Physics

Potential difference across 2k $\Omega$  is 5V, thus current through it,

  $i=\frac{5}{2*10^3}=25*10^{-4}A.$          

Yes. Mechanical energy is the sum of both kinetic and potential energies. Kinetic energy cannot be a negative value since mass is always positive and velocity square leads to a positive value. However, potential energy can be negative depending on the reference point.

Not in all cases. This happens due to the interruption of non conservative forces like friction, air resistance, etc. which are responsible for decreasing the mechanical energy by converting it into other forms of energy such as heat, noise, etc.

Conservation of Mechanical Energy is linked to newton's second law of motion (F = ma) in combination with work energy theorem. The work energy theorem states that total work done on an object equals to change in kinetic energy. This theorem is used to prove that the sum of kinetic energy and potential energy remains constant.

$\frac{GM}{{\left(3R/2\right)}^2}=\frac{GM}{R^3}*r$

$\Rightarrow OA=\frac{4R}{9}=r$

$AB=R-\frac{4R}{9}=\frac{5R}{9}\Rightarrow OA:AB=4:5=x:y\Rightarrow x=4$

With the help of conservation of volume, we can write

$27*\frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3\Rightarrow R=3r.......\left(1\right)$

With the help of conservation of charge, we can write

Q = 27 q.(2)

Potential energy of single drop = U1 = $\frac{q^2}{8\pi {\epsilon }_{0}r}$ $\frac{{}^{}}{{}_{}}$

Potential energy of bigger drop = $U_2=\frac{Q^2}{8\pi {\epsilon }_{0}R}=\frac{27*27*q^2}{8\pi {\epsilon }_0\left(3r\right)}=243\left(\frac{q^2}{8\pi {\epsilon }_{0}r}\right)=243U_1$ ${}_{}\frac{{}^{}}{{}_{}}\frac{{}^{}}{{}_{}}\frac{{}^{}}{{}_{}}{}_{}$

$\Rightarrow \frac{U_2}{U_1}=243$

$l=\frac{90-30}{4000}=15mA$

$I_1=\frac{30}{5000}=6mA\&l_2=9mA$

$v=\omega \sqrt[]{A^2-s^2}\Rightarrow \frac{A\omega }{2}=\omega \sqrt[]{A^2-s^2}\Rightarrow s=\frac{\sqrt[]{3}A}{2}\Rightarrow x=3$

$\Delta Q=\Delta U+\Delta W$

^{$\Rightarrow Q=\Delta U+\frac{Q}{5}\Rightarrow \Delta U=\frac{4Q}{5}=n{C}_v\Delta T\Rightarrow \frac{4Q}{5}=\frac{5R}{2}\Delta T\Rightarrow \Delta T=\frac{8Q}{25R}$}

^{$Q=nc\Delta T=1*C*\frac{80}{25R}\Rightarrow C=\frac{25R}{8}\Rightarrow x=25$}

Band Width = 2 * n * the highest modulation frequency

$\Rightarrow n=\frac{90k{H}_z}{2*5kHz}=9$