Applied Mechanics and Graphic Statics The shape of a suspended cable for a uniformly distributed load over it is Catenary Circular Parabolic Cubic parabola Catenary Circular Parabolic Cubic parabola ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics Minimum potential energy of a system will be in the position of Stable equilibrium Neutral equilibrium Unstable equilibrium All of these Stable equilibrium Neutral equilibrium Unstable equilibrium All of these ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics The following is not a law of static friction: The magnitude of the limiting friction bears a constant ratio to the normal reaction between two surfaces The force of friction is dependent upon the area of contact The force of friction always acts in a direction opposite to that in which the body tends to move The force of friction depends upon the roughness of the surface The magnitude of the limiting friction bears a constant ratio to the normal reaction between two surfaces The force of friction is dependent upon the area of contact The force of friction always acts in a direction opposite to that in which the body tends to move The force of friction depends upon the roughness of the surface ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics A particle moves with a velocity of 2 m/sec in a straight line with a negative acceleration of 0.1 m/sec2. Time required to traverse a distance of 1.5 m, is 40 sec 20 sec 30 sec 15 sec 40 sec 20 sec 30 sec 15 sec ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics If ‘v’ and ‘ω’ are linear and angular velocities, the centripetal acceleration of a moving body along the circular path of radius ‘r’, will be r/ω² v²/r ω²/r r/v² r/ω² v²/r ω²/r r/v² 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 = 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 ANSWER DOWNLOAD EXAMIANS APP
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