Physics

$v_1=\sqrt[]{2g10}$

$v_2=\sqrt[]{2g5}$

$\overrightarrow{l}=\Delta \overrightarrow{p}$

$l=0.1\left\{\sqrt[]{2g10}+\sqrt[]{2g5}\right\}$

$=0.1\left\{10\sqrt[]{2}+10\right\}$

$=\left(\sqrt[]{2}+1\right)Ns$

$i_{battery}=\frac{\left(20-5\right)}{200}=\frac{15}{200}A$

$i_{300\Omega }=\frac{5}{300}A$

$i_{zener}=\frac{15}{200}-\frac{5}{300}$  

= 58.33 mA

By conservation of mechanical energy

mg (1) =  $\frac{1}{2}$ mv² + mg (0.5)

v² = 10

v =  $\sqrt[]{10}$ m/s

$i=\frac{7}{3.5k+0.9k\Omega }=\frac{7}{4.4k}$

$V_0=i*700\Omega =\frac{7}{4.4k}*7k=\frac{9.9}{4.4}=1.1V$

Kinetic energy: Potential energy = 1 : –2

Velocity at maximum height = vcoss30°

L = m (vcos30) H

$=mv\left(\frac{\sqrt[]{3}}{2}\right)*\frac{v^{2}si{n}^{2}30}{2g}$

$=\sqrt[]{3}\frac{m{v}^3}{16g}$

For 2 kg block

T – 2g sin37 = 2a        . (i)

For 4 kg block

4g – 2T =  $\frac{4a}{2}$

2g – T = a                    . (ii)

T = (2g – a)

2g – a – 2g *  $\frac{3}{5}$  = 2a

3a = 2g *  $\frac{2}{5}$

$a=\frac{4g}{15}$

$r=R_0{\left(192\right)}^{\frac{1}{3}}$

$\frac{r}{2}=R_0{\left(m\right)}^{\frac{1}{3}}$

$m=\frac{192}{8}=24$

-> $V_{esc}=\sqrt[]{\frac{2GM}{R}}$

$Ratio=\sqrt[]{\frac{81}{9}}=3$