Physics

  $\frac{v^2}{2a}=s$ (a = mg)

v² = 2 * mgs

v² = 2 * (0.4) * 10 * 10

v² = 80

W_f = Dk

= $-\frac{1}{2}*100*80$  

W_f = – 4000

a = $\frac{3}{9}g$ , For C, 3g – T = 3a = (3) $\frac{3}{g}$ g

T = 2g = 20 N

$\frac{\sigma }{\in }=Y$

$\frac{\sigma }{Y}=\in \Rightarrow \frac{20}{5*10^{-8}*2*10^{11}}$

$=2*10^{-3}$  

$=20*10^{-4}\Rightarrow 20$  

Charge on C₁ = CV

            Charge on C₂ = 4CV

When connected in parallel

$V_C=\frac{5V}{3}$  

$Q{\text{'}}_1=\frac{5}{3}CV$ $Q{\text{'}}_2=\frac{10}{3}CV$

$?$ $E=\frac{1}{2}C{V}^2$

$E\text{'}=\frac{25}{18}C{V}^2+\frac{25}{9}C{V}^2$  

$\frac{25}{6}C{V}^2=\frac{50E}{6}$

x = 50

$-0.85=\frac{-13.6}{n^2}$  

n = 4

Number of transitions = $\frac{4*3}{2}=6$  

Current in circuit (I) = $\frac{8-2}{6}$  = 1 A.

So, V_C – 4 (1) – 2 + 8 – 2 (1) = V_A

V_C – 6 – 2 + 8 = V_A

V_C – V_A = 0 V

COAM gives Iw₀ = 2Iw

$\omega =\frac{{\omega }_0}{2}$

Loss in KE =  $\frac{1}{2}I{\omega }_0^2-\frac{1}{2}\left(2I\right){\left(\frac{{\omega }_0}{2}\right)}^2$

$=\frac{1}{2}I{\omega }_0^2$

$=\frac{1}{4}*5*\frac{2}{2}*100=250J$  

Dv = a (t + 1 – t)

a = 50 m/s²

$s=u+\frac{a}{2}\left(2\left(t+1\right)-1\right)$  

$120=u+\frac{50}{2}\left(2t+1\right)$  

u = 120 – 25 (2t +1)

In (t + 2)^th second

$s\text{'}=u+\frac{a}{2}\left(2\left(t+2\right)-1\right)$

= u + 25 (2t + 3)

= 120 – 25 (2t) – 25 + 25 (2t) + 75

S' = 170 m

$\left|m\right|=\left|\frac{v}{u}\right|=2$  

|v| = |2u|

n + 2n = 45

n = 15 cm

u = –15

v = 30

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$  

$\frac{1}{30}+\frac{1}{-15}=\frac{1}{f}$  

$\frac{1-2}{30}=\frac{-1}{30}=\frac{1}{f}$  

f = 30 cm

Block is not slipping, so

f = mrw² = 2 * 1 * 16 = 32

$S=\frac{F}{L}$  

[S] = [MT^–2]

$F=nA\frac{dv}{dx}$  

$\eta =\frac{ML{T}^{-2}\cdot T}{L^2}=M{L}^{-1}{T}^{-1}$

$\overrightarrow{L}=\overrightarrow{r}*\overrightarrow{p}$

L º [ML²T^–1]

$KE=\frac{1}{2}l{\omega }^2$  

KE º ML²T⁻²