Answer: f = 927.55Hz
Explanation: Since the the tube is open-closed, the length of air and the wavelength of sound passing through the tube is given below
L = λ/4 where λ = wavelength.
speed of sound in air = v = 343m/s.
fundamental frequency of open closed tube = 315Hz
λ = 4L.
v = fλ
343 = 315 * 4L
343 = 1260 * L
L = 343/ 1260
L = 0.27m
In the same tube of length L = 0.27m but different medium ( helium), the speed of sound is 1010m/s.
The length of tube and wavelength are related by the formulae below
L = λ/4, λ=4L
λ = 4 * 0.27
λ = 1.087m.
v = fλ
1010 = f * 1.087
f = 1010/1.807
f = 927.55Hz
Benthos
Option b is the answer
Answer: An organ pipe is open at both ends. It is producing sound at its third harmonic, the frequency of which is 262 Hz. The speed of sound is 343 m/s. What is the length of the pipe?
Explanation: thanks for asking
Answer:
33 Celsius is 306.15 in absolute temperature
Ok, so you've got to figure out a force F and you have the speed in which the boxer punches on determinate time and the mass of the sheet of paper.
So based on the formula that says that the Force is equal to the mass multiplied by the acceleration => F=ma.
You look at it and see that you only have mass which is measured on KG so there is no problem.
then you have the acceleration which is measured on meters and is defined by: a = Δv/Δt
So now you can replace the velocity and the time you have there
⇒ a 25m/s / 0.05s
you have computing that ⇒ 50m because the seconds were cancelled out.
and then you plug the meters into the force equation.
F=(0.005kg)(50)
F=0.25N
so the boxer will have a force of 0.25 Newton's.
