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(B) Given,$F_x = 5 \, N$ and $F_y = 5 \, N$.The magnitude of the resultant force $F$ is given by:$|\vec{F}| = \sqrt{F_x^2 + F_y^2} = \sqrt{5^2 + 5^2}

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$5\sqrt{2}, \pi/4$

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(B) Given,$F_x = 5 \, N$ and $F_y = 5 \, N$.The magnitude of the resultant force $F$ is given by:$|\vec{F}| = \sqrt{F_x^2 + F_y^2} = \sqrt{5^2 + 5^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} \, N$.The direction $	heta$ with the $X$-axis is given by:$	an 	heta = \frac{F_y}{F_x} = \frac{5}{5} = 1$.Since $	an 	heta = 1$,we have $	heta = \pi/4$ radians (or $45^\circ$).Thus,the magnitude is $5\sqrt{2} \, N$ and the direction is $\pi/4$.

If two forces of $5 \, N$ each are acting along $X$ and $Y$ axes,then the magnitude and direction of the resultant is

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