An airplane is flying overhead at a constant elevation of 3000ft. A man is viewing the plane from a position 4000ft from the bas
e of a radio tower. The airplane is flying horizontally away from the man. If the plane is flying at a rate of 500ft/s, at what rate is the distance between the man and the plane increasing when the plane passes over the radio tower?
1 answer:
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The wavelength of the wave is 0.055 m
Explanation:
The relationship between speed, frequency and wavelength of a wave is given by the wave equation:
where
v is the speed
f is the frequency
is the wavelength
For the sound wave in this problem we have
v = 340 m/s is the speed
f = 6,191 Hz is the frequency
Solving for , we find the wavelength:
Learn more about waves and wavelength:
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Answer:
Work done = 13605.44
Explanation:
Data provided in the question:
For elongation of 2.1 cm (0.021 m) work done by the spring is 3.0 J
The relation between Energy (U) and the elongation (s) is given as:
U = ................(1)
where,
k is the spring constant
on substituting the valeus in the above equation, we get
3.0 =
or
k = 13605.44 N/m
now
for the elongation x = 2.1 + 4.1 = 6.2 cm = 0.062 m
using the equation 1, we have
U =
or
U = 26.149 J
Also,
Work done = change in energy
or
W = 26.149 - 3.0 = 23.149 J