Physics

The SI unit of the electrical power is Watt which is symbolized as W. One watt is equal to one joule per second (J/s). In the context of current and voltage, one watt is the power consumed when one ampere of the current flows through the potential difference of one volt.

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.