Two forces F_1 = 3 N at 0^o and F_2 = 5 N a | vedclass

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(C) The resultant force $R$ of two vectors $F_1$ and $F_2$ with an angle $	heta$ between them is given by the formula: $R = \sqrt{F_1^2 + F_2^2 + 2F_

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$7$

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(C) The resultant force $R$ of two vectors $F_1$ and $F_2$ with an angle $	heta$ between them is given by the formula: $R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \cos 	heta}$.Given $F_1 = 3 \ N$,$F_2 = 5 \ N$,and $	heta = 60^o$.Substituting the values: $R = \sqrt{3^2 + 5^2 + 2(3)(5) \cos 60^o}$.Since $\cos 60^o = 0.5$,we have $R = \sqrt{9 + 25 + 30(0.5)} = \sqrt{34 + 15} = \sqrt{49} = 7 \ N$.$A$ force that balances these two forces must be equal in magnitude and opposite in direction to the resultant force.Therefore,the magnitude of the balancing force is $7 \ N$.

Two forces $F_1 = 3 \ N$ at $0^o$ and $F_2 = 5 \ N$ at $60^o$ act on a body. Then a single force that would balance the two forces must have a magnitude of .......... $N$

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