Physics

Let, Rain drop moving with terminal velocity vt in the air Fb (Bnoyancy force) = $\frac{4}{3}\pi r^3{\rho }_{air}g$ $\frac{}{}{}^{}{}_{}$

$mg=\frac{4}{3}\pi r^3{\rho }_{air}g$

$F_v=\left(visconsforce\right)=6\pi \eta r{v}_T$

$F_b+F_v=mg$

$\Rightarrow \frac{4}{3}\pi r^3{\rho }_{air}g+6\pi \eta r{v}_T=\frac{4}{3}\pi r^3\rho g$

$v_T=\frac{2}{9}\frac{r^2}{\eta }\left(\rho -{\rho }_{air}\right)$

$v_T\propto r^2$

  $P=\overrightarrow{F}.\overrightarrow{v}=v\frac{dP}{dt}=v^2\left(\frac{dm}{dt}\right)=0.5*25=12.5W$

V = u + at -> v = -100 - 10 * 10 = -200 m/s

$g^1=\frac{GM}{{\left(R+h\right)}^2}$

$g^1=\frac{g}{{\left(1+\frac{h}{R}\right)}^2}$

Given h = D = 2R

$g^1=\frac{g}{{\left(1+\frac{2R}{R}\right)}^2}$

$g^1=\frac{g}{9}$

^{$\overrightarrow{r}=2\widehat{i}+\widehat{j}+2\widehat{k}$}

^{$\overrightarrow{\tau }=\overrightarrow{r}*\overrightarrow{F}$}

^{$=\left(2\widehat{i}+\widehat{j}+2\widehat{k}\right)*\left(3\widehat{i}+4\widehat{j}-2\widehat{k}\right)=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill -\widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill 2\hfill \\ \hfill 3\hfill & \hfill 4\hfill & \hfill -2\hfill \end{array}\right|$}

^{$\overrightarrow{\tau }=-10\widehat{i}+10\widehat{j}+5\widehat{k}$}

  $\left|\widehat{A}+\widehat{B}\right|=\sqrt[]{A^2+B^2+2ABcos\theta }$ $\stackrel{}{}\stackrel{}{}\text{exception 25:}$

$=\sqrt[]{1+1+2cos\theta }$

$=\sqrt[]{2\left(1+cos\theta \right)}$

= $\sqrt[]{2*2co{s}^2\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ $\text{exception 25:}$

$$ $\left|\widehat{A}+\widehat{B}\right|=2cos\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$  -(1)

$\left|\widehat{A}-\widehat{B}\right|=\sqrt[]{A^2+B^2-2ABcos\theta }$

$=\sqrt[]{1+1-2cos\theta }$

$\raisebox{1ex}{$$}\!\left/ \!\raisebox{-1ex}{$$}\right.$ $=2sin\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$  -(2)

(2) $÷$ $$ (1)

$\frac{\left|\widehat{A}-\widehat{B}\right|}{\left|\widehat{A}+\widehat{B}\right|}=\frac{2sin\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{2cos\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

$\Rightarrow \left|\widehat{A}-\widehat{B}\right|=\left|\widehat{A}+\widehat{B}\right|tan\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

$\overrightarrow{A}*\overrightarrow{A}=AAsin\theta \widehat{n}$

$=AAsin0°\widehat{n}$

= 0 [since Angle between the vectors are zero degree]

$\overrightarrow{A}*\overrightarrow{A}=0$

$z=\frac{A^2{B}^3}{C^4}$

The relative error in z is given by

$\frac{\Delta z}{z}=\frac{2\Delta A}{A}+\frac{3\Delta B}{B}+\frac{4\Delta C}{C}$

Voltage gain = $\frac{I_C{R}_0}{I_B{R}_i}$  

R₀ ->Output Resistance

R_i ->Input Resistance

I_C ->Collector current

I_{B ->}Base current

Voltage gain = $\frac{I_C}{I_B}\frac{R_0}{R_i}=\frac{\left(5*10^{-3}\right)}{\left(100*10^{-6}\right)}*\frac{60}{200}=15$  

In 1D kinematics, you use scalar equations for one direction. In 2D, position, velocity, and acceleration become vectors with x and y components. You apply the same kinematic equations independently to each dimension. Just remember to treat horizontal and vertical motions as separate 1D problems to be solved simultaneously.