A flywheel of diameter 1.2 m has a constant angular acceleration of 5.0 rad/s2. the tangential acceleration of a point on its ri
1 answer:
You might be interested in
Answer:
The most correct option is;
B. 10 km
Explanation:
Where:
y = Distance between the two headlights
d = Aperture of observers eye
λ = Wavelength of light
L = Distance between the observer and the headlight
Therefore, from the above solution, the distance between the observer and the headlights is 9386.066 km which is approximately 10 km.
Also we have
sinθ = y/L = 1.22 (λ/d)
sinθ = 1.22×10⁻⁴ rad
As the question is about changing in frequency of a wave for an observer who is moving relative to the wave source, the concept that should come to our minds is "Doppler's effect."
Now the general formula of the Doppler's effect is:
-- (A)
Note: We do not need to worry about the signs, as everything is moving towards each other. If something/somebody were moving away, we would have the negative sign. However, in this problem it is not the issue.
Where,
g = Speed of sound = 340m/s.
= Velocity of the receiver/observer relative to the medium = ?.
= Velocity of the source with respect to medium = 0 m/s.
= Frequency emitted from source = 400 Hz.
= Observed frequency = 408Hz.
Plug-in the above values in the equation (A), you would get:
Solving above would give you,
= 6.8 m/s
The correct answer = 6.8m/s
Now the general formula of the Doppler's effect is:
Note: We do not need to worry about the signs, as everything is moving towards each other. If something/somebody were moving away, we would have the negative sign. However, in this problem it is not the issue.
Where,
g = Speed of sound = 340m/s.
Plug-in the above values in the equation (A), you would get:
Solving above would give you,
The correct answer = 6.8m/s