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.
Answer:
option A
Explanation:
given,
height of the drop of stone = 9.44 m
speed of the stone = ?
As the stone is dropped the energy of the stone will be conserved.
using conservation of energy.
Potential energy = Kinetic energy

v = 13.60 m/s
Hence, the correct answer is option A
Answer:
q=3.5*10^-4
Explanation:
<u>concept:</u>
The force acting on both charges is given by the coulomb law:
F=kq1q2/r^2
the centripetal force is given by:
Fc=mv^2/r
The kinetic energy is given by:
KE=1/2mv^2
<u>The tension force:</u>
<u><em>when the plane is uncharged </em></u>
T=mv^2/r
T=2(K.E)/r
T=2(50 J)/r
T=100/r
<u><em>when the plane is charged </em></u>
T+k*|q|^2/r^2=2(K.E)charged/r
100/r+k*|q|^2/r^2=2(53.5 J)/r
q=√(2r[53.5 J-50 J]/k) √= square root on whole
q=√2(2)(53.5 J-50 J)/8.99*10^9
q=3.5*10^-4
Force can alter its direction,slow or stop it you could say it can change its velocity