Physics

$\frac{T}{T\text{'}}=\frac{2\pi \sqrt[]{\frac{\mathcal{l}}{g}}}{2\pi \sqrt[]{\frac{\mathcal{l}}{\frac{16}{g}}}}$

$\Rightarrow \frac{T}{T\text{'}}=\frac{1}{\left(\frac{1}{4}\right)}=4$

$\Rightarrow T\text{'}=\frac{T}{4}$

W = mgh

$=80*9.8*\left(\frac{80}{100}\right)$

$=627.2J$

$\begin{array}{l}\Rightarrow w_f+w_g=0\\ \Rightarrow w_f=-627.2J\end{array}$  

$E=\frac{hc}{\lambda }=\frac{1242eV-nm}{621nm}=2eV$

minimum 2eV is required to emit.

For capacitor : current leads emf

by 90°.

$l=\frac{1}{2}C{\epsilon }_0{E}_0^2=\frac{1}{2}C{\epsilon }_0{\left(C{B}_0\right)}^2=\frac{1}{2}{C}^3{\epsilon }_0{B}_0^2$

$B_0=\sqrt[]{\frac{2l}{{\epsilon }_0{C}^3}}$

$=\sqrt[]{\frac{2*0.092}{8.85*10^{-12}*27*10^{24}}}$

$=2.27*10^{-8}T$

Using conservation of momentum

=> 180 V₁ = 4V₂

=> V₂ = 45 V₁

  $Q=K{E}_{180}+K{E}_{\alpha }$              

$Q=\frac{1}{2}180V_1^2+\frac{1}{2}4V_2^2$        

= $\frac{1}{2}*180*{\left(\frac{V_2}{45}\right)}^2+\frac{1}{2}4V_2^2$

$K{E}_{\alpha }=\frac{5.5MeV}{\left(\frac{46}{45}\right)}=5.38MeV$

Component of $\overrightarrow{A}an\overrightarrow{B}$ $\stackrel{}{}\stackrel{}{}$

$=\left(\overrightarrow{A}.\widehat{B}\right)\widehat{B}$

$\left[\left(\widehat{i}+\widehat{j}+\widehat{k}\right).\frac{\left(\widehat{i}+\widehat{j}\right)}{\sqrt[]{2}}\right]\frac{\left(\widehat{i}+\widehat{j}\right)}{\sqrt[]{2}}$

$=\frac{1}{2}\left(1+1\right)\left(\widehat{i}+\widehat{j}\right)$

$=\widehat{i}+\widehat{j}$

$\lambda =\frac{h}{mv}=\frac{h}{\sqrt[]{2km}}=\frac{h}{2\left(ev\right)m}$

$\lambda \alpha \frac{1}{\sqrt[]{m}}\Rightarrow \frac{{\lambda }_e}{{\lambda }_p}=\sqrt[]{\frac{m_p}{m_e}}$

Using conservation of energy :

$?\frac{?GMm}{R}+\frac{1}{2}m{\left(\sqrt[]{\frac{2GM}{R}}\right)}^2=\frac{?GMm}{r}+\frac{1}{2}m{v}^2$

$?\frac{1}{2}m{v}^2=\frac{GMm}{r}$

$?{\displaystyle \underset{R}{\overset{R+h}{?}}{r}^{1/2}dr=\sqrt[]{2GM}{\displaystyle \underset{0}{\overset{t}{?}}dt}}$

$?\frac{2}{3}{R}^{3/2}\left[{\left(1+\frac{h}{R}\right)}^{3/2}?1\right]=\sqrt[]{2GM}t$

$t=\frac{1}{3}\sqrt[]{\frac{2R}{g}}\left[{\left(1+\frac{h}{R}\right)}^{3/2}?1\right]$

On a heating curve, plateaus occur at the melting and boiling points because the heat energy supplied goes into latent heat (phase change) rather than raising the temperature.