Answer:
a) ![f_1=5.587Hz](https://tex.z-dn.net/?f=f_1%3D5.587Hz)
b) ![f_{n+1}-f_n=5.587Hz](https://tex.z-dn.net/?f=f_%7Bn%2B1%7D-f_n%3D5.587Hz)
Explanation:
The frequency of the
harmonic of a vibrating string of length <em>L, </em>linear density
under a tension <em>T</em> is given by the formula:
![f_n=\frac{n}{2L} \sqrt{\frac{T}{\mu}](https://tex.z-dn.net/?f=f_n%3D%5Cfrac%7Bn%7D%7B2L%7D%20%5Csqrt%7B%5Cfrac%7BT%7D%7B%5Cmu%7D)
a) So for the <em>fundamental mode</em> (n=1) we have, substituting our values:
![f_1=\frac{1}{2(347m)} \sqrt{\frac{65.4\times10^6N}{4.35kg/m}}=5.587Hz](https://tex.z-dn.net/?f=f_1%3D%5Cfrac%7B1%7D%7B2%28347m%29%7D%20%5Csqrt%7B%5Cfrac%7B65.4%5Ctimes10%5E6N%7D%7B4.35kg%2Fm%7D%7D%3D5.587Hz)
b) The <em>frequency difference</em> between successive modes is the fundamental frequency, since:
![f_{n+1}-f_n=\frac{n+1}{2L} \sqrt{\frac{T}{\mu}}-\frac{n}{2L} \sqrt{\frac{T}{\mu}}=(n+1-n)\frac{1}{2L} \sqrt{\frac{T}{\mu}}=\frac{n}{2L} \sqrt{\frac{T}{\mu}}=f_1=5.587Hz](https://tex.z-dn.net/?f=f_%7Bn%2B1%7D-f_n%3D%5Cfrac%7Bn%2B1%7D%7B2L%7D%20%5Csqrt%7B%5Cfrac%7BT%7D%7B%5Cmu%7D%7D-%5Cfrac%7Bn%7D%7B2L%7D%20%5Csqrt%7B%5Cfrac%7BT%7D%7B%5Cmu%7D%7D%3D%28n%2B1-n%29%5Cfrac%7B1%7D%7B2L%7D%20%5Csqrt%7B%5Cfrac%7BT%7D%7B%5Cmu%7D%7D%3D%5Cfrac%7Bn%7D%7B2L%7D%20%5Csqrt%7B%5Cfrac%7BT%7D%7B%5Cmu%7D%7D%3Df_1%3D5.587Hz)
Answer:
I_syst = 278.41477 kg.m²
Explanation:
Mass of platform; m1 = 117 kg
Radius; r = 1.61 m
Moment of inertia here is;
I1 = m1•r²/2
I1 = 117 × 1.61²/2
I1 = 151.63785 kg.m²
Mass of person; m2 = 62.5 kg
Distance of person from centre; r = 1.05 m
Moment of inertia here is;
I2 = m2•r²
I2 = 62.5 × 1.05²
I2 = 68.90625 kg.m²
Mass of dog; m3 = 28.3 kg
Distance of Dog from centre; r = 1.43 m
I3 = 28.3 × 1.43²
I3 = 57.87067 kg.m²
Thus,moment of inertia of the system;
I_syst = I1 + I2 + I3
I_syst = 151.63785 + 68.90625 + 57.87067
I_syst = 278.41477 kg.m²
The sun is a clear example of objects releasing radiation in nature
Answer:
The box will be moving at 0.45m/s. The solution to this problem requires the knowledge and application of newtons second law of motion and the knowledge of linear motion. The vertical component of the force Fp acts vertically upwards against the directio of motion. This causes a constant upward force of 23sin45° to act on the box. Fhe frictional force of 13N also acts vertically upwards and so two forces act upwards against rhe force of gravity resulting un a net force of 0.7N acting kn the box. This corresponds to an acceleration of 0.225m/s². So in w.0s after i start to push v = 0.45m/s.
Explanation:
Forces are exerted I believe : all of the above
The action force might be Tyler throwing the ball
I don't know the last one