In fig shown block B moves down with a velocity 10m/s.The velocity of A in the position shown is

Question

Samuel Garry · Answer

Its answer is 12.5 m/sec till body A will remain at surface The ideal weight of a tempting the sets of question is breaking the component of velocity in the direction of string and then equating it equal to velocity of body B Net velocity of body A be v Take its component in the direction of string as v cos 37 So 10 = 4/5 v V = 12.5

Avi · Answer

If velocity of A is V then velocity of B is 2V Velocity of A= 2× velocity of B V cos37° = 2×10 V×4÷5 = 20 V = 20×5÷4 V=25m/s

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In fig shown block B moves down with a velocity 10m/s.The velocity of A in the position shown is

In fig shown block B moves down with a velocity 10m/s.The velocity of A in the position shown is

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