one-half of what it was.

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3 years ago