Suppose a rectangular piece of aluminum has a length D, and its square cross section has the dimensions W XW, where D (W x W) to
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Answer:
a) the values of the angle α is 45.5°
b) the required magnitude of the vertical force, F is 41 lb
Explanation:
Applying the free equilibrium equation along x-direction
from the diagram
we say
∑Fₓ = 0
Pcosα - 425cos30° = 0
525cosα - 368.06 = 0
cosα = 368.06/525
cosα = 0.701
α = cos⁻¹ (0.701)
α = 45.5°
Also Applying the force equation of motion along y-direction
∑Fₓ = ma
Psinα + F + 425sin30° - 600 = (600/32.2)(1.5)
525sin45.5° + F + 212.5 - 600 = 27.95
374.46 + F + 212.5 - 600 = 27.95
F - 13.04 = 27.95
F = 27.95 + 13.04
F = 40.99 ≈ 41 lb
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