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:Esotericism
Explanation:
it’s something that’s in intentional out of body experience
The last one.
One grey sphere, Four white spheres, and One red spheres
Answer:
Dr = 263 10⁻⁶ m
Explanation:
The diffraction pattern for constructive interference is described by
a sin θ = m λ
in this it indicates that the order of diffraction is m = 1
Let's use a direct proportion rule to find the separation of two slits. If there are 600 lines in 1 me, what is the distance between 2 slits
a = 2 lines 1/600
a = 2/600
a = 3.33 10⁻³ mm = 3.33 10⁻⁴ cm
let's use trigonometry
tan θ = y / L
as the measured angles are small
tan θ = sin θ / cos θ sin θ
sin θ = y / L
we substitute
a y/L = λ
y = λ L / a
for λ = 400 10-9 m
I = 400 10⁻⁹ 2.9 / 3.33 10⁻³
i = 346.89 10⁻⁶ m
f
or λ = 700 nm
y_f = 700 10⁻⁻⁹ 2.9 / 3.33 10⁻³
y_f = 609.609 10⁻⁶ m
the separation of this spectrum
Δr = v_f - i
Dr = (609.609 - 346) 10 ⁻⁶
Dr = 263 10⁻⁶ m
