Answer:
Time period between the successive beats will be 0.1703 sec
Explanation:
We have given speed of the sound v = 349 m/sec
Wavelength of piano 
Wavelength of piano 
So frequency of piano A 
Frequency of piano B 
So beat frequency f = 455.61 - 449.74 = 5.87 Hz
So time period 
So time period between the successive beats will be 0.1703 sec
You need to post a picture so someone can help
Answer:
0.34 m
Explanation:
From the question,
v = λf................ Equation 1
Where v = speed of sound, f = frequency, λ = Wave length
Make λ the subject of the equation
λ = v/f............... Equation 2
Given: v = 340 m/s, f = 500 Hz.
Substitute these values into equation 2
λ = 340/500
λ = 0.68 m
But, the distance between a point of rarefaction and the next compression point, in the resulting sound is half wave length
Therefore,
λ/2 = 0.68/2
λ/2 = 0.34 m
Hence, the distance between a point of rarefaction and the next compression point, in the resulting sound is 0.34 m
The pressure at 100 meters below the surface of sea water with a density of 1150kg is 145.96 psi.
We know that arc length (x(t)) is given with the following formula:
Where r is the radius of the barrel. We must keep in mind that as barrel rolls its radius decreases because less and less tape is left on it.
If we say that the thickness of the tape is D then with every full circle our radius shrinks by d. We can write this down mathematically:
When we plug this back into the first equation we get:
We must solve this quadratic equation.
The final solution is:
It is rather complicated solution. If we asume that the tape has no thickness we get simply:
