The resultant of these forces $\overrightarrow{O P}, \overrightarrow{O Q}, \overrightarrow{O R}, \overrightarrow{O S}$ and $\overrightarrow{{OT}}$ is approximately $\ldots \ldots {N}$.
[Take $\sqrt{3}=1.7, \sqrt{2}=1.4$ Given $\hat{{i}}$ and $\hat{{j}}$ unit vectors along ${x}, {y}$ axis $]$
$9.25 \hat{{i}}+5 \hat{{j}}$
$3 \hat{{i}}+15 \hat{{j}}$
$2.5 \hat{i}-14.5 \hat{{j}}$
$-1.5 \hat{{i}}-15.5 \hat{{j}}$
Two forces with equal magnitudes $F$ act on a body and the magnitude of the resultant force is $F/3$. The angle between the two forces is
An object of $m\, kg$ with speed of $v\, m/s$ strikes a wall at an angle $\theta$ and rebounds at the same speed and same angle. The magnitude of the change in momentum of the object will be
$ABC$ is an equilateral triangle. Length of each side is $a$ and centroid is point $O$. then $\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}=.......$
Mark the correct statement :-
The resultant of $\overrightarrow A + \overrightarrow B $ is ${\overrightarrow R _1}.$ On reversing the vector $\overrightarrow {B,} $ the resultant becomes ${\overrightarrow R _2}.$ What is the value of $R_1^2 + R_2^2$