Question
a 1-kilogram block slides down a frictionless inclined plane with an angle of 35 degrees and 1.0 meters high. what is its velocity at the bottom of the plane?
Answers
V^2 = Vo^2 + 2g*h
V^2 = 0 + 19.6*1 = 19.6
V = 4.43 m/s.
V^2 = 0 + 19.6*1 = 19.6
V = 4.43 m/s.
2.03
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