Physics

$l=\frac{m{\mathcal{l}}^2}{3}$  

Energy conservation Low

$mg\mathcal{l}=\frac{1}{2}\frac{m{\mathcal{l}}^2}{3}{\omega }^2.....\left(i\right)$                

And speed V =    $\omega r=\omega \mathcal{l}$

then    $V=\sqrt[]{6g\mathcal{l}}=\sqrt[]{6*10*.6}$

->6 m/s

for smooth surface

$a=gsin30°=\frac{g}{2}$

$S_1=ut+\frac{1}{2}a{t}^2$         

$S_1=\frac{1}{2}\frac{g}{2}{t}^2=\frac{g}{4}{t}^2.......\left(i\right)$               

for rough Surface

$S=\frac{1}{2}\frac{g}{2}\left(1-\mu \sqrt[]{3}\right){\alpha }^2{t}^2........\left(ii\right)$               

By (i) and (ii)

$\mu =\frac{1}{\sqrt[]{3}}\left(\frac{{\alpha }^2-1}{{\alpha }^2}\right)x=\sqrt[]{3}$            

For A satellite T₁ = 1hour

$So{\omega }_1=2\pi red/hour$                

for B satellite T₂ = 8 hour

given R₁ = 2 * 10³ Km

Relative  $\omega =\frac{V_1-V_2}{R_2-R_1}=\frac{2\pi *10^3}{6*10^3}$  

$\Rightarrow \frac{\pi }{3}$ rad/hour

 x = 3

K.E. energy of electron = eV

Translational K.E. of N₂ =   $\frac{3}{2}KT$

So eV = $\frac{3}{2}KT$  

$1.6*10^{-19}*0.1=\frac{3}{2}*1.38*10^{-23}*T$                

T = 773 – 273 = 500°C

Since process is isochoric

So    $\Delta U=n{C}_v\Delta T$

$\Delta U=n\left(\frac{5}{2}R\right)\Delta T--(i)\left[C_V=\frac{5}{2}R\right]$      

And external work

$\Delta W=nR\Delta T--\left(ii\right)$    

$\frac{5}{2}=\frac{x}{10}\Rightarrow x=25.00$                

In first condition R₁ = 36 $\Omega$  

In second condition R₂ = 18  $\Omega$

$P_1=\frac{V^2}{R_1}=\frac{{\left(240\right)}^2}{36}$               

$P_2=\frac{V^2}{R_2}+\frac{V^2}{R_2}=\frac{{\left(240\right)}^2}{18}+\frac{{\left(240\right)}^2}{18}$               

$P_2=\frac{{\left(240\right)}^2}{9}$               

So   $\frac{P_1}{P_2}=\frac{{\left(240\right)}^2/36}{{\left(240\right)}^2/9}=\frac{1}{4}$

x = 4

Force  = Aya   $\Delta T$

Force $=\left(10*10^{-4}\right)*\left(2*10^{11}\right)*10^{-5}*400$  

$F=8*10^{5}N$              

  $x*10^5=8*10^5$              

x = 8

Amp. a slit with

Intensity a (Amp)² a (slit width)²

  $\frac{l_1}{l_2}={\left(\frac{3}{1}\right)}^2=\frac{9}{1}\Rightarrow l_1=9I_2$

$\frac{l_{min}}{l_{max}}=\frac{{\left(\sqrt[]{l_1}-\sqrt[]{l_2}\right)}^2}{{\left(\sqrt[]{l_1}+\sqrt[]{l_2}\right)}^2}=\frac{{\left(3-1\right)}^2}{{\left(3+1\right)}^2}$            

  $\Rightarrow \frac{1}{4}=\frac{x}{4}$

x = 1.00

Formula for Power = F·v cos $$$\theta$

According to the question, force is perpendicular.

So $\theta$ = 90

and cos 90 = 0

Hence, the result turns out to be 0 i.e. no power generated inspite of force being applied.

In simple words,

Energy = total work done on an object.

Power = how fast the work is done on an object.

For example: Writing an article vs how fast you write the article. This sums up the exact meaning of both these terms.