What is the net force acting on a 50-kg object that accelerates at a rate of +3.5 m/s/s?
1 answer:
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Answer:
(1) 120°
Explanation:
The wheel is purely rolling, which means it rotates about an axis through the point where it touches the ground (see Figure 11-6).
The net velocity at point P on the periphery, relative to the bottom of the wheel, is:
v = ωr
where r is the distance from the point at the bottom of the wheel to P (see diagram).
To find r, we need to use some geometry. From Inscribed Angle Theorem, we know the inscribed angle is half the arc angle. And from Thales' Theorem, we know an angle inscribed across a diameter is a right angle.
Therefore:
cos (θ/2) = r / (2R)
r = 2R cos (θ/2)
So the net velocity at P is:
v = 2Rω cos (θ/2)
We want this to equal the velocity at the center of the circle, which is Rω.
Rω = 2Rω cos (θ/2)
1 = 2 cos (θ/2)
cos (θ/2) = 1/2
θ/2 = 60°
θ = 120°
The original frequency of horn of Car A is 1071 Hz.
Explanation:
Doppler effect describes the change in the frequency of sound waves with respect to the observer. As the sound waves emitted from a source need to travel the air medium to reach observer, it will undergo loss in energy. So there will be change in its frequency compared to original frequency. Depending upon the direction of travel of source and observer the shifting of frequency will vary.
Here vo is the observer velocity and vs is the velocity of the source. So Vo = 15 m/s as car B is the observer and Vs = 35 m/s as car A is the source. And f is the frequency of sound wave at source that is car A.
Similarly, the doppler shift in frequency is the frequency of sound heard by car B which is f' = 1140 Hz. And v is the speed of sound that is v = 343 m/s
1140 =
f = 1140/1.0649= 1071 Hz.
Thus, the original frequency of horn of Car A is 1071 Hz.