4. a particle is acted upon by two forces of 12 n and 9 n, with an angle of 135° between them. find the…

Explanation:

Step1: Recall the formula for resultant of two - vectors

The formula for the magnitude of the resultant $\vec{R}$ of two vectors $\vec{A}$ and $\vec{B}$ with an angle $\theta$ between them using the parallelogram law is $R=\sqrt{A^{2}+B^{2}+2AB\cos\theta}$.

Step2: Identify the values of $A$, $B$ and $\theta$

Here, $A = 12\ N$, $B=9\ N$ and $\theta = 135^{\circ}$, and $\cos135^{\circ}=-\frac{\sqrt{2}}{2}$.

Step3: Substitute the values into the formula

[ \begin{align*} R&=\sqrt{12^{2}+9^{2}+2\times12\times9\times\cos135^{\circ}}\ &=\sqrt{144 + 81+216\times(-\frac{\sqrt{2}}{2})}\ &=\sqrt{144 + 81-108\sqrt{2}}\ &=\sqrt{225-108\sqrt{2}}\ &=\sqrt{225 - 108\times1.414}\ &=\sqrt{225-152.712}\ &=\sqrt{72.288}\ &\approx8.5\ N \end{align*} ]

Answer:

Approximately $8.5\ N$