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14 days ago

12

The loudness of a sound is inversely proportional to the square of your distance from the source of the sound. if your friend is

right next to the speakers at a loud concert and you are four times as far away from the speakers, how does the loudness of the music at your position compare to the loudness at your friend's position?

2 answers:

Yuliya22 [1.1K]14 days ago

5 0

Define loudness as L, distance as d, and k as the variation constant. The appropriate equation representing the relationship is:
L = k / d^2
For Situation 1,
L1 = k / d1^2
For Situation 2,
L2 = k / (d1 / 4)^2
To have equal k, it follows that L2 = 16 L1.
Thus, the loudness your friend experiences is 16 times greater than yours.

Sav [1.1K]14 days ago

3 0

Answer:

The loudness at your location amounts to just 1/16 of the loudness at your friend’s spot.

Explanation:

Loudness diminishes as the distance from the speaker increases

and it varies inversely with the square of the distance.

So, we have

$L = \frac{k}{r^2}$

Comparing loudness at two different distances gives

$\frac{L_1}{L_2} = (\frac{r_2}{r_1})^2$

Knowing that,

$r_2 = 4 r_1$

and thus

$\frac{L_1}{L_2} = 4^2$

$L_2 = \frac{L_1}{16}$

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Complete Question

An aluminum "12 gauge" wire measures a diameter of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field E in the wire varies over time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is recorded in seconds.

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Answer:

The charge is $Q =2.094 C$

Explanation:

The question indicates that

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The radius of the wire is $r = \frac{0.00205}{2} = 0.001025 \ m$

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The charge is effectively given by the equation

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$Q = 120 \int\limits^{t}_{0} { E(t) } \, dt$

By substituting values

$Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt$

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.$

The question states that t = 5 seconds

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.$

$Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }$

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