Engineering Mechanics The power developed by a body acted upon by a torque 'T' Newton meter (N - m) and revolving at ω radian/s is given by T.ω (in watts) T.ω/4500 (in kilowatts) T.ω/75 (in kilowatts) T.ω/60 (in watts) T.ω (in watts) T.ω/4500 (in kilowatts) T.ω/75 (in kilowatts) T.ω/60 (in watts) ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The acceleration of a particle moving with simple harmonic motion, at any instant is given by ω2.y ω3.y ω.y ω2/y ω2.y ω3.y ω.y ω2/y ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The loss of kinetic energy during inelastic impact, is given by (where m1 = Mass of the first body,m2 = Mass of the second body, and u1 and u2 = Velocities of the first and second bodies respectively.) [2(m₁ + m₂)/m₁ m₂] (u₁² - u₂²) [m₁ m₂/2(m₁ + m₂)] (u₁ - u₂)² [m₁ m₂/2(m₁ + m₂)] (u₁² - u₂²) [2(m₁ + m₂)/m₁ m₂] (u₁ - u₂)² [2(m₁ + m₂)/m₁ m₂] (u₁² - u₂²) [m₁ m₂/2(m₁ + m₂)] (u₁ - u₂)² [m₁ m₂/2(m₁ + m₂)] (u₁² - u₂²) [2(m₁ + m₂)/m₁ m₂] (u₁ - u₂)² ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The velocity ratio for the third system of pulleys is 2n - 1 n² n 2n 2n - 1 n² n 2n ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics If ‘P’ is the force acting on the body, ‘m’ is the mass of the body and ‘a’ is the acceleration of the body, then according to Newton's second law of motion, P - m.a = 0 P × m.a = 0 P/m.a = 0 P + m.a = 0 P - m.a = 0 P × m.a = 0 P/m.a = 0 P + m.a = 0 ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The term 'Centroid' is The same as centre of gravity The point of suspension The point of application of the resultant of all the forces tending to cause a body to rotate about a certain axis None of the listed here The same as centre of gravity The point of suspension The point of application of the resultant of all the forces tending to cause a body to rotate about a certain axis None of the listed here ANSWER DOWNLOAD EXAMIANS APP
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