Applied Mechanics and Graphic Statics Two parallel forces 20 kg and 15 kg act. In order that the distance of the resultant from 20 kg force may be the same as that of the former resultant was from 15 kg, the 20 kg force is diminished by 8.75 kg 6.25 kg 10.5 kg 5.5 kg 8.75 kg 6.25 kg 10.5 kg 5.5 kg ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics Effect of a force on a body depends upon its Position All of these Direction Magnitude Position All of these Direction Magnitude ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics Free body diagram is an Isolated joint with only body forces acting on it Isolated joint with all the forces, internal as well as external, acting on it None of these Isolated joint with internal forces acting on it Isolated joint with only body forces acting on it Isolated joint with all the forces, internal as well as external, acting on it None of these Isolated joint with internal forces acting on it 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 = 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 e = v₁ - v₂/u₂ - u₁ ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics A ball which is thrown upwards, returns to the ground describing a parabolic path during its flight Kinetic energy of the ball remains constant Speed of the ball remains constant Vertical component of velocity remains constant Horizontal component of velocity remains constant Kinetic energy of the ball remains constant Speed of the ball remains constant Vertical component of velocity remains constant Horizontal component of velocity remains constant ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics If two bodies of masses M1 and M2(M1 > M2) are connected by alight inextensible string passing over a smooth pulley, the tension in the string, will be given by T = g(M2 - M1)/(M1 + M2) T = g(M1 - M2)/(M1 + M2) T = g(M1 + M2)/(M1 × M2) T = g(M2 + M1)/(M2 - M1) T = g(M2 - M1)/(M1 + M2) T = g(M1 - M2)/(M1 + M2) T = g(M1 + M2)/(M1 × M2) T = g(M2 + M1)/(M2 - M1) ANSWER DOWNLOAD EXAMIANS APP
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