Frequency of a sound wave is commonly referred to as pitch. That is the specialized name for frequency of a sound wave.
Just remember it as pitch.
Answer: Newton's third law
Explanation:
Another way of saying this is for every action, there is an equal but opposite reaction. This means that when you push on a wall, the wall pushes back on you with a force equal in strength to the force you exerted.
Answer:
A. something pushes or pulls it to stop.
Explanation:
Newtons first law. hope this helps
Setting up an integral of
rotation is used as a method of of calculating the volume of a 3D object formed
by a rotated area of a 2D space. Finding the volume is similar to finding the
area, but there is one additional component of rotating the area around a line
of symmetry.
<span>First the solid of revolution
should be defined. The general function
is y=f(x), on an interval [a,b].</span>
Then the curve is rotated
about a given axis to get the surface of the solid of revolution. That is the
integral of the function.
<span>It all depends of the
function f(x), which must be known in order to calculate the integral.</span>
To solve this exercise we will use the expression of the displacement in the aromatic movement in the two given points. Thus we will find the value of time as a function of frequency. In the end we will replace that value and find the time taken,
The expression for the displacement of the object in simple harmonic motion is
![x = Acos(2\pi ft)](https://tex.z-dn.net/?f=x%20%3D%20Acos%282%5Cpi%20ft%29)
At the initial position the value of x is zero,
![0 = Acost (2\pi ft_1)](https://tex.z-dn.net/?f=0%20%3D%20Acost%20%282%5Cpi%20ft_1%29)
![cost (2\pi ft_1) = 0](https://tex.z-dn.net/?f=cost%20%282%5Cpi%20ft_1%29%20%3D%200)
![2\pi ft_1 = \frac{\pi}{2}](https://tex.z-dn.net/?f=2%5Cpi%20ft_1%20%3D%20%5Cfrac%7B%5Cpi%7D%7B2%7D)
![t _1 = \frac{1}{4f}](https://tex.z-dn.net/?f=t%20_1%20%3D%20%5Cfrac%7B1%7D%7B4f%7D)
Now the expression for the displacement of the object in simple harmonic motion is
![x = Acos (2\pi ft)](https://tex.z-dn.net/?f=x%20%3D%20Acos%20%282%5Cpi%20ft%29)
At the final position the value of x is -1.8cm
Replacing we have that
![-1.8=1.8cos(2\pi ft_2)](https://tex.z-dn.net/?f=-1.8%3D1.8cos%282%5Cpi%20ft_2%29)
![-1 = cos(2\pi ft_2)](https://tex.z-dn.net/?f=-1%20%3D%20cos%282%5Cpi%20ft_2%29)
![t_2 = \frac{1}{2f}](https://tex.z-dn.net/?f=t_2%20%3D%20%5Cfrac%7B1%7D%7B2f%7D)
The total change in time will be
![t = t_2-t_1](https://tex.z-dn.net/?f=t%20%3D%20t_2-t_1)
![t = \frac{1}{2f}-\frac{1}{4f}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B1%7D%7B2f%7D-%5Cfrac%7B1%7D%7B4f%7D)
![t = \frac{1}{4f}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B1%7D%7B4f%7D)
Replacing the value of the frequency we have that
![t = \frac{1}{4*5}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B1%7D%7B4%2A5%7D)
![t = 0.05s](https://tex.z-dn.net/?f=t%20%3D%200.05s)
Therefore the time taken for the displacement in the interval of x=0 to x=-1.8 is 0.05s