Physics

Using parallel axis theorem:

$l=2*\left[l_{cm}+m{d}^2\right]$  

$l=2*\left[\frac{2}{5}\left(1.5\right){\left(0.5\right)}^2+\left(1.5\right){\left(2.5\right)}^2\right]$

l = 19.05 kgm²

The basic unit of energy is a Joule (J), which is equal to one watt of power expended for one second. In the day-to-day scenarios such as household electricity consumption, joule is too small a unit to be convenient. For billing and metering purpose, it is measured in kilowatt-hours (kWh). Electrical energy is measured in kilowatt-hours (kWh) as the commercial unit. 1kWh = 1000 W * 3600 s 

For calculating the units consumed, we will be using the following formula 

Units = [Power (W) x Time (h)]/1000

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