Applied Mechanics and Graphic Statics When a body falls freely under gravitational force, it possesses Maximum weight No effect on its weight Minimum weight No weight Maximum weight No effect on its weight Minimum weight No weight ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics A body A of mass 6.6 kg which is lying on a horizontal platform 4.5 m from its edge is connected to the end of a light string whose other end is supporting a body of mass 3.2 kg as shown in below figure. If the friction between the platform and the body A is 1/3, the acceleration is 0.5 m/sec² 1.25 m/sec² 1.00 m/sec² 0.75 m/sec² 0.5 m/sec² 1.25 m/sec² 1.00 m/sec² 0.75 m/sec² ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics The following is in unstable equilibrium A satellite encircling the earth A solid cube resting on one edge A uniform solid cone resting on its base on a horizontal plane A uniform solid cone resting on a generator on a smooth horizontal plane A satellite encircling the earth A solid cube resting on one edge A uniform solid cone resting on its base on a horizontal plane A uniform solid cone resting on a generator on a smooth horizontal plane ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics A ladder of weight 'w' rests against a smooth vertical wall, and rests on rough horizontal ground, the coefficient of friction between the ladder and the ground being 1/4. The maximum angle of inclination of the ladder to the vertical, if a man of weight 'w' is to walk to the top of it safely, is tan'1 x, where x is 4 3 43834 43833 4 3 43834 43833 ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics The vertical reaction at the support ‘A’ of the structure shown in below figure, is 3.5 t 2 t 3 t 1 t 3.5 t 2 t 3 t 1 t 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 = v₁ - v₂/u₂ + u₁ e = v₂ - v₁/u₁ - u₂ e = u₂ - u₁/v₁ - v₂ e = v₁ - v₂/u₂ - u₁ e = v₁ - v₂/u₂ + u₁ e = v₂ - v₁/u₁ - u₂ e = u₂ - u₁/v₁ - v₂ e = v₁ - v₂/u₂ - u₁ ANSWER DOWNLOAD EXAMIANS APP
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