$A$ particle is moving with a constant speed of $\sqrt{2} \, m/s$ on a circular path of radius $10 \, cm$. Find the magnitude of the average velocity when it has covered $\frac{3}{4}$ of the circular path.

$A$ particle of mass $M$ is moving in a circle of fixed radius $R$ in such a way that its centripetal acceleration at time $t$ is given by $n^2Rt^2$ where $n$ is a constant. The power delivered to the particle by the force acting on it,is

The second's hand of a watch has $6\, cm$ length. The speed of its tip and the magnitude of the difference in velocities of its tip at any two perpendicular positions will be respectively:

$A$ particle moving in a circle of radius $R$ with uniform speed takes time $T$ to complete one revolution. If this particle is projected with the same speed at an angle $\theta$ to the horizontal,the maximum height attained by it is equal to $4 R$. The angle of projection $\theta$ is then given by :

$A$ projectile is thrown upward with a velocity $v_0$ at an angle $\alpha$ to the horizontal. The change in velocity of the projectile when it strikes the same horizontal plane is

$A$ particle of mass $1 \ kg$ is projected with an initial velocity $10 \ ms^{-1}$ at an angle of projection $45^{\circ}$ with the horizontal. The average torque acting on the projectile,between the time at which it is projected and the time at which it strikes the ground,about the point of projection in $Nm$ is: