Physics

from newtan's law of cooling :

$\frac{dT}{dt}=-k\left(T-T_0\right)$

Using average value :

$\frac{75-65}{5}=-k\left[\frac{\left(75+65\right)2}{2}-25\right]$

$\frac{\left(65-T\right)}{5}=-k\left[\frac{\left(65+T\right)}{2}-25\right]$

$\Rightarrow \frac{10}{65-T}=\frac{45}{\frac{65+T}{2}-25}$

$\Rightarrow T=57°C$

$T=2\pi \sqrt[]{\frac{\mathcal{l}}{g}}$

$g=4{\pi }^2\frac{\mathcal{l}}{T^2}$

$\Rightarrow \frac{\Delta g}{g}=\frac{\Delta \mathcal{l}}{\mathcal{l}}+\frac{2\Delta T}{T}$

Student 1 will get least error because it has more number of oscillation

$V_P=\sqrt[]{V_0^2+V_0^2}=\sqrt[]{2}{V}_0$

At minimum deviation :

i = i'

r = r'

$r+r\text{'}=A\Rightarrow r=\frac{A}{2}$

Snell's law :

1 sin I = $\sqrt[]{3}sinr=\sqrt[]{3}sin\frac{i}{2}$ $\text{exception 25:}\text{exception 25:}\frac{}{}$

$\Rightarrow 2cos\frac{i}{2}=\sqrt[]{3}$

$\Rightarrow i=60°$

$\begin{array}{l}i=2r\Rightarrow r=30°\\ \Rightarrow A=60°\end{array}$

Using gauss law in differential form

$\overrightarrow{\nabla }.\overrightarrow{E}=\frac{\rho }{{\epsilon }_0}$

$\overrightarrow{\nabla }.\left({\epsilon }_0\overrightarrow{E}\right)=\rho$

$\Rightarrow \overrightarrow{\nabla }.\overrightarrow{D}=\rho$

$\Rightarrow \left(\frac{\partial }{\partial x}\widehat{i}+\frac{\partial }{\partial y}\widehat{j}+\frac{\partial }{\partial z}\widehat{k}\right).\overrightarrow{D}=\rho$

$\begin{array}{l}\Rightarrow \rho =-e^{-x}siny+e^{-x}siny+2=2\\ \Rightarrow Q=2\left(Volume\right)=4*10^{-9}c=4nc\end{array}$

$F=\left[\frac{2k\lambda }{r_1}-\frac{2k\lambda }{r_2}\right]q=2k\lambda q\left[\frac{1}{r_1}-\frac{1}{r_2}\right]$

$2*9*10^9*3*10^{-6}\left[\frac{1000}{10}-\frac{1000}{12}\right]q$

$\begin{array}{l}\Rightarrow 4=9*10^{5}q\\ \Rightarrow q=4.44\mu C\end{array}$

$R_1={?}_1\frac{\mathcal{l}}{A}{R}_2={?}_2\frac{\mathcal{l}}{A}$

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{A}{\mathcal{l}}\left[\frac{1}{{?}_1}+\frac{1}{{?}_2}\right]$

$=\frac{3*10^{?6}}{\frac{1}{4}}\left[\frac{1}{1.7*10^{?8}}+\frac{1}{2.6*10^{?8}}\right]$

$R=0.858*10^{?3}?$

Using conservation of energy:

$mgh=\frac{1}{2}{l}_{cm}{\omega }^2+\frac{1}{2}m{v}^2=\frac{1}{2}{l}_{cm}\frac{V^2}{r^2}+\frac{1}{2}m{v}^2$

$v^2=\frac{mgh}{\frac{1}{2}\frac{l_{cm}}{r^2}+m{v}^2}$

So, sphere has the greatest velocity of COM and ring has least.

Total energy of monoatomic gas at equilibrium

$=\frac{3}{2}{K}_{B}T$

x (t) = A sin $\omega$ $$ t + B cos $\omega t$ $$

$\Rightarrow x\left(t\right)=\sqrt[]{A^2+B^2}sin\left(\omega t+\delta \right)=\sqrt[]{A^2+B^2}cos\left(\omega t-\left(\frac{\pi }{2}-\delta \right)\right)$

At t = 0:

$x\left(0\right)=\sqrt[]{A^2+B^2}sin\left(\delta \right)$

$v\left(0\right)=\omega \sqrt[]{A^2+B^2}cos\left(\delta \right)$

$\Rightarrow {\left(x\left(0\right)\right)}^2+{\left(\frac{v\left(0\right)}{\omega }\right)}^2=A^2+B^2$

$tan?=tan\left(\frac{\pi }{2}-\delta \right)=cot\delta =\frac{v\left(0\right)}{x\left(0\right)\omega }$

$\Rightarrow ?=ta{n}^{-1}\left(\frac{v\left(0\right)}{x\left(0\right)\omega }\right)$