Physics

Both torque and angular momentum have direction. We determine their direction based on the right-hand rule. That makes them vector quantities in rotational motion. 

Yes, an object can have angular momentum without torque. The physics behind it is that if no external force act on an object, its angular momentum is constant. That is based on the law of conservation of momentum. 

Torque is the measure of rotational force on an object (rigid or extended body). It changes the object's angular momentum. To calculate in physics, torque equals the rate of change of angular momentum. The formula for that is

$?=\frac{dL}{dt}$

Structure of Bithinol is;

C = 500   $\mu F,$ V = 100 v, L = 50 mH                                                        

In this LC – oscillation

q = q₀ cos   $\omega t$

$i=\frac{-dq}{dt}=q_0\omega sin\omega t\omega =\frac{1}{\sqrt[]{2c}}=\frac{1}{\sqrt[]{50*10^{-3}*5*10^{-4}}}$

=  $\frac{1000}{5}=200$  

So,     i_max =   $q_0\omega =500*10^6*100*200$

= 10A

  ${\lambda }_A=25\lambda ,$ ${\lambda }_B=16\lambda$      

At t = 0 Þ N_A = N_B = N₀

after t =  $\frac{1}{a\lambda }:-\frac{N_B}{N_A}=\frac{N_0{e}^{-16\lambda t}}{N_0{e}^{-25\lambda t}}=e^{\left(25\lambda -16\lambda \right)t}$  

->e =  $e^{\left(9\lambda \frac{1}{a\lambda }\right)}$  

$\Rightarrow \frac{9}{a}=1\Rightarrow a=9$      

$i_{\left(zener-max\right)}=25mA$                                                                

 20 – i_max R – 8 = 0

i_max R = 12At minimum zener current $\left(\mu A\right):-$  

$20-i_{min}R-i_{min}{R}_L=0$  

  $\frac{R}{R_L}=\frac{12}{8}=\frac{3}{2}$

${\mathcal{l}}_{min}R=12$

  $i_{min}{R}_L=8$

At max^m zener current –

$20-i_{max}R-8=0$  

$i_L=O\left\{as{i}_{z}ma{x}^m=25mA\right\}$             

i_maxR = 12v

25 * 10^-3 R = 12

$R=\frac{12*10^3}{25}=12*40=480\Omega$

From conservation of Energy

$\frac{1}{2}m{v}_0^2=mgh$

$v_0=10\sqrt[]{2}$

For A to B

$a=-gsin45°=\frac{-10}{\sqrt[]{2}}$

T = t₁ + t₂

$\begin{array}{l}=2\sqrt[]{2}+2\\ =2\left(\sqrt[]{2}+1\right)\end{array}$

g_eff = g –   $\left(\frac{{\rho }_w}{{\rho }_b}\right)g$

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

$T\text{'}=2\pi \sqrt[]{\frac{\mathcal{l}}{g_{eff}}}$

$\Rightarrow T\text{'}=\sqrt[]{\frac{5}{4}}T$

=    $\sqrt[]{\frac{5}{4}}*10$

=    $5\sqrt[]{5}sec$

So, x = 5

.From conservation of energy

mgh = $\frac{1}{2}m{v}^2+\frac{1}{2}l{\omega }^2$ $\frac{}{}{}^{}\frac{}{}{}^{}$

$\Rightarrow mgh=\frac{1}{2}m{v}^2+\frac{1}{2}\frac{m{R}^2}{2}$

10 h = $\frac{16}{2}+\frac{16}{4}$ $\frac{}{}\frac{}{}$

h = 1.2 m = 120 cm