PLEASE HELP ME WITH THIS ONE QUESTION A tuning fork of 500 Hz is struck in a room with a speed of sound of 340 m/s. What is the distance between a point of rarefaction and the next compression point, in the resulting sound?

0.34 m

Explanation:

From the question,

v = λf Equation 1

Where v = speed of sound, f = frequency, λ = Wave length

Make λ the subject of the equation

λ = v/f Equation 2

Given: v = 340 m/s, f = 500 Hz.

Substitute these values into equation 2

λ = 340/500

λ = 0.68 m

But,  the distance between a point of rarefaction and the next compression point, in the resulting sound is half wave length

Therefore,

λ/2 = 0.68/2

λ/2 = 0.34 m

Hence, the distance between a point of rarefaction and the next compression point, in the resulting sound is 0.34 m

d. air pressure decreases as altitude increases.

this is an example of "committed" time.

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