Answer:
Explanation:
1.
vf = vi + a(t)
vi = 0 m/s
a = 17/13
= 1.3 m/s^2
2.
F = M * a
= 15 * 1.3
= 19.62 N
3.
Normal force, Fn = m * g
= 15 * 9.8 m/s²
= 147 N.
Then find the maximum force of friction, knowing that μs = 0.76
Ff = Fn × μs
= 147 * 0.76
= 111.83 N
Maximum acceleration,
Ff = m × a
111.83 = 15 * a
a = 7.4556 m/s²
4.
In order to find the acceleration for the box, you need to know the net force of the box moving in the x direction and the frictional force, and you will end up with the Force of the vehicle minus the Frictional force (Ff) between the box and the vehicle, resulting your net force in the x direction.
F = m*a and Ff = μ * N
m*a = μk * N
m*a = μk * m * g
a = μk * g
a = 0.61 * 9.81
5.98 m/s^2 for the acceleration of the box on top of the vehicle.
5.
Assuming the box has re settled and is no longer sliding when braking begins and the surface remains horizontal, the maximum negative acceleration will again be
a = -7.4556 m/s².
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C)Wavelength
It has a low frequency electromagnetic radiation that's used to transmit information.
Answer:
Lowest resonant frequency: ![F_0=114Hz\\](https://tex.z-dn.net/?f=F_0%3D114Hz%5C%5C)
Wave Speed: ![v=198.36 \; m/s](https://tex.z-dn.net/?f=v%3D198.36%20%5C%3B%20m%2Fs)
Explanation:
The lowest resonant frequency
For a fixed string a resonant mode is always a integer multiple of fundamental frequency (the lowest resonant frequency):
![F_n =nF_0](https://tex.z-dn.net/?f=F_n%20%3DnF_0)
Is the frequency for a resonant mode
is fundamental frequency (the lowest resonant frequency).
Due to problem says that there's not intermediate resonant frequencies between 684 Hz and 798 Hz, i.e, they are consecutive resonant modes, equations are as follows:
![684=nF_0 \; (1)\\798=(n+1)F_0 \; (2)](https://tex.z-dn.net/?f=684%3DnF_0%20%5C%3B%20%281%29%5C%5C798%3D%28n%2B1%29F_0%20%5C%3B%20%282%29)
Solving for n from equation (1) and replacing it in the equation (2)
![790=(\frac{684}{F_0} +1)F_0](https://tex.z-dn.net/?f=790%3D%28%5Cfrac%7B684%7D%7BF_0%7D%20%2B1%29F_0)
And finally solving for ![F_0](https://tex.z-dn.net/?f=F_0)
![789=684+F_0 =>F_0=114 \;Hz](https://tex.z-dn.net/?f=789%3D684%2BF_0%20%3D%3EF_0%3D114%20%5C%3BHz)
Wave speed
For a fixed string the wave speed can be calculated by using
![F_0=\frac{V}{2L}](https://tex.z-dn.net/?f=F_0%3D%5Cfrac%7BV%7D%7B2L%7D)
Where V is the wave speed and L is the string length.
So, using this equation and solving for V:
![V=2*L*F_0 => V=2*(0.871 \; m)*114 \; Hz\\V=198.36 \;m/s](https://tex.z-dn.net/?f=V%3D2%2AL%2AF_0%20%3D%3E%20V%3D2%2A%280.871%20%5C%3B%20m%29%2A114%20%5C%3B%20Hz%5C%5CV%3D198.36%20%5C%3Bm%2Fs)
<em>It Is very important convert the string length unit to meters</em>