Physics

Kindly go through the solution

p = i²R

P' = 0.64 i²R

$I=\left(\frac{2}{5}M{R}^2+M{R}^2\right)*2$

$=\frac{14}{5}*2*\frac{9}{16}$

$=\frac{63}{20}$          

= 3.15 kg m²

By Brewsters law

m = tani

m =   $\sqrt[]{3}$

$1*\frac{\sqrt[]{3}}{2}=\sqrt[]{3}*sinr$

$sinr=\frac{1}{2}$    

r = 30°

[A] = L²

$B=\frac{x^2}{tE}\equiv \frac{L^2}{TM{L}^2{T}^{-2}}=\frac{1}{M{T}^{-1}}$            

[B] = M^–1T

[AB] = [M^–1L²T]

$P=\overrightarrow{F}\cdot \overrightarrow{v}$

= 18t⁴ – 18t²

->P (t = 2) =18 [16 − 4] = 216 W

f_K = μmg cosq

= 0.1 * $\frac{50*\sqrt[]{3}}{2}$

$=2.5\sqrt[]{3}N$

 F₁    = mg sinq + f_K

= 25 + 2.5 $\sqrt[]{3}$  

F₂    = mg sinq – f_K

= 25 – 2.5 

F₁ – F₂ = 5 $\sqrt[]{3}$  N

${\left(C_v\right)}_{mix}=\left(\frac{2*\frac{3}{2}+6*\frac{5}{2}}{2+6}\right)R$

$=\frac{\left(3+15\right)R}{8}=\frac{9}{4}R$

$?=\frac{h}{p}$

$?=\frac{h}{m{v}_p}$

and   $2?=\frac{h}{4m{v}_{?}}$

$\frac{1}{2}=\frac{4v_{?}}{v_p}$

$\frac{v_p}{v_{?}}=8$

For 3 ->2 transitions

            DE = 13.6   $\left(\frac{1}{4}?\frac{1}{9}\right)$

            = 13.6 *   $\frac{5}{36}$

            = 1.88 eV

            E =   $\frac{Stress}{Strain}$

As temperature increases, strain increases

Elasticity decreases