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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 of the following is true concerning nuclear fusion?

Zielflug [23.3K]

Choice-'a' is a slippery, misleading, ambiguous statement,
but it's less wrong than any of the other choices on this list.

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3 years ago

Based on the following equation, answer the questions below. ρ = (2γϕ + ψ)/rg where ρ [=] moles per cubic foot [mol/ft3] γ [=] j

AlekseyPX

1) Fundamental units of $\Psi$ are $[\frac{mol}{m\cdot s^2}]$

2) Fundamental units of $\Phi$ are $[\frac{mol}{m^3}]$

Explanation:

The equation for the variable $\rho$ is

$\rho =\frac{2\gamma \Phi+\Psi}{rg}$

where we have:

$\rho$ measured in $[\frac{mol}{ft^3}]$

$\gamma$ measured in $[\frac{J}{kg}]$

$r$ measured in $[in]$

$g$ measured in $[\frac{m}{s^2}]$

We can re-write the equation as

$\rho rg = 2\gamma \Phi + \Psi$

And we notice that the units of the term on the left must be equal to the units of the term on the right.

This means that:

1) First of all, $\Psi$ must have the same units of $\rho r g$ . So,

$[\rho r g]=[\frac{mol}{ft^3}][in][\frac{m}{s^2}]$

However, both ft (feet) and in (inches) are not fundamental dimensions: this means that they can be expressed as meters. Therefore, the fundamental units of $\Psi$ are

$[\Psi]=[\frac{mol}{m^3}][m][\frac{m}{s^2}]=[\frac{mol}{m\cdot s^2}]$

2)

The term $2\gamma \Phi$ must have the same units of $\Psi$ in order to be added to it. Therefore,

$[\gamma \Phi] = [\frac{mol}{m\cdot s^2}]$

We also know that the units of $\gamma$ are $[\frac{J}{kg}]$ , therefore

$[\frac{J}{kg}][\Phi]= [\frac{mol}{m\cdot s^2}]$

And so, the fundamental units of $\Phi$ are

$[\Phi]= [\frac{mol\cdot kg}{J\cdot m\cdot s^2}]$

However, the Joules can be written as

$[J]=[kg][\frac{m^2}{s^2}]$

Therefore

$[\Phi]= [\frac{mol\cdot kg}{(kg \frac{m^2}{s^2})\cdot m\cdot s^2}]=[\Phi]= [\frac{mol}{m^3}]$

#LearnwithBrainly

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3 years ago

At a certain instant an object is moving to the right with speed 1.0 m/s and has a constant acceleration to the left of 1.0 m/s2

professor190 [17]

Explanation:

Given that,

Initial speed of the object, u = 1 m/s

Acceleration of the object, $a=-1\ m/s^2$

We need to find the time when the object comes at rest (v=0). Let it is given by :

$t=\dfrac{v-u}{a}$

$t=\dfrac{0-1\ m/s}{-1\ m/s^2}$

t = 1 second

So, the object will comes to rest at 1 second. Hence, this is the required solution.

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3 years ago

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timurjin [86]

Answer:

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3 years ago

Aleksandr-060686 [28]

Answer:

A factor of 2*4 = 8

Explanation:

F_g = (G*m1*m2)/r^2

where m1 and m2 are the two masses, G is Newton's gravitational constant, and r is the distance between the center of mass of the two objects.

So, if you double m1 and quadruple m2:

m1' = 2*m1

m2' = 4*m2

Then F_g' = (G*m1'*m2')/r^2 = (G*2*m1*4*m2)/r^2 = 8*(G*m1*m2)/r^2 = 8*F_g

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3 years ago