Answer:
a. Speed = 1.6 m/s
b. Amplitude = 0.3 m
c. Speed = 1.6 m/s
Amplitude = 0.15 m
Explanation:
a.
The frequency of the wave must be equal to the reciprocal of the time taken by the boat to move from the highest point to the highest point again. This time will be twice the value of the time taken to travel from the highest point to the lowest point:
frequency =
= 0.25 Hz
The wavelength of the wave is the distance between consecutive crests of wave. Therefore,
Wavelength = 6.4 m
Now, the speed of the wave is given as:
Speed = (Frequency)(Wavelength)
Speed = (0.25 Hz)(6.4 m)
<u>Speed = 1.6 m/s</u>
<u></u>
b.
Amplitude is the distance between the mean position of the wave and the extreme position. Hence, it will be half the distance between the highest and lowest point:
Amplitude = (0.5)(0.6 m)
<u>Amplitude = 0.3 m</u>
<u></u>
c.
frequency =
= 0.25 Hz
Speed = (Frequency)(Wavelength)
Speed = (0.25 Hz)(6.4 m)
<u>Speed = 1.6 m/s</u>
<u></u>
Amplitude = (0.5)(0.3 m)
<u>Amplitude = 0.15 m</u>
Answer:
Total displacement = 0
Explanation:
He lives his front porch and still returns to his front porch.
Now, displacement is a vector quantity and as such, it is the distance between the initial point of movement and the final point of movement.
In this case the initial point is the same as the final point and thus the displacement is zero.
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.
Using safe research practices would be a very good common-sense practice in a lab environment.