Physics

Since $a=\frac{gsin?}{1+\frac{I}{M{R}^2}}$

$a=\frac{g*\frac{\sqrt[]{3}}{2}}{1+\frac{1}{2}}=\frac{g\frac{\sqrt[]{3}}{2}}{\frac{3}{2}}$

$a=\frac{g}{\sqrt[]{3}}$

f = 15

$m=-\frac{v}{u}=+\frac{1}{2}$            

  $v=-\frac{u}{2}$           

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$            

$\frac{2}{-u}+\frac{1}{u}=\frac{1}{f}$            

u = – f = – 15cm

PV = nRT

->Pµn

->Ratio= $\frac{32}{2}$ =16

$\left|\Delta E_0\right|=\left(-136\left\{1-\frac{1}{4}\right\}\right)eV$

|DE₀| = –10.2

$\lambda =\frac{12400}{10.2}*10^{-10}m$

$\rho =\frac{h}{\lambda }=\frac{6.63*10^{-34}*10.2}{12400*10^{-10}}$                  

$?$ $mv=\frac{h}{\lambda }$            

  $1.8*10^{-27}$           

$v=\frac{6.63*10.2*10^{-34}}{12400*10^{-10}}$

$v=\frac{6.63*10.2}{12400*1.8}*10^3$            

$=\frac{6.63*102}{124*1.8}=3.02$ ]

= 3 m/s

From Gauss law

$?=\frac{q_{enclosed}}{{\epsilon }_0}=\frac{5q}{{\epsilon }_0}$

KE =  $\frac{1}{2}$ m W² (A²– n²)

TE = $\frac{1}{2}$ mw²A²

^{$\frac{KE}{TE}=\frac{A^2-n^2}{A^2}=\frac{1-\frac{1}{9}}{1}=\frac{8}{9}$}

Since Gand S are in parallel

-> V_G = V_S

->3mA * 10 = 8A * R_S

->R_S = 3.75mW

i_battery= $\frac{10?3}{1000}$  = 7mA

i_2kW = $\frac{3}{2000}$  =1.5 mA

i_z = (7 – 1.5) mA

= 5.5mA

For maximum value of s, initially, electron must move away from plate.

ut + $\frac{1}{2}$ at² = s

t = 1u = 1m/s               s = –1 m

1 * 1 – $\frac{1}{2}$ a * 1² = – 1

-> a = 4m/s²

$\frac{qE}{m}=4$      

  $\frac{q\sigma }{2{\epsilon }_{0}m}=4$     

$\sigma =\frac{4*2*9*10^{-12}*9*10^{-31}}{1.6*10^{-19}}$

$=\frac{8*81}{1.6}*10^{-24}$      

= 4.05 * 10^–22C/m²

  $y=xtan\theta -\frac{g{x}^2}{2v^{2}co{s}^2\theta }$

(2.5, 2.5) must lie on this

-> $1=tan\theta -\frac{g*2.5}{2v^{2}co{s}^2\theta }$

-> $\frac{25}{2v^{2}co{s}^2\theta }=tan\theta -1$

-> $v^2=\frac{25}{2}\left\{\frac{1+ta{n}^2\theta }{tan\theta -1}\right\}$   

-> $v_{min}=5\sqrt[]{\sqrt[]{2}+1}$  

[Happens when tan θ = $\sqrt[]{2}$  + 1]