On the kelvin scale what is the freezing point of water
1 answer:
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Answer:
841.5 Hz
Explanation:
Given
y = 50 cm = 0.5 m
d = 5.00 m
L = 12.0 m away from the wall
v = speed of sound = 343 m/s
The image of the scenario is presented in the attached image.
When destructive interference is being experienced from 50 cm (0.5 m) parallel to the wall, the path difference between the distance of the two speakers from the observer is equal to half of the wavelength of the wave.
Let the distance from speaker one to the observer's new position be d₁
And the distance from the speaker two to the observer's new position be d₂
(λ/2) = |d₁ - d₂|
d₁ = √(12² + 3²) = 12.3693 m
d₂ = √(12² + 2²) = 12.1655 m
|d₁ - d₂| = 0.2038 m
(λ/2) = |d₁ - d₂| = 0.2038
λ = 0.4076 m
For waves, the velocity (v), frequency (f) and wavelength (λ) are related thus
v = fλ
f = (v/λ) = (343/0.4076) = 841.5 Hz
Hope this Helps!!!
