Answer:
the frequency heard by the observer is equal to 2677 Hz
Explanation:
given,
velocity of the observer = 17 m/s
speed of the sound = 343 m/s
velocity of the source = 0 m/s
frequency emitted from the source = 2550 Hz

velocity of observer is negative as it is approaching the source. f = 2676.38 Hz ≈ 2677 Hz
hence, the frequency heard by the observer is equal to 2677 Hz
Answer:
Second drop: 1.04 m
First drop: 1.66 m
Explanation:
Assuming the droplets are not affected by aerodynamic drag.
They are in free fall, affected only by gravity.
I set a frame of reference with the origin at the nozzle and the positive X axis pointing down.
We can use the equation for position under constant acceleration.
X(t) = x0 + v0 * t + 1/2 * a *t^2
x0 = 0
a = 9.81 m/s^2
v0 = 0
Then:
X(t) = 4.9 * t^2
The drop will hit the floor when X(t) = 1.9
1.9 = 4.9 * t^2
t^2 = 1.9 / 4.9

That is the moment when the 4th drop begins falling.
Assuming they fall at constant interval,
Δt = 0.62 / 3 = 0.2 s (approximately)
The second drop will be at:
X2(0.62) = 4.9 * (0.62 - 1*0.2)^2 = 0.86 m
And the third at:
X3(0.62) = 4.9 * (0.62 - 2*0.2)^2 = 0.24 m
The positions are:
1.9 - 0.86 = 1.04 m
1.9 - 0.24 = 1.66 m
above the floor
Answer:OBSERVING IS WHATCHING A OBJECT VERY CLOSELY AND EXPIERIMENTING IS WER YOUTEST ON A CERTAIN THING OR CREATURE OR MASS OR ELEMENT
Explanation:IM THE MYSTERY MAN WHOOSH
Answer:
76.74 Hz
Explanation:
Given:
Wave velocity ( v ) = 330 m / sec
wavelength ( λ ) = 4.3 m
We have to calculate Frequency ( f ):
We know:
v = λ / t [ f = 1 / t ]
v = λ f
= > f = v / λ
Putting values here we get:
= > f = 330 / 4.3 Hz
= > f = 3300 / 43 Hz
= > f = 76.74 Hz
Hence, frequency of sound is 76.74 Hz.