Physics, asked by Anonymous, 3 months ago

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

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Answered by Anonymous
1

ʜᴏʟᴀ ᴍᴀᴛᴇ !

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ᴀɴsᴡᴇʀ

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

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

Answered by Anonymous
0

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

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

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