Applied Mechanics and Graphic Statics The centre of gravity of the trapezium as shown in below figure from the side is at a distance of (h/3) × [(b + 2a)/(b + a)] (h/3) × [(2b + a)/(b + a)] (h/2) × [(b + 2a)/(b + a)] (h/2) × [(2b + a)/(b + a)] (h/3) × [(b + 2a)/(b + a)] (h/3) × [(2b + a)/(b + a)] (h/2) × [(b + 2a)/(b + a)] (h/2) × [(2b + a)/(b + a)] ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics Newton’s Law of Motion is: The rate of change of momentum is directly proportional to the impressed force, and takes place in the same direction, in which the force acts All listed here Everybody continues in its state of rest or of uniform motion in a straight line, unless it is acted upon by some external force To every action, there is always an equal and opposite reaction The rate of change of momentum is directly proportional to the impressed force, and takes place in the same direction, in which the force acts All listed here Everybody continues in its state of rest or of uniform motion in a straight line, unless it is acted upon by some external force To every action, there is always an equal and opposite reaction ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics A cube on a smooth horizontal surface Can be in any of these states Cannot be in stable equilibrium Cannot be in neutral equilibrium Cannot be in unstable equilibrium Can be in any of these states Cannot be in stable equilibrium Cannot be in neutral equilibrium Cannot be in unstable equilibrium ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics In a simple screw jack, the pitch of the screw is 9 mm and length of the handle operating the screw is 45 cm. The velocity ratio of the system is 25 5 314 1.5 25 5 314 1.5 ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics One Joule is equivalent to 1 kg wt metre 1 Newton metre 1 dyne metre 9.81 Newton metre 1 kg wt metre 1 Newton metre 1 dyne metre 9.81 Newton metre 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 40 kg 60 kg 10 kg 100 kg 40 kg 60 kg 10 kg 100 kg ANSWER DOWNLOAD EXAMIANS APP
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