Class 11th

for bullet

$l=\Delta p=mv-0$

$=\frac{4}{1000}*50=\frac{20}{100}=0.2kg-m/s$

Using conservation of momentum :

Pi = Pt

$\Rightarrow \frac{4}{1000}*50=4*v$

v = 0.05 m/s

f : R -> R.

$f\left(x\right)=[\begin{array}{l}\frac{x^3}{{\left(1-cos2x\right)}^2}log\left(\frac{1+2x{e}^{-2x}}{{\left(1-x{e}^{-x}\right)}^2}\right)x\ne 0\\ \alpha x=0\end{array}$                

As f is continuous at x = 0

$\therefore \alpha =\underset{x\to 0}{Lim}f\left(x\right)$              

$=\underset{x\to 0}{Lim}\frac{1}{2}\left[e^{-2x}+e^{-x}\right]=\frac{1}{2}*2=1$            

$\therefore \alpha =1$              

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

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

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

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

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.