Applied Mechanics and Graphic Statics The shape of a suspended cable under its own weight, is Parabolic Circular Elliptical Catenary Parabolic Circular Elliptical Catenary ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics The unit of impulse, is kg.m/sec kg.m/sec kg.m/sec² kg.m/sec kg.m/sec kg.m/sec kg.m/sec² kg.m/sec ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics The dimensions of power are. M’L’r2 M’L2T2 M’L’T3 M’L-‘T* M’L’r2 M’L2T2 M’L’T3 M’L-‘T* ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics A weight of 100 kg is supported by a string whose ends are attached to pegs ‘A’ and ‘B’ at the same level shown in below figure. The tension in the string is 120 kg 50 kg 100 kg 750 kg 120 kg 50 kg 100 kg 750 kg 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 = v₁ - v₂/u₂ + u₁ e = v₁ - v₂/u₂ - u₁ e = u₂ - u₁/v₁ - v₂ e = v₂ - v₁/u₁ - u₂ e = v₁ - v₂/u₂ + u₁ e = v₁ - v₂/u₂ - u₁ e = u₂ - u₁/v₁ - v₂ 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(M2 + M1)/(M2 - M1) T = g(M1 + M2)/(M1 × M2) T = g(M1 - M2)/(M1 + M2) T = g(M2 - M1)/(M1 + M2) T = g(M2 + M1)/(M2 - M1) T = g(M1 + M2)/(M1 × M2) T = g(M1 - M2)/(M1 + M2) ANSWER DOWNLOAD EXAMIANS APP
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