Questions
A particle of mass 0.5 kg is projected horizontally with velocity 10 m/s from the top of a tower of height 40 m. The angular momentum of the particle about the base of the tower just after projection is
detailed solution
Correct option is D
(4)L=mvh=0.5 × 10 × 40 kg−m2/s =200 kg−m2/sTalk to our academic expert!
Similar Questions
A particle of mass m = 5 units is moving with a uniform speed , units in the XOY plane along the line y = x + 4. The magnitude of the angular momentum of the Particle about the origin is
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