Physics

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 *  $\frac{2}{5}$

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

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

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

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

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

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

Kindly go through the solution 

 $L=mucos\theta \frac{u^{2}si{n}^2\theta }{2g}$

$=m{u}^2\frac{1}{4\sqrt[]{2}g}\Rightarrow x=8$

Kindly go through the solution

R_eq = r

$P=\frac{V^2}{r}=\frac{4}{2}=2W$

  $k_{eq}=\frac{2k\cdot k}{3k}+k=\frac{5k}{3}$

Angular frequency of oscillation $\left(\omega \right)=\sqrt[]{\frac{k_{eq}}{m}}$  

$\omega =\sqrt[]{\frac{5k}{3m}}$            

Period of oscillation (t) = $\frac{2\pi }{\omega }=2\pi \sqrt[]{\frac{3m}{5k}}$  

$=\pi \sqrt[]{\frac{12m}{5k}}$                                                   

$T=2\pi \sqrt[]{\frac{I}{g}}$

$g=\frac{4{\pi }^{2}I}{T^2}$

$\frac{\Delta g}{g}=\frac{\Delta I}{I}+\frac{2\Delta T}{T}$

$=\frac{0.2}{20}+2\left(\frac{1}{40}\right)$

= 6%

$\frac{1}{2}{\epsilon }_0{E}^2=\frac{B^2}{2{\mu }_0}$

$?E=CBandC=\frac{1}{{\mu }_0{\epsilon }_0}$          

The magnitude of resultant vector (R)

$=\sqrt[]{a^2+b^2+2abcos\theta }$            

here a = b = A

then R = $\sqrt[]{A^2+A^2+2A^{2}cos\theta }$  

$=A\sqrt[]{2}\sqrt[]{1+cos\theta }$

$=\sqrt[]{2}A\sqrt[]{2co{s}^2\frac{\theta }{2}}$

$=2Acos\frac{\theta }{2}$