Waves in a fish bowl jostled by the thingamajigger move to the sides at an average velocity of 0.50 m/s. if they occur once every 0.25 seconds, what is the distance between crests of the waves? show your work!

0.125 m

Explanation:

In this problem, we have:

v = 0.50 m/s is the average velocity of the wave

T = 0.25 s is the period of the wave

We can find the frequency of the wave, which is equal to the reciprocal of the period:

The problem is asking us to find the distance between two crests of the wave: this is equivalent to the wavelength. The wavelength is related to the average velocity and the frequency by the formula:

Substituting the numerical values, we find

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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