Physics

In Class 11 Physics, we typically talk about two main categories of forces. One is contact forces, and the second is non-contact forces. Contact forces are friction, tension, normal force, and spring force. On the other hand, non-contact forces consist of gravitational, electrostatic, and magnetic forces. These types of forces help us in understanding Newton's laws of motion, free-body diagrams (FBDs), and the concept of equilibrium.

Using snell's law –                        

sin I = $?sin\text{?}r$  

->sin45° =  $?sin30°$  

$?=\frac{sin45}{sin30}=\frac{1}{\sqrt[]{2}}*2=\sqrt[]{2}$  

 B_v = 6 * 10^-5 T                                              

B_v = B_N sin37°

$\therefore B_N=\frac{B_v}{sin37°}=\frac{6*10^{-5}}{3/5}=10^{-4}T$

W_total = W_D®E + W_E->F = Area of triangle DEF

= $\frac{1}{2}*3*300=450J.$

$\overrightarrow{F}=5\widehat{i}+3\widehat{j}-7\widehat{k}$

$\overrightarrow{r}=2\widehat{i}+2\widehat{j}+\widehat{k}$

$\overrightarrow{\tau }=\left|\overrightarrow{r}*\overrightarrow{F}\right|=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 2\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill 5\hfill & \hfill 3\hfill & \hfill -7\hfill \end{array}\right|\begin{array}{cc}\hfill =\widehat{i}\left(-k_1-3\right)-\widehat{j}\left(-14-5\right)+\widehat{k}\left(6-10\right)\hfill & \hfill =-17\widehat{i}+19\widehat{j}-4\widehat{k}\hfill \end{array}$  

$\frac{\Delta K}{K_i}*100=\frac{K_f-K_i}{K_i}*100$                

$=\frac{\frac{P_f^2}{2m}-\frac{P_i^2}{2m}}{\frac{P_i^2}{2m}}*100=\frac{P_f^2-P_i^2}{P_i^2}*100$               

$=\frac{{\left(1.2P_i\right)}^2-{\left(P_i\right)}^2}{P_i^2}*100=\frac{1.44P_i^2-P_i^2}{P_i^2}*100$               

 = 44%

m₁ = 5kg                                                                      

m₂ = 3kg

As m₁ is in rest -

T = m_gg = 30

N = m₁ g cosq

T =    $m_{1}gsin\theta \Rightarrow 50sin\theta =30$

->sinθ  $=\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\Rightarrow \theta =37°$  

$\therefore N=50cos\theta =50cos37°=50*\frac{4}{5}=40N$

m = 1kg ;  $\Delta U=\frac{-G{m}_{e}m}{\left(R+h\right)}-\left(\frac{G{m}_{e}m}{R}\right)$  

$\Delta U=\frac{GMm}{R}-\frac{GMm}{R+h}$

$=m{g}_{0}R-m{g}_0\frac{R}{\left(1+\frac{h}{R}\right)}=m{g}_0\left\{1-\frac{1}{\left(1+\frac{3R}{R}\right)}\right\}$

$=m{g}_{0}R\left(\frac{3}{4}\right)$

$=\frac{3}{4}*1*10*6400km$

$=48000*10^{3}J.$

    $=48MJ.$