Class 11th

Yes. Mechanical energy is the sum of both kinetic and potential energies. Kinetic energy cannot be a negative value since mass is always positive and velocity square leads to a positive value. However, potential energy can be negative depending on the reference point.

Not in all cases. This happens due to the interruption of non conservative forces like friction, air resistance, etc. which are responsible for decreasing the mechanical energy by converting it into other forms of energy such as heat, noise, etc.

x – y = 0 . (i)

x + 2y = 3. (ii)

2x + y = 6 . (iii)

Solving (i) & (ii), we get (1, 1)

Solving (ii) & (iii), we get (3, 0)

Solving (iii) & (i), we get (2, 2)     

$AB=\sqrt[]{5,}BC=\sqrt[]{5},CA=\sqrt[]{2}$

$\therefore AB=BC$ Isosceles triangle

Hence, option (B) is correct answer.

Conservation of Mechanical Energy is linked to newton's second law of motion (F = ma) in combination with work energy theorem. The work energy theorem states that total work done on an object equals to change in kinetic energy. This theorem is used to prove that the sum of kinetic energy and potential energy remains constant.

Al < Mg < Si < S < P

1^st I. E increase along period with exception on moving from group 2 to group 13 and group 15 to group 16

$\frac{GM}{{\left(3R/2\right)}^2}=\frac{GM}{R^3}*r$

$\Rightarrow OA=\frac{4R}{9}=r$

$AB=R-\frac{4R}{9}=\frac{5R}{9}\Rightarrow OA:AB=4:5=x:y\Rightarrow x=4$

$v=\omega \sqrt[]{A^2-s^2}\Rightarrow \frac{A\omega }{2}=\omega \sqrt[]{A^2-s^2}\Rightarrow s=\frac{\sqrt[]{3}A}{2}\Rightarrow x=3$

By properties,

OA.AB = PA.AQ, OA=1, AB = 12

12 = a²

$a=2\sqrt[]{3}$

$\therefore AreaofPQB=\frac{1}{2}*12*2.2\sqrt[]{3}=24\sqrt[]{3}$ sq. units

$\Delta Q=\Delta U+\Delta W$

^{$\Rightarrow Q=\Delta U+\frac{Q}{5}\Rightarrow \Delta U=\frac{4Q}{5}=n{C}_v\Delta T\Rightarrow \frac{4Q}{5}=\frac{5R}{2}\Delta T\Rightarrow \Delta T=\frac{8Q}{25R}$}

^{$Q=nc\Delta T=1*C*\frac{80}{25R}\Rightarrow C=\frac{25R}{8}\Rightarrow x=25$}

$V=k{T}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

$T{V}^{-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=K$

$\Rightarrow \gamma -1=-3/2$

$\Rightarrow \gamma =-1/2$

Work done = $\frac{nR\Delta T}{-y+1}=\frac{1*R*90}{3/2}=6R$

n = 60