Class 11th

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)

$\Rightarrow (ii)÷(i)$

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

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

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

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

$F=100N$

The period T is the time taken to complete one full revolution (2? radians). In uniform circular motion, it relates to angular velocity by? =2? /T and to tangential speed by v=? r=2? r/T .

Using equation of continuity

$\begin{array}{rr}& A_1{V}_1=A_2{V}_2\\ & \frac{V_1}{V_2}=\frac{A_2}{A_1}={\left(\frac{d_2}{d_1}\right)}^2={\left(\frac{4.8}{6.4}\right)}^2=\frac{9}{16}\end{array}$

Photochemical smog generally does not content SO₂ gas.

$B{r}_2+5F_2\left(excess\right)\to 2Br{F}_5$

$Br{F}_5+3H_{2}O\to HBr{O}_3+5HF$

$2B{F}_3+6NaH\to B_2{H}_6+6NaF$

$B_2{H}_6+2\ddot{N}{\left(Me\right)}_3\to 2H_{3}B\leftarrow :N{\left(Me\right)}_3$