The frequency of the human ear canal is 2.92 kHz.
Explanation:
As the ear canal is like a tube with open at one end, the wavelength of sound passing through this tube will propagate 4 times its length of the tube. So wavelength of the sound wave will be equal to four times the length of the tube. Then the frequency can be easily determined by finding the ratio of velocity of sound to wavelength. As the velocity of sound is given as 339 m/s, then the wavelength of the sound wave propagating through the ear canal is
Wavelength=4*Length of the ear canal
As length of the ear canal is given as 2.9 cm, it should be converted into meter as follows:

Then the frequency is determined as
f=c/λ=339/0.116=2922 Hz=2.92 kHz.
So, the frequency of the human ear canal is 2.92 kHz.
To look for displacement, just draw a vector from your beginning stage to your last position and settle for the length of this line. So we begin by drawing a line to the north which is 30 ft, since it is north, the line is going up, then it move 5 ft to the south, so put a line going down, so we are in 25 ft, North so that would be the answer.
Answer:
Resultant is 152 N at 28.5 degrees south to the 100 N force
Explanation:
<span>Because P = W ÷ t, and W = F*t, you can substitute (W) for (F*t). Then substitute (F) for (m*a). This will leave you with P = (m*a*d)/t. Since you need velocity, youd want to solve for a so you can use v = a*t. a = (P*t)/(m*d) therefore, substituting a in v = a*t, v = (P*t*t)/(m*d)</span>