One of the harmonics of a column of air in a tube that is open at one end and closed at the other has a frequency of 448 hz, and the next higher harmonic has a frequency of 576 hz. how long is the tube? the speed of sound in air is 343 m/s. one of the harmonics of a column of air in a tube that is open at one end and closed at the other has a frequency of 448 hz, and the next higher harmonic has a frequency of 576 hz. how long is the tube? the speed of sound in air is 343 m/s. 1.00 m 2.68 m 1.34 m 0.335 m 0.670 m

1.34 m

Explanation:

For an open-end tube, the frequency difference between two consecutive harmonics is equal to the fundamental frequency of the tube:

In this case, we have

so, the fundamental frequency is

For an open-end tube, the fundamental frequency is also given by:

where v is the speed of sound and L the length of the tube.

Since we know v = 343 m/s, we can solve the formula for L:

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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no. dot product   =0 when vecs at right angles