Physics

Since process is isochoric

So    $\Delta U=n{C}_v\Delta T$

$\Delta U=n\left(\frac{5}{2}R\right)\Delta T--(i)\left[C_V=\frac{5}{2}R\right]$      

And external work

$\Delta W=nR\Delta T--\left(ii\right)$    

$\frac{5}{2}=\frac{x}{10}\Rightarrow x=25.00$                

In first condition R₁ = 36 $\Omega$  

In second condition R₂ = 18  $\Omega$

$P_1=\frac{V^2}{R_1}=\frac{{\left(240\right)}^2}{36}$               

$P_2=\frac{V^2}{R_2}+\frac{V^2}{R_2}=\frac{{\left(240\right)}^2}{18}+\frac{{\left(240\right)}^2}{18}$               

$P_2=\frac{{\left(240\right)}^2}{9}$               

So   $\frac{P_1}{P_2}=\frac{{\left(240\right)}^2/36}{{\left(240\right)}^2/9}=\frac{1}{4}$

x = 4

Force  = Aya   $\Delta T$

Force $=\left(10*10^{-4}\right)*\left(2*10^{11}\right)*10^{-5}*400$  

$F=8*10^{5}N$              

  $x*10^5=8*10^5$              

x = 8

Amp. a slit with

Intensity a (Amp)² a (slit width)²

  $\frac{l_1}{l_2}={\left(\frac{3}{1}\right)}^2=\frac{9}{1}\Rightarrow l_1=9I_2$

$\frac{l_{min}}{l_{max}}=\frac{{\left(\sqrt[]{l_1}-\sqrt[]{l_2}\right)}^2}{{\left(\sqrt[]{l_1}+\sqrt[]{l_2}\right)}^2}=\frac{{\left(3-1\right)}^2}{{\left(3+1\right)}^2}$            

  $\Rightarrow \frac{1}{4}=\frac{x}{4}$

x = 1.00

Formula for Power = F·v cos $$$\theta$

According to the question, force is perpendicular.

So $\theta$ = 90

and cos 90 = 0

Hence, the result turns out to be 0 i.e. no power generated inspite of force being applied.

In simple words,

Energy = total work done on an object.

Power = how fast the work is done on an object.

For example: Writing an article vs how fast you write the article. This sums up the exact meaning of both these terms.

No. Even if the power is higher, the energy can be inefficient. This happens when much of the energy generated gets wasted or isn't utilized as effectively as it should have been. In such cases, high power can still lead to inefficiency of energy.

Answer (1)

$eE=e\frac{dV}{dr}=m{\omega }^{2}r$

$?V=\int dV=\underset{R}{\overset{2R}{\int }}\frac{m}{e}{\omega }^{2}rdr$

$=?V=\frac{3m{\omega }^2{R}^2}{2e}$

$l=\frac{120-60}{4000}=0.015A$

Thus l₂ = l - L_L

= 0.015 – 0.006

= 0.009 = 9mA

For first line of Lyman

$\frac{1}{\lambda }=R\left(1-\frac{1}{4}\right)=R\frac{3}{4}$

$\lambda =\frac{4}{3R}$

3rd line (Paschen)

$\frac{1}{{\lambda }_3}=R\left(\frac{1}{3^2}-\frac{1}{6^2}\right)=\frac{R}{9}*\frac{3}{4}$

2nd line (B almer)

${\frac{1}{\lambda }}_2=R\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=\frac{R}{4}*\frac{3}{4}$

$a\left(\frac{4}{3R}\right)=\frac{20}{3R}\Rightarrow a=5$