A cannonball and a marble roll smoothly from rest down an incline. Is the cannonball’s(a) time to the bottom and(b) translationa
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A. 90.1 m
The wavelength of a wave is given by:
where
v is the speed of the wave
f is its frequency
For the sound emitted by the whale, v = 1531 m/s and f = 17.0 Hz, so the wavelength is
B. 102 kHz
We can re-arrange the same equation used previously to solve for the frequency, f:
where for the dolphin:
v = 1531 m/s is the wave speed
is the wavelength
Substituting into the equation,
C. 13.6 m
Again, the wavelength is given by:
where
v = 340 m/s is the speed of sound in air
f = 25.0 Hz is the frequency of the whistle
Substituting into the equation,
D. 4.4-8.7 m
Using again the same formula, and using again the speed of sound in air (v=340 m/s), we have:
- Wavelength corresponding to the minimum frequency (f=39.0 Hz):
- Wavelength corresponding to the maximum frequency (f=78.0 Hz):
So the range of wavelength is 4.4-8.7 m.
E. 6.2 MHz
In order to have a sharp image, the wavelength of the ultrasound must be 1/4 of the size of the tumor, so
And since the speed of the sound wave is
v = 1550 m/s
The frequency will be