$A$ body moves due East with a velocity of $20 \, km/h$ and then due North with a velocity of $15 \, km/h$. The resultant velocity is .......... $km/h$.

The sum of the magnitudes of two forces acting at a point is $16 \,N$. If their resultant is normal to the smaller force and has a magnitude of $8 \,N$, then the forces are:

The speed of a particle changes from $\sqrt{5} \ m/s$ to $2\sqrt{5} \ m/s$ in a time $t$. If the magnitude of change in its velocity is $5 \ m/s$,the angle between the initial and final velocities of the particle is (in $^{\circ}$)

If $|A|=2$ and $|B|=4$ and the angle between them is $60^{\circ}$,then $|A-B|$ is

The vector sum of the two forces $\vec{A}$ and $\vec{B}$ is perpendicular to their vector difference. Hence, the forces $\vec{A}$ and $\vec{B}$ are

Two vectors $\vec{A}$ and $\vec{B}$ have equal magnitudes. Then the vector $\vec{A} + \vec{B}$ is perpendicular to: