"Maritime" air . . . . . always high relative humidity
"Polar" air . . . . . always cold
"Maritime polar" air . . . . . choice-A
Answer:
distance and length are the same quantity
As the source approaches you, the sound waves are compressed, so
the pitch of the sound is higher than what the source is actually emitting.
Then, after it passes you and begins moving away, the sound waves
are stretched, so the pitch of the sound is lower than what the source
is actually emitting.
The frequency of the human ear canal is 2.92 kHz.
Explanation:
As the ear canal is like a tube with open at one end, the wavelength of sound passing through this tube will propagate 4 times its length of the tube. So wavelength of the sound wave will be equal to four times the length of the tube. Then the frequency can be easily determined by finding the ratio of velocity of sound to wavelength. As the velocity of sound is given as 339 m/s, then the wavelength of the sound wave propagating through the ear canal is
Wavelength=4*Length of the ear canal
As length of the ear canal is given as 2.9 cm, it should be converted into meter as follows:
Then the frequency is determined as
f=c/λ=339/0.116=2922 Hz=2.92 kHz.
So, the frequency of the human ear canal is 2.92 kHz.
Your answer is C)
a)t=2.78 sec
b)R=835.03 m
c)
Explanation:
Given that
h= 38 m
u=300 m/s
here given that
The finally y=0
So
t=2.78 sec
The horizontal distance,R
R= u x t
R=300 x 2.78
R=835.03 m
The vertical component of velocity before the strike