Physics

From Guass's Law

${\displaystyle ?\overrightarrow{E}.\overrightarrow{dA}=\frac{q_{in}}{{\in }_0}}$            

$E.4\pi r^2=\frac{{\displaystyle \underset{0}{\overset{r}{\int }}\left(4\pi r^2.dr\right)\left(\alpha r\right)}}{{\in }_0}$            

$E=\left(\frac{\alpha r^2}{4{\in }_0}\right)$            

$y\left(x,t_0\right)=A\mathrm{s}\mathrm{i}\mathrm{n}?\left(\frac{x}{b}\right)$

$y(x,0)=A\mathrm{s}\mathrm{i}\mathrm{n}?\left(\frac{x+v{t}_0}{b}\right)$

$y(x,t)=A\mathrm{s}\mathrm{i}\mathrm{n}?\left(\frac{x+v{t}_0-vt}{b}\right)$

$\therefore y(x,t)=A\mathrm{s}\mathrm{i}\mathrm{n}?\left. [\frac{x-v\left(t-t_0\right)}{b}\right]$

$\mathrm{}{v}_{\mathrm{m}\mathrm{a}\mathrm{x}}=A\omega =A\sqrt[]{\frac{k}{m}}$

Momentum conservation   $m{v}_{\text{max}}+0=m{v}_A+2m{v}_B$

$v_A+2v_B=v_{\mathrm{m}\mathrm{a}\mathrm{x}}$

$v_2-v_1=u_1-u_2\Rightarrow v_B-v_A=v_{\mathrm{m}\mathrm{a}\mathrm{x}}$

By (1) & (2) $v_A=-\frac{A\omega }{3}$

$\left|v_A\right|=A_{\text{new}}\omega$

$\frac{A\omega }{3}=A_{\text{new}}\omega \mathrm{}\therefore \mathrm{}{A}_{\text{new}}=\frac{A}{3}\mathrm{};\mathrm{}f=\frac{1}{3}$

${\stackrel{?}{B}}_1=\frac{E_1}{c}\mathrm{c}\mathrm{o}\mathrm{s}?(kz-\omega t){\stackrel{?}{B}}_1$    where   ${\stackrel{?}{B}}_1={\hat{c}}_1*{\stackrel{?}{E}}_1=\hat{k}*\hat{i}={\stackrel{?}{B}}_1=\hat{j}$

${\stackrel{?}{B}}_2=\frac{E_2}{c}\mathrm{c}\mathrm{o}\mathrm{s}?(kz+\omega t){\stackrel{?}{B}}_2$ where ${\stackrel{?}{B}}_2={\hat{c}}_2*{\stackrel{?}{E}}_2=-\hat{k}*\hat{j}={\stackrel{?}{B}}_2=\hat{i}$

$\therefore \mathrm{}\stackrel{?}{B}={\stackrel{?}{B}}_1+{\stackrel{?}{B}}_2$

$\stackrel{?}{B}=\frac{E_1}{c}\hat{j}\mathrm{c}\mathrm{o}\mathrm{s}?(kz-\omega t)+\frac{E_2}{c}\hat{i}\mathrm{c}\mathrm{o}\mathrm{s}?(kz+\omega t)$

$V_1=\frac{k{Q}_3}{3R}+\frac{k{Q}_2}{2R}+\frac{k{Q}_1}{R}=30$

$V_2=\frac{k{Q}_3}{3R}+\frac{k{Q}_2}{2R}+\frac{k{Q}_1}{2R}=15$

$V_3=\frac{k{Q}_3}{3R}+\frac{k{Q}_2}{3R}+\frac{k{Q}_1}{3R}=10$

Eq. (1) $\&$  (2) $\to \frac{k{Q}_1}{R}=30$  

Eq. (2) $\&$  (3) $\to \frac{k{Q}_2}{6R}+\frac{k{Q}_1}{6R}=5\Rightarrow \frac{k{Q}_2}{R}=-\frac{k{Q}_1}{R}+30=0$

By (3) $\to \frac{k{Q}_3}{R}=30-\frac{k{Q}_2}{R}-\frac{k{Q}_1}{R}=0$

Potential of innermost shell (earthed) $=\frac{k{Q}_1^{\mathrm{\text{'}}}}{R}+\frac{k{Q}_2}{2R}+\frac{k{Q}_3}{3R}=0$

$\Rightarrow \mathrm{}\frac{k{Q}_1^{\mathrm{\text{'}}}}{R}=0\mathrm{}\therefore \mathrm{}{Q}_1^{\mathrm{\text{'}}}=0$

The Earth's magnetic field makes a magnetic inclination (dip) or angle between Earth's overall magnetic field and the horizontal plane at a particular location. It shows how much the magnetic field is tilted from the horizontal.

tan I = B_V / B_H,

where, B_V is the vertical component and B_H is the horizontal component of the Earth's magnetic field.

According to Gauss's law for magnetism, the net magnetic flux through any closed surface is zero.

$?\mathbf{B}\cdot dA=0$

It states that all magnetic field lines must enter and exit from a surface. In this way, it differs from electric charges, which can exist independently as positive and negative.

Magnetism is the force exerted by a magnet. It comes when the magnet attracts or repulses an object. For example, a magnet sticks to certain objects because the magnetic force or magnetism pulls it.

1R + 0.5 = 1.5
1R = 1
R = 1/10 = 100Ω

(Q? - Q? )/Q? = (T? - T? )/T? ; Q = (300/600) * 1600 = 800 J
work done = Q? - Q? = 1600 – 800 = 800J