$A$ force $F = Be^{-Ct}$ acts on a particle of mass $m$,which is at rest at $t = 0$. Its terminal velocity is:

$A$ force $\vec{F}=(40 \hat{i}+10 \hat{j}) \, N$ acts on a body of mass $5 \, kg$. If the body starts from rest,its position vector $\vec{r}$ at time $t=10 \, s$ will be -

$5\, kg$ દળ ધરાવતા એક પદાર્થ પર અચળ બળ $\overrightarrow F = {F_x}\hat i + {F_y}\hat j$ લાગે છે. $t = 0\, s$ સમયે તેનો વેગ $\overrightarrow v = (6\hat i - 2\hat j)\, m/s$ છે અને $t = 10\, s$ સમયે તેનો વેગ $\overrightarrow v = 6\hat j\, m/s$ છે. તો બળ $\overrightarrow F$ શોધો.

The magnitude of the external force acting on moving bodies can be known directly by

What is dynamics?

$A$ body of mass $2 \ kg$ moving with velocity of $\overrightarrow{v}_{in} = 3 \hat{i} + 4 \hat{j} \ ms^{-1}$ enters into a constant force field of $6 \ N$ directed along the positive $z$-axis. If the body remains in the field for a period of $\frac{5}{3} \ s$, then the velocity of the body when it emerges from the force field is: