Class 11th

For combustion of Mg:

Mg (s) + $\frac{1}{2}{O}_2\left(g\right)\to MgO\left(s\right)$

Here,   $\Delta ng=-\frac{1}{2}$

Now using

$\Delta H=\Delta U+\Delta ngRT$

-601.7 =   $\Delta U+\left(-\frac{1}{2}\right)*8.3*10^{-3}*300$

$\Delta U=-600.45kJ$

So; magnitude of   $\Delta U$ is 600 kJ (the nearest integer).

$g=\frac{9}{{\left(1+\frac{h}{R}\right)}^2}$

$\Rightarrow \frac{g}{3}=\frac{g}{{\left(1+\frac{h}{R}\right)}^2}$

$\Rightarrow {\left(1+\frac{h}{R}\right)}^2=3$

1 + $\frac{h}{R}=\sqrt[]{3}$ $\frac{}{}\text{exception 25:}$

$\Rightarrow \frac{h}{R}=\sqrt[]{3}-1$

$\frac{h}{R}=1.732-1$

$\frac{h}{R}=0.732$

h = 0.732 * 6400

= 4684.8 km » 4685 km

Using : PV = nRT

1.5 * 416 = n * 0.083 * 300

n = 25mol =  $\frac{W}{M}$

25 =  $\frac{100}{M}$

So, molar mass, M = 4 g/mol.

$\overrightarrow{F}=4x\widehat{i}+3y^2\widehat{j}$

$\begin{array}{l}d\overrightarrow{r}=d\overrightarrow{s}=dx\widehat{i}+dy\widehat{j}\\ {\displaystyle \int dw={\displaystyle \int \overrightarrow{F}.d\overrightarrow{r}}}\\ ={\displaystyle \int \left(4x\widehat{i}+3y^2\widehat{j}\right).\left(dx\widehat{i}+dy\widehat{j}\right)}\\ ={\displaystyle \underset{1}{\overset{2}{\int }}4xdx+{\displaystyle \underset{2}{\overset{3}{\int }}3y^{2}dy}}\\ =\frac{4}{2}{\left[x^2\right]}_1^2+\frac{3}{3}{\left[y^3\right]}_2^3\\ =2[4-1]+[27-8]\end{array}$

= 6 + 19 = 25 J

By N L M : T sin θ = ${}^{}$  $m{\omega }^{2}R$ - (1)

& R = $\mathcal{l}$ $$ sin θ - (2)

From (1) & (2)

$\Rightarrow Tsin\theta =m{\omega }^2\mathcal{l}sin\theta$

$\Rightarrow T=m{\omega }^2\mathcal{l}$

$\Rightarrow 80=\left(\frac{100}{1000}\right){\omega }^2*2$

$\Rightarrow \frac{800}{2}={\omega }^2$

$\Rightarrow {\omega }^2=400$

$\omega =20rad/sec$

And, $\omega =\frac{2\pi N}{2\pi }$ $\frac{}{}$

$N=\frac{20*60}{2\pi }$

$N=\frac{600}{\pi }rpm=\frac{k}{\pi }$

a µg = 0.5 * 9.8 = 4.9 m/sec²

u = 9.8 m/sec

S =?

v = 0

v² = u² + 2as

  ^{$\Rightarrow 0=9.8^2-2*4.9s$}

^{$\Rightarrow 0=9.8*9.8-9.8s$}

s = $\frac{9.8*9.8}{9.8}=9.8m$  

T = 2t = $\frac{}{}\frac{}{}$ $\frac{2usin\theta }{g}\Rightarrow t=\frac{usin\theta }{g}$

R = u cos θ (T)

$\Rightarrow R=ucos\theta \left(\frac{2usin\theta }{g}\right)$

$\Rightarrow R=ucos\theta 2t$

$\Rightarrow cos\theta =\frac{R}{50t}*\frac{t}{t}$

$\Rightarrow cos\theta =\frac{Rt}{50t^2}$

$\Rightarrow cos\theta =\frac{R}{50t^2}*\frac{usin\theta }{g}$

$\Rightarrow cot\theta =\frac{R}{50t^2}*\frac{25}{10}$

cot θ = $\frac{R}{20t^2}$ $\theta =co{t}^{-1}\left(\frac{R}{20t^2}\right)$ $\frac{}{{}^{}}$

$\Delta Q=ms\Delta T\Rightarrow s=\frac{\Delta Q}{m\Delta T}\Rightarrow \left[s\right]=\frac{M{L}^2{T}^{-2}}{MQ}=L^2{T}^{-2}{Q}^{-1}$

$\Delta Q=mL\Rightarrow \left[L\right]=\frac{M{L}^2{T}^{-2}}{M}=L^2{T}^{-2}$

We know, $\Delta P=-B\frac{\Delta v}{v}$ $\left[?B=-\Delta P\left(\frac{v}{\Delta v}\right)\right]$

$\Rightarrow 100*\Delta P=-Bx\left[\frac{\Delta v}{v}*100\right]$

$\Rightarrow 100+\Delta P=-3*10^{10}\left(-2\right)$

$\Delta P=\frac{6*10^{10}}{100}$

$\Delta P=6*10^{8}N{m}^{-2}$ $\frac{}{}$ 

The SI unit of wave speed is metres per second (m/s).

This unit tells us the distance a wave travels in one second. 

Since wave speed (v) is calculated using the formula:

$v=f \lambda$

(where f is frequency in hertz (Hz) and lambda is wavelength in metres),

multiplying Hz (1/s) by metres (m) gives the result in m/s.