Question

Resolve a weight of 10 N in a direction parallel to a slope inclined at 45° to the horizontal.​

Answer 1

Answer:

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Explanation:

Let the force 10N makes an angle θ with the inclined line

Let the force 10N makes an angle θ with the inclined lineparallel component = 10×sinθ N ;

Let the force 10N makes an angle θ with the inclined lineparallel component = 10×sinθ N ; Perpendicular component = 10×cosθ N

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Answer 2

Explanation:

Let the force 10N makes an angle θ with the inclined line

parallel component = 10×sinθ N ; perpendicular component = 10×cosθ N

I hope it will help you ✌️✌️❣️