Physics

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

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{4}+\frac{1}{4}$  

 R_eq = 2

$\frac{\Delta R_{eq}}{R_{eq}^2}=\frac{\Delta R_1}{R_1^2}+\frac{\Delta R_2}{R_2^2}$

$\begin{array}{l}\frac{\Delta R_{eq}}{4}=\frac{0.8}{16}+\frac{0.4}{16}+\frac{1.2}{16}\\ R_{eq}=0.3\end{array}$                

$Y=\frac{Mg{L}^3}{4b{d}^3\delta }$

For significant error in Y

$=\frac{\Delta M}{M}+\frac{3\Delta L}{L}+\frac{\Delta b}{b}+3\frac{\Delta d}{d}+\frac{\Delta \delta }{\delta }$              

$=\frac{1*10^{-3}}{2}+\frac{3*10^{-3}}{1}+\frac{10^{-2}}{4}+3*\frac{0.01*10^{-1}}{0.4}+\frac{10^{-2}}{5}$             

= 0.0155

$Y=\frac{Mg{L}^3}{4b{d}^3?}$              

For significant error in Y

$=\frac{?M}{M}+\frac{3?L}{L}+\frac{?b}{b}+3\frac{?d}{d}+\frac{??}{?}$           

$=\frac{1*10^{?3}}{2}+\frac{3*10^{?3}}{1}+\frac{10^{?2}}{4}+3*\frac{0.01*10^{?1}}{0.4}+\frac{10^{?2}}{5}$              

= 0.0155

Direction of E in the direction of y axes so flux is only due to top and bottom surface for bottom surface y = 0 E = 0

and for top surface y = 0.5 m      So

$E=150*{\left(0.5\right)}^2=\frac{150}{4}$

$fluxflowing?=EA=\frac{150}{4}*{\left(0.5\right)}^2$              

$=\frac{150}{16}$             

Gausses law $?=\frac{q}{{\epsilon }_0}$  

  $\frac{150}{16}=\frac{q}{{\epsilon }_0}$            

$=8.3*10^{-11}C$