Applied Mechanics and Graphic Statics The centre of gravity of a homogeneous body is the point at which the whole Weight of the body is assumed to be concentrated Area of the surface of the body is assumed to be concentrated Volume of the body is assumed to be concentrated All listed here Weight of the body is assumed to be concentrated Area of the surface of the body is assumed to be concentrated Volume of the body is assumed to be concentrated All listed here ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics A car goes round a curve of radius 100 m at 25 m/sec. The angle to the horizontal at which the road must be banked to prevent sideways friction on the car wheels is tan"1 x, where x is (Assume g = 10 m/sec²) 43898 43959 43832 44079 43898 43959 43832 44079 ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics A trolley wire weighs 1 kg per metre length. The ends of the wire are attached to two poles 20 m apart. If the horizontal tension is 1000 kg, the central dip of the cable is 3 cm 5 cm 2 cm 4 cm 3 cm 5 cm 2 cm 4 cm ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics A train weighing 196 tonnes experiences a frictional resistance of 5(11/22) per tonne. The speed of the train at the top of a down gradient 1 in 78.4 is 36 km/hour. The speed of the train after running 1 km down the slope, is 5 √3 m/sec 5 √10 m/sec 10 √5 m/sec 3 √5 m/sec 5 √3 m/sec 5 √10 m/sec 10 √5 m/sec 3 √5 m/sec ANSWER DOWNLOAD EXAMIANS APP
Applied Mechanics and Graphic Statics To double the period of oscillation of a simple pendulum Its length should be doubled The mass of its bob should be doubled The mass of its bob should be quadrupled Its length should be quadrupled Its length should be doubled The mass of its bob should be doubled The mass of its bob should be quadrupled Its length should be quadrupled 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 = u₂ - u₁/v₁ - v₂ 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 ANSWER DOWNLOAD EXAMIANS APP
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