Physics

From observation, we can say that right of $-5\mu c$ charge, net electric force (or electric field) can be zero.

$So,\frac{k\left(5\mu c\right)q}{x^2}=\frac{k\left(20\mu c\right)q}{{\left(5+x\right)}^2}$

$\Rightarrow \frac{1}{x^2}=\frac{4}{{\left(5+x\right)}^2}$

$\Rightarrow \frac{5+x}{x}=2$

x = 5 cm

(a) Torque ® ML²T^-2 ↑ (iii)

(b) Impulse ® MlT^-1 ↑ (i)

(c) Tension ↑ MLT^-2 ↑ (iv)

(d) Surface Tension ↑  $\frac{ML{T}^{-2}}{L}=M{T}^{-2}$ (ii)

$\overrightarrow{L}=\overrightarrow{r}*\overrightarrow{p}\Rightarrow \overrightarrow{L}=mvr\left(\widehat{k}\right)$

Direction & magnitude both remain same

 for particle moving with constant speed.

$\frac{dP}{dV}=-aP,atV=V_0,P=P_0$

T_max =? for n = 1

$\frac{dP}{dV}=-aP$ (Given)

On integrating, we get

${\displaystyle \underset{P_0}{\overset{P}{\int }}\frac{dP}{P}}=-a{\displaystyle \underset{0}{\overset{v}{\int }}dV}$

$T_{max}=\frac{P_0}{aeR}?n=1$

$m=\frac{f}{f+u}+\frac{-f}{-f-f}=+\frac{1}{2}$

$v=\frac{uf}{u+f}\Rightarrow v=\frac{-f*-f}{-f-f}=\frac{f^2}{-2f}=\frac{-f}{2}$

 

$\therefore$ dist of image from optical cnetre = $\frac{f}{2}$

$A\underset{}{\overset{{\lambda }_A}{\to }}B\underset{}{\overset{{\lambda }_B}{\to }}C$

Amount of B,

$N_B=\frac{{\lambda }_A{N}_{A_0}}{{\lambda }_B-{\lambda }_A}\left[e^{-{\lambda }_{A}t}-e^{-{\lambda }_{B}t}\right]$

where K = $\frac{{\lambda }_A}{{\lambda }_B-{\lambda }_A}{N}_{A_0}$

Input are :

(0, 0) ; (0, 1); (1, 0); (1, 1).

Thus, the output y is : (1, 0) s

$M{V}_0=M{V}_1+m{V}_2....\left(i\right)$

$\therefore M{V}_1=m{V}_2$  . (ii)

$M{V}_0^2=M{V}_1^2+m{V}_2^2$ . (iii)

$M{V}_0^2=M{V}_1^2+m{\left(\frac{M{V}_1}{m}\right)}^2\left[usinge{q}^n\left(ii\right)\right]$

$\left(1+\frac{M}{m}\right)=4$

${\left(\frac{M}{m}\right)}_{max}=3$

The main difference lies in the direction. We should know that in a longitudinal wave, the displacement of particles is parallel to the direction of wave propagation. In a transverse wave, however, the displacement is perpendicular to the direction of wave propagation.

The units of angular wave number and angular frequency are slightly different.

• Angular wave number k has units of radian per metre (rad/m).
• Angular frequency has units of radian per second (rad/s).