Class 11th

You can check the below given table for the comarative differences between valence bond theory and molecular orbital theory.

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)

$\Rightarrow (ii)÷(i)$

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

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.