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horrorfan [7]
4 years ago
7

Two speakers, one directly behind the other, are each generating a 240-Hz sound wave. What is the smallest separation distance b

etween the speakers that will produce destructive interference at a listener standing in front of them?
Physics
2 answers:
olga nikolaevna [1]4 years ago
6 0

Answer:

Explanation:

frequency, f = 240 hz

speed of sound, v = 343 m/s

Let λ is the wavelength.

v = f x λ

343 = 240 x λ

λ = 1.43 m

The smallest separation for the destructive interference is given by

d = λ / 2

d = 1.43 / 2 = 0.71 m

AveGali [126]4 years ago
3 0

Answer:

The smallest separation distance between the speakers is 0.71 m.

Explanation:

Given that,

Two speakers, one directly behind the other, are each generating a 240-Hz sound wave, f = 240 Hz

Let the speed of sound is 343 m/s in air. The speed of sound is given by the formula as :

v=f\lambda\\\\\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{343\ m/s}{240\ Hz}\\\\\lambda=1.42\ m

To produce destructive interference at a listener standing in front of them,

d=\dfrac{\lambda}{2}\\\\d=\dfrac{1.42}{2}\\\\d=0.71\ m

So, the smallest separation distance between the speakers is 0.71 m. Hence, this is the required solution.

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HA → H⁺  +  A⁻

If 1 mol of HA produces 1 mol of H⁺  and  A⁻, 1.00 * 10⁻⁴ mol of HA will produce 1.00 * 10⁻⁴ mol of  H⁺  and  A⁻.

Since only 16% dissociation occurs = 0.16

Number of moles of  H⁺ produced = 0.16 *  1.00 * 10⁻⁴

Number of moles of  H⁺ produced = 1.6 * 10⁻⁵mols

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Number of moles of  A⁻ produced = 1.6 * 10⁻⁵mols

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Concentration = num of moles/volume

Let the volume of the final solution be V

Conc of HA = 8.4 * 10⁻⁵/V

Conc of H⁺ = 1.6 * 10⁻⁵/V

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To calculate the dissociation constant

k_{a} = [H^{+} ][A^{-} ]/[HA]

k_{a}= [1.6 * 10^{-5} /V][1.6 * 10^{-5} /V]/[8.4 * 10^{-5} /V]\\k_{a}= 3.05 * 10^{-6} /V\\k_{a} = 1.00 * 10^{-4}\\ 1.00 * 10^{-4} = 3.05 * 10^{-6} /V\\V= 3.05 * 10^{-6}/ 1.00 * 10^{-4}\\V=3.05 * 10^{-2}\\V=0.0305 L

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