The answer is <span>higher than.
</span><span>A sound-producing object is moving toward an observer. The sound the observer hears will have a frequency higher than that actually being produced by the object.
Why?
</span>As the source of the waves is moving toward the observer, each of the successive wave crest<span> is emitted from a position closer to the observer than the previous wave.
Thus each wave takes slightly less time to reach the observer than the previous wave. So, the time between the arrival of successive wave crests at the observer is reduced, increasing the frequency. </span>
Distance/ time = 100/5 = 20 km per hour
The average speed at 5hrs is 20 km per hour.
Answer:
Change in momentum will be -4.4 kgm/sec
So option (A) is correct option
Explanation:
Mass of the ball is given m = 0.10 kg
Initial velocity of ball 
And velocity after rebound 
We have to find the change in momentum
So change in momentum is equal to 
( here negative sign shows only direction )
So option (A) will be correct answer
Answer:
(A) 0.2306 m
(B) 1.467 Hz
(C) 0.1152 m
Explanation:
spring constant (K) = 16.4 N/m
mass (m) = 0.193 kg
acceleration due to gravity (g) = 9.8 m/s^{2}
(A) force = Kx, where x = extension
mg = Kx
0.193 x 9.8 = 16.4x
x = 0.1153 m
now the mass actually falls two times this value before it gets to its equilibrium position ( turning point ) and oscillates about this point
therefore
2x = 0.2306 m
(B) frequency (f) = \frac{1}{2π} x 
frequency (f) = \frac{1}{2π} x 
frequency = 1.467 Hz
(C) the amplitude is the maximum position of the mass from the equilibrium position, which is half the distance the mass falls below the initial length of the spring
= \frac{0.2306}{2} = 0.1152 m
This is a Doppler effect. Generally, if you move to a frequency source, you would detect an increase in frequency and when you move away from a source you would detect a decrease.
For this question, before you pass them, you are actually approaching them, so you would hear a higher frequency than the constant 300 Hz they are playing at.
Using the condensed formula:
f ' = ((v <u>+</u> vd)/(v <u>+</u> vs)) * f
Where: vd = Velocity of the detector.
vs = Velocity of the frequency source.
v = Velocity of sound in air.
f ' = Apparent frequency.
f = Frequency of source.
v = 343 m/s, vd = detector = 27.8 m/s, vs = velocity of the source =0. (the flautists are not moving).
f = 300 Hz.
There would be an overall increase in frequency, so we maintain a plus at the numerator and a minus at the denominator.
f ' = ((v + vd)/(v - vs)) * f
f ' = ((343+ 27.8)/(343 - 0)) * 300
= (370.8/343)* 300 = 324.3
Therefore frequency before passing them = 324.3 Hz.
Cheers.
