Physics

E = (50 N/s) sin (t – x/c)

Energy 5.5 * 10^-12 J., vol v =?

$u=\frac{1}{2}{\epsilon }_0{E}^2\to$ energy density

$?u=\frac{U}{v}\Rightarrow U=uv$

= $\frac{1}{2}{\epsilon }_0{E}^2.v$

$v=\frac{5.5*10^{-12}*2}{8.55*2500*10^{-12}}*10^{6}c{m}^3$

= 497

$\cong 500c{m}^3$

(a) $Calamine:ZnC{O}_3$  

(b) Malachite : $CuC{O}_3.Cu{\left(OH\right)}_2$  

(c) Siderite : FeCO₃

(d) Sphalerite : ZnS

$?\frac{P_i=P_f}{m*40}=\frac{m}{2}v\text{'}+\frac{m}{2}*60$

v’ = 20 m/s

$K{E}_f=\frac{1}{2}\frac{m}{2}\left[v^2+60^2\right]$

$K{E}_f=\frac{1}{2}\frac{m}{2}\left[v^2+60^2\right]$

$\frac{\Delta KE}{KE}=\frac{\frac{1}{2}m\left(2000\right)-\frac{1}{2}m\left(1600\right)}{\frac{1}{2}m\left(1600\right)}$

$\therefore x=1$

Sound waves and water waves are mechanical waves, which always require a medium to travel. Light, however, does not require a medium to propagate, as it is electromagnetic.  Light can travel through vacuum, while sound and water waves cannot travel through vacuum.   

$x=Asin\omega t$   (Eq. of a particle executing SHM)

When KE = PE

$\frac{1}{2}m{v}^2=\frac{1}{2}k{x}^2$

$\frac{1}{2}m{\omega }^2\left(A^2-x^2\right)=\frac{1}{2}k{x}^2\left[?k=m{\omega }^2\right]$

A2 – x2 = x2

 $So,\frac{A}{\sqrt[]{2}}=Asin\omega t$

$\frac{1}{\sqrt[]{2}}=sin\omega t$

$\therefore t=\frac{T}{8}$

$P=-B\left(\frac{\Delta v}{v}\right)$

or  $B=\frac{-P}{\left(\frac{\Delta v}{v}\right)}$

$B=\frac{-\rho gh}{\left(\frac{-0.005V_0}{V_0}\right)}\Rightarrow h=\frac{908*10^8*0.005}{10^3*9.8}$

$\Rightarrow h=500m$

$\mu =9*10^{-4}kg/m$

T = 900 N

$f_0=500Hz\to$ resonance frequency

$f_0^{\text{'}}=550Hz\to$ Next higher frequency

$v=\sqrt[]{\frac{T}{\mu }}=\sqrt[]{\frac{900N}{9*10^{-4}}}=1000m/s$

$\therefore 500Hz=\frac{10*1000}{2*\mathcal{l}}$

$\mathcal{l}=10m$

For $W_1,$  

$\frac{\left(m+10+20\right)*10}{8*10^{-7}}\le 1.25*10^9$              

$\Rightarrow m\le 70kg$              

For W₂,

$\frac{\left(m+10\right)*10}{4*10^{-7}}\le 1.25*10^9$              

-> $m\le 40kg$               

Thus, $m\le 40kg$  

$cos{?}_1=\frac{R}{\sqrt[]{R^2+{\left(3R\right)}^2}}=\frac{1}{\sqrt[]{10}}$

$cos{?}_2=\frac{R}{\sqrt[]{R^2+{\left(3R-R\right)}^2}}=\frac{1}{\sqrt[]{2}}$

$\Rightarrow \frac{cos{?}_2}{cos{?}_1}=\frac{\sqrt[]{10}}{\sqrt[]{2}}=\sqrt[]{5}\Rightarrow x=1$

Velocity of bob after collision,

v_{1 $=\sqrt[]{5gR}=\sqrt[]{5*10*0.5}=5m/s$}

From Conservation of Momentum,

$\frac{10}{1000}*v=\frac{-10}{1000}*100+1*5$  

->v = 400 m/s