Applied Mechanics and Graphic Statics The centre of gravity of a quadrant of a circle lies along its central radius at a distance of 0.6 R 0.2 R 0.4 R 0.3 R 0.6 R 0.2 R 0.4 R 0.3 R ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics The shape of a suspended cable under its own weight, is Catenary Elliptical Circular Parabolic Catenary Elliptical Circular Parabolic ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics Instantaneous center is at infinity when the angular velocity is Maximum Constant Zero Minimum Maximum Constant Zero Minimum ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics A uniform rod 9 m long weighing 40 kg is pivoted at a point 2 m from one end where a weight of 120 kg is suspended. The required force acting at the end in a direction perpendicular to rod to keep it equilibrium, at an inclination 60° with horizontal, is 100 kg 10 kg 60 kg 40 kg 100 kg 10 kg 60 kg 40 kg ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics A stone was thrown vertically upwards from the ground with a velocity of 50 m/sec. After 5 seconds another stone was thrown vertically upwards from the same place. If both the stones strike the ground at the same time, then the velocity with which the second stone was thrown should be (Assume g = 10 m/sec²) 40 m/sec 50 m/sec 15 m/sec 25 m/sec 40 m/sec 50 m/sec 15 m/sec 25 m/sec ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics ‘u₁’ and ‘u₂’ are the velocities of approach of two moving bodies in the same direction and their corresponding velocities of separation are ‘v₁’ and ‘v₂’. As per Newton's law of collision of elastic bodies, the coefficient of restitution (e) is given by e = u₂ - u₁/v₁ - v₂ e = v₁ - v₂/u₂ - u₁ e = v₂ - v₁/u₁ - u₂ e = v₁ - v₂/u₂ + u₁ e = u₂ - u₁/v₁ - v₂ e = v₁ - v₂/u₂ - u₁ e = v₂ - v₁/u₁ - u₂ e = v₁ - v₂/u₂ + u₁ ANSWER DOWNLOAD EXAMIANS APP
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