Physics, asked by jusjessi2607, 1 year ago

A block is kept on a inclined plane of inclination θ of length l. The velocity of particle at the bottom of inclined is (the coefficient of friction is μ)(a) [2gl(\mu\cos\theta-\sin\theta)]^{\frac{1}{2}}(b) \sqrt{2gl(\sin\theta-\mu\cos\theta)}(c) \sqrt{2gl(\sin\theta+\mu\cos\theta)}(d) \sqrt{2gl(\cos\theta+\mu\sin\theta)}

Answers

Answered by guduuu
2
hey dear ur answer is c part hope it helps u
Similar questions