Answer:
Speed at which the ball passes the window’s top = 10.89 m/s
Explanation:
Height of window = 3.3 m
Time took to cover window = 0.27 s
Initial velocity, u = 0m/s
We have equation of motion s = ut + 0.5at²
For the top of window (position A)

For the bottom of window (position B)


We also have

Solving

So after 1.11 seconds ball reaches at top of window,
We have equation of motion v = u + at

Speed at which the ball passes the window’s top = 10.89 m/s
Answer:
20.7 s
Explanation:
The equation to calculate the velocity for a uniform acceleration a, time t and initial velocity v₀:
v = a*t + v₀
Solve for t:
t = (v - v₀)/a
An analog signal carries information by copying an original sound
When we speak through the microphone, it turns our sound into some sort of electronic wave.
This electronic wave is caught by a recording device and later could be replicated into Mp3 file that we usually listen to
we know
T=2.pie sqr root(l/g)
l= (sqrof 7/22)*9.8
L= .99 or approx 1m
Answer:
(a) f= 622.79 Hz
(b) f= 578.82 Hz
Explanation:
Given Data
Frequency= 600 Hz
Distance=1.0 m
n=120 rpm
Temperature =20 degree
Before solve this problem we need to find The sound generator moves on a circular with tangential velocity
So
Speed of sound is given by
c = √(γ·R·T/M)
............in an ideal gas
where γ heat capacity ratio
R universal gas constant
T absolute temperature
M molar mass
The speed of sound at 20°C is
c = √(1.40 ×8.314472J/molK ×293.15K / 0.0289645kg/mol)
c= 343.24m/s
The sound moves on a circular with tangential velocity
vt = ω·r.................where
ω=2·π·n
vt= 2·π·n·r
vt= 2·π · 120min⁻¹ · 1m
vt= 753.6 m/min
convert m/min to m/sec
vt= 12.56 m/s
Part A
For maximum frequency is observed
v = vt
f = f₀/(1 - vt/c )
f= 600Hz / (1 - (12.56m/s / 343.24m/s) )
f= 622.789 Hz
Part B
For minimum frequency is observed
v = -vt
f = f₀/(1 + vt/c )
f= 600Hz / (1 + (12.56m/s / 343.24m/s) )
f= 578.82 Hz