Physics

$a_{cp}=\frac{v^2}{r}=\frac{{\left(\sqrt[]{3gl}\right)}^2}{l}=\frac{3gl}{l}=3g$

Kinetic energy $=\frac{1}{2}I{\omega }^2$ , and for ring   $I=m{r}^2$

Hence $KE=\frac{1}{2}m{r}^2{\omega }^2$

From ampere's circuital law

$?\stackrel{?}{B}.\stackrel{?}{dl}={\mu }_0{i}_{enc}$

$={\mu }_0\left(3+1\right)=4{\mu }_0$

The relation between $\alpha$ and $\beta$  is given as

$\beta =\frac{\alpha }{1-\alpha }$

${\beta }_1=\frac{\frac{20}{21}}{1-\frac{20}{21}}=20$

${\beta }_2=\frac{\frac{100}{101}}{1-\frac{100}{101}}=100$

$M=0.5kg$

$V=1.5m/s$

$K=50N/m$

Maximum Compression of the spring

$\frac{1}{2}m{v}^2=\frac{1}{2}k{x}^2$

$k{x}^2=m{v}^2$
$x=v\sqrt[]{\frac{m}{k}}=1.5\sqrt[]{\frac{0.5}{50}}.$

$x=0.15m$

In a number without decimal, trailing zeros are insignificant. If a number has a decimal, then preceding zeros are insignificant.

Hence, in 40.40 all zeros are significant

$kx=mg-F_{up}$

$kx=mg-\frac{md}{D}g$

$x=\frac{mg}{k}\left(1-\frac{d}{D}\right)$

Rate of burning of fuel $\left(\frac{dm}{dt}\right)=1kg/s$  and velocity of ejected fuel

$\left(v\right)=60km/s$

$=60*10^{3}m/s.$  

Forcev =  Rate of change of momentum

$=\frac{dp}{dt}=\frac{d\left(m\nu \right)}{dt}=v\frac{dm}{dt}=\left(60*10^3\right)*1$

  $=60000$ N.

$\stackrel{?}{F}=e{E}_0\stackrel{?}{j}$

${\stackrel{?}{a}}_y=\frac{e{E}_0}{m}\stackrel{?}{j}$

$=v_0\stackrel{?}{i}+\left(\frac{e{E}_{0}t}{m}\right)\stackrel{?}{j}$

$\left|\stackrel{?}{v}\right|=\sqrt[]{v_0^2+{\left(\frac{e{E}_{0}t}{m}\right)}^2}=v_0\sqrt[]{1+{\left(\frac{e{E}_{0}t}{m}\right)}^2}$

$\lambda =\frac{{\lambda }_0}{\sqrt[]{1+{\left(\frac{e{E}_{0}t}{m}\right)}^2}}$

Force $\left(F\right)=10N$  and acceleration