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kobusy [5.1K]
3 years ago
6

Two speakers create identical

Physics
1 answer:
Inessa [10]3 years ago
4 0

Answer:

2.185 m

Explanation:

We are given that

Frequency of identical sound wave=f=240 Hz

Distance of speaker from person 1=L_1=1.47 m

Velocity of sound wave,v=343 m/s

We have to find the minimum distance to speaker 2 for there to be destructive interference at that spot.

We know that

v=f\lambda

\lambda=\frac{v}{f}=\frac{343}{240}

\lambda=1.43 m

We know that

L_2-L_1=(n-\frac{1}{2})\lambda

For minimum distance

n=1

Substitute the value

L_2-1.47=(1-\frac{1}{2})\times 1.43

L_2-1.47=\frac{1}{2}\times 1.43

L_2-1.47= 0.715

L_2=0.715+1.47=2.185 m

Hence, the minimum distance to speaker 2 for to be destructive interference at that spot=2.185 m

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The electric field due to the 4.0 μC charge is E = kq/r² where

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<h3>What is the electric field due to the -4.0 μC charge?</h3>

The electric field due to the -4.0 μC charge is E = kq'/r² where

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= -2kq[1/x]₂⁰

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Answer:

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The two equations become:

m₂a = T - m₁a

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