Question

a force of magnitude 10N acts on a partical at angle 60 with the horizontal component this force is
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Answer 1

Answer:

$\huge\mathfrak\blue{Answer}$

$\sqrt[10]{3n}$

The resultant force acts due north.

⇒ The horizontal components of the two forces cancel

$10 = p \: \sin(30 ) = \frac{p}{2} = 10 = p = 20n$

Resultant

$r = p \cos(30) = 20 \times \frac{ \sqrt{} 3}{2} = \sqrt[10]{3n}$

Answer 2

Answer:

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