Answer:
I'm not a genius ok?
Explanation:
1. Radar communication, Analysis of the molecular and atomic structure, telephone communication
2. c
Answer:
v=s/t
s=vt
t=s/v
t=(120×10‐³)/172.8
(the distance meters has been changed to kilometres)
t=1/1440 hrs
Given ,
(a) ![x = \frac{m_2L}{m_1+m_2}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7Bm_2L%7D%7Bm_1%2Bm_2%7D)
<u>Explanation:</u>
Given:
Moment of Inertia of m₁ about the axis, I₁ = m₁x²
Moment of Inertia of m₂ about the axis. I₂ = m₂ (L - x)²
Kinetic energy is rotational.
Total kinetic energy is ![E = \frac{1}{2} I_1w_0^2 + \frac{1}{2}I_2w_0^2 = \frac{1}{2} w_0^2(m_1x^2 + m_2(L-x)^2)](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20I_1w_0%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7DI_2w_0%5E2%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20w_0%5E2%28m_1x%5E2%20%2B%20m_2%28L-x%29%5E2%29)
Work done is change in kinetic energy.
To minimize E, differentiate wrt x and equate to zero.
![m_1x - m_2(L-x) = 0\\\\x = \frac{m_2L}{m_1+m_2}](https://tex.z-dn.net/?f=m_1x%20-%20m_2%28L-x%29%20%3D%200%5C%5C%5C%5Cx%20%3D%20%5Cfrac%7Bm_2L%7D%7Bm_1%2Bm_2%7D)
Alternatively, work done is minimum when the axis passes through the center of mass.
Center of mass is at ![\frac{m_2L}{m_1 + m_2}](https://tex.z-dn.net/?f=%5Cfrac%7Bm_2L%7D%7Bm_1%20%2B%20m_2%7D)