Class 11th

  $S_{10}=530\Rightarrow 5\left[2a+9d\right]=530$

2a + 9d = 106 . (i)

$S_5=140\Rightarrow \frac{5}{2}\left[2a+4d\right]=140$              

a + 2d = 28 . (ii)

Solving (i) & (ii) a = 88 d = 10

$\therefore S_{20}-S_6=14a+175d=1862$            

W = mgh

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

$=627.2J$

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

L : 2x + y = k.

$y=-2x+k.istangent\frac{x^2}{3}-\frac{y^2}{3}=1$                

$\Rightarrow k^2=3{\left(-2\right)}^2-3=9\Rightarrow k=3ask>0$              

y = -2x + 3 is also tangent to y² = $4\left(\frac{\alpha }{4}\right)x$  

$\Rightarrow 3=\frac{\alpha /4}{-2}$ -> a = -24

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

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

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

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

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

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.

Change of state refers to a substance transitioning between solid, liquid, and gas phases, during which energy is absorbed or released at constant temperature. The latent heat is the energy required for that transition without changing the temperature.

It works best for small to moderate temperature differences; large differences may require more complex heat transfer models.

It's applied in forensics for estimating time of death, in engineering for cooling rate analysis, and in food Science to predict cooling times.

The ideal gas equation fails at high pressures, low temperatures, and for gases with strong intermolecular forces or large molecular volumes.