Physics

Angular momentum is an axial vector along axis of rotation.

We cannot call the trailing zeroes as significant numbers. If you remember from the rules, it needs to have a decimal point somewhere to be considered as one. What is recommended is that one can use the measure as to the power of 10.  

  $F_x=\frac{-dv}{dx}$           

  $\overrightarrow{F}=-3x\widehat{i}-5\widehat{j}-6\widehat{k}$          

  ${\left|\overrightarrow{F}\right|}_{x=6}=\sqrt[]{18^2+5^2+6^2}$          

$=\sqrt[]{324+25+36}$

= $\sqrt[]{385}$    

= 19.62N

%error= $\left(\frac{dV}{V}+\frac{dR}{R}\right)*100$  

$=\left(\frac{5}{200}+\frac{0.2}{20}\right)*100$                          

= 3.5

Since $a=\frac{gsin?}{1+\frac{I}{M{R}^2}}$

$a=\frac{g*\frac{\sqrt[]{3}}{2}}{1+\frac{1}{2}}=\frac{g\frac{\sqrt[]{3}}{2}}{\frac{3}{2}}$

$a=\frac{g}{\sqrt[]{3}}$

f = 15

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

  $v=-\frac{u}{2}$           

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$            

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

u = – f = – 15cm

PV = nRT

->Pµn

->Ratio= $\frac{32}{2}$ =16

$\left|\Delta E_0\right|=\left(-136\left\{1-\frac{1}{4}\right\}\right)eV$

|DE₀| = –10.2

$\lambda =\frac{12400}{10.2}*10^{-10}m$

$\rho =\frac{h}{\lambda }=\frac{6.63*10^{-34}*10.2}{12400*10^{-10}}$                  

$?$ $mv=\frac{h}{\lambda }$            

  $1.8*10^{-27}$           

$v=\frac{6.63*10.2*10^{-34}}{12400*10^{-10}}$

$v=\frac{6.63*10.2}{12400*1.8}*10^3$            

$=\frac{6.63*102}{124*1.8}=3.02$ ]

= 3 m/s

From Gauss law

$?=\frac{q_{enclosed}}{{\epsilon }_0}=\frac{5q}{{\epsilon }_0}$

KE =  $\frac{1}{2}$ m W² (A²– n²)

TE = $\frac{1}{2}$ mw²A²

^{$\frac{KE}{TE}=\frac{A^2-n^2}{A^2}=\frac{1-\frac{1}{9}}{1}=\frac{8}{9}$}