Physics, asked by Anonymous, 4 months ago

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

Answers

Answered by Anonymous
2

ʜᴏʟᴀ ᴍᴀᴛᴇ !

 \: \: \: \: \: \: \: \: \: \: \:

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 itscottoncandy06
0

Answer:

hy❤

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

hope this helps

have a NYC day

keep smiling

Candy

Similar questions