Answer to Question #15534 in Mechanics | Relativity for Dirghayu Raichura
the answer is 0.75 but i want the solution the answer is given in the back side of my book . i.e. yugbodh physics class 11th.
reduced by a factor 4. (That is because S = a * t^2 / 2 remains the
The friction is therefore 3/4 of the component of the weight
m * g * cos 45 * k = (3/4) * m * g * sin 45, where k is the
k = 3/4 = 0.75
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