912.
outer ear:
pinna
ear canal
middle ear:
ossicles and ear drum
inner ear:
semcircular canals
cochlea
auditory nerve
13.
frequency = wavespeed ÷ wavelength
14.
if frequency increases you would experience a higher pitch in sound
15.
humans can hear 20Hz to 20kHz
16.
The Doppler effect is the change in frequency or wavelength of a wave for an observer who is moving relative to the wave source. Can be used for machines measuring speed via doppler effect.
17.
Doppler in hospitals can be used for ultrasound to provide images for diagnosis and monitoring.
The wavelength of the note is
![\lambda = 39.1 cm = 0.391 m](https://tex.z-dn.net/?f=%5Clambda%20%3D%2039.1%20cm%20%3D%200.391%20m)
. Since the speed of the wave is the speed of sound,
![c=344 m/s](https://tex.z-dn.net/?f=c%3D344%20m%2Fs)
, the frequency of the note is
![f= \frac{c}{\lambda}=879.8 Hz](https://tex.z-dn.net/?f=f%3D%20%5Cfrac%7Bc%7D%7B%5Clambda%7D%3D879.8%20Hz%20)
Then, we know that the frequency of a vibrating string is related to the tension T of the string and its length L by
![f= \frac{1}{2L} \sqrt{ \frac{T}{\mu} }](https://tex.z-dn.net/?f=f%3D%20%5Cfrac%7B1%7D%7B2L%7D%20%5Csqrt%7B%20%5Cfrac%7BT%7D%7B%5Cmu%7D%20%7D%20%20)
where
![\mu=0.550 g/m = 0.550 \cdot 10^{-3} kg/m](https://tex.z-dn.net/?f=%5Cmu%3D0.550%20g%2Fm%20%3D%200.550%20%5Ccdot%2010%5E%7B-3%7D%20kg%2Fm)
is the linear mass density of our string.
Using the value of the tension, T=160 N, and the frequency we just found, we can calculate the length of the string, L:
30 km/h * 17 h = 30*17 km/h *h
= 510 km
This is a non testable question because it cannot be answered by doing an experiment. But it could be modified for example Dogs are more obedient then cats.
The correct answer to this is (A. Units Only).
It shows that there is a velocity of 35, but the units are missing.