A body slides down a smooth inclined plane of angle terha and takes time 't'. If the plane is rough with coifficient of fricton 'u' it takes time 'Pt'(P>1). Find u
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1
Answer:
General case s = (1/2)at2 ⇒ t √2s/a Smooth case Acceleration a = sinθ = g/√2 t1 = √2√2/g Rough case Acceleration a = g sinθ - μg cosθ = (1 - μ)g/√2 Read more on Sarthaks.com - https://www.sarthaks.com/109867/when-slides-from-rest-along-smooth-inclined-plane-making-angle-with-horizontal-takes-time
Answered by
1
Answer:
√2s
1.414s
12–√s
2s
Answer :
B
Solution :
t=2lgsinθ−−−−−−√
Explanation:
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