Class 11th

Group?17, elements have very high ionisation enthalpy due to increase in atomic size, ionization enthalpy decreases down the group.

Sodium carbonate is a white crystalline solid which exists as a decahydrate,  This is also called washing soda.

Due to its high dielectric constant, water is a good solvent.

Bond angles give some idea regarding the distribution of orbitals around the central atom in a molecule/complex ion and hence it helps us in determining its shape

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

T = 2g = 20 N

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

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

$=2*10^{-3}$  

$=20*10^{-4}\Rightarrow 20$  

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

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