$A$ person holds a ball of mass $0.25 \ kg$ in his hand and throws it,so that it leaves his hand with a speed of $12 \ m \ s^{-1}$. In this process,if his hand moved through a distance of $0.9 \ m$,then the net force acted on the ball is (in $N$)

$A$ body,under the action of a force $\vec F = 6\hat i - 8\hat j + 10\hat k$,acquires an acceleration of $1\, m/s^2$. The mass of this body must be

$A$ constant force $F$ acts on two different masses $m_1$ and $m_2$ independently and produces accelerations $A_1$ and $A_2$ respectively. When the same force $F$ acts on their combined mass $(m_1 + m_2)$,the acceleration $A$ produced is:

The position vector of a particle related to time $t$ is given by $\overrightarrow{r} = (10t \hat{i} + 15t^2 \hat{j} + 7 \hat{k}) \text{ m}$. The direction of the net force experienced by the particle is:

The magnitude and direction of the acceleration produced in a body of mass $5 \text{ kg}$ when two mutually perpendicular forces $8 \text{ N}$ and $6 \text{ N}$ act on it, are respectively:

$A$ body of mass $5\, kg$ starts from the origin with an initial velocity $\vec{u} = 30\hat{i} + 40\hat{j}\,m/s$. If a constant force $\vec{F} = -(\hat{i} + 5\hat{j})\,N$ acts on the body,the time in which the $y$-component of the velocity becomes zero is ......... $s$.