Physics

Assume both the block is moving with common acceleration 'a' then

$a=\frac{F}{m+M}.......(i)$

N = mg

N = 2g

$f=ma=\frac{mF}{m+M}$

For no slipping

$\Rightarrow \frac{2F}{10}\le g$

$F\le 5g$

$\begin{array}{l}F\le 5*9.8\\ F\le 49.0N\end{array}$

Given, m = 0.6 $\pm$ $$ 0.006

$\Rightarrow p=\frac{m}{v}=\frac{m}{\pi r^2\mathcal{l}}$

$\frac{\Delta \rho }{\rho }*100=\frac{\Delta m}{m}*100+2\frac{\Delta r}{r}*100+\frac{\Delta \mathcal{l}}{\mathcal{l}}*100$

= 1 + 2 * 1 + 1

From the physics perspective, in uniform circular motion, speed and angular velocity are constant. That makes the angular acceleration and tangential acceleration are zero. Non-uniform motion, on the other hand, has varying speed and angular velocity, where angular acceleration is non-zero. 

Let $V_{R{G}_r}$ ${}_{{}_{}}$ (velocity of Rain w.r.t. ground or w.r.t to standing Girl)

${\overrightarrow{V}}_{R{G}_r}=V_x\widehat{i}+V_y\widehat{j}$

$tan\left(270+45°\right)=-cot45°=\frac{V_y}{V_x}=-1$

Now, velocity of Rain w.r.t. moving Girl

${\overrightarrow{V}}_{R{G}_m}=\left(V_x-15\sqrt[]{2}\right)\widehat{i}+V_y\widehat{j}=\left(15\sqrt[]{2}-15\sqrt[]{2}\right)\widehat{i}+V_y\widehat{j}=V_y\widehat{j}$

${\overrightarrow{V}}_{R{G}_m}=-15\sqrt[]{2}\widehat{j}\Rightarrow {\overrightarrow{V}}_{R{G}_m}=\frac{-30}{\sqrt[]{2}}\widehat{j}\Rightarrow \left|{\overrightarrow{V}}_{R{G}_m}\right|=\frac{30}{\sqrt[]{2}}$

We know that horizontal speed will remain constant

20 cos = v cos ……… (i)

Along y-axis

$U_y=20sin\alpha ,V_y=vsin\beta$

$V_y=V_y+a_{y}t$

$$ $Vsin\beta =20sin\alpha -10*10$  ……. (ii)

$\Rightarrow (ii)÷(i)$

$\Rightarrow tan\beta =tan\alpha -5sec\alpha$

In forward bias diode act as a short circuit wire. Hence, the equivalent circuit is now.

So,  ${}_{}\frac{}{}$ $V_{ab}=\frac{30}{5+10}*5=10\mathrm{V}$

$\mathrm{}50.00$

Net force on motor will be

$F_{\text{min}}=[920+68(10\left)\right]g+6000$

$=22000\mathrm{N}$

So, required power for motor

$\begin{array}{rr}{P}_{\text{min}}& ={\stackrel{?}{F}}_{\text{min}}\cdot \stackrel{?}{v}\\ & =22000*3\\ & =66000\text{watt}\end{array}$

Relaxation time $\left(\tau \right)\propto \frac{\mathrm{V}}{\sqrt[]{\mathrm{T}}}$

and $\mathrm{}\mathrm{T}\propto \frac{1}{{\mathrm{V}}^{\gamma -1}}$

$\begin{array}{rr}& \tau \propto {\mathrm{V}}^{1+\frac{\gamma -1}{2}}\mathrm{}\tau \propto {\mathrm{V}}^{\frac{1+\gamma }{2}}\\ & \frac{{\tau }_f}{{\tau }_{\mathrm{i}}}={\left(\frac{2\mathrm{V}}{\mathrm{V}}\right)}^{\frac{1+\gamma }{2}}\mathrm{}\frac{{\tau }_{\mathrm{f}}}{{\tau }_{\mathrm{i}}}=(2{)}^{\frac{1+\gamma }{2}}\end{array}$

Electric field at $\mathrm{t}=0$ & $(\mathrm{x},\mathrm{y},\mathrm{z})=\left(\mathrm{0,0},\frac{\pi }{\mathrm{k}}\right)$  is

$\begin{array}{rr}& \stackrel{?}{E}=-\left(\frac{\hat{i}+\hat{j}}{\sqrt[]{2}}\right)E_0\text{and}\stackrel{?}{F}=\stackrel{?}{E}q\\ & \therefore \mathrm{}\stackrel{?}{F}=-\left(\frac{\hat{i}+\hat{j}}{\sqrt[]{2}}\right)q\end{array}$

Which is antiparallel to $\left(\frac{\stackrel{\mathrm{ˆ}}{\mathrm{i}}+\stackrel{\mathrm{ˆ}}{\mathrm{j}}}{\sqrt[]{2}}\right)$

$\frac{\mathrm{T}}{\sqrt[]{2}}=100\mathrm{}\mathrm{}\frac{\mathrm{T}}{\sqrt[]{2}}=\mathrm{F}$

$F=100N$