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Engineering sceince - Math Problem Example

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ENGINEERING SCIENCE By Presented to Institution Monday, June 01, 2015 Task 1 Question 1 Where T is the torque, m is mass and r is the radius Since Then T = I Effective diameter is (external diameter – wall thickness) = (54 – 3) mm =51mm = 0.051m Thus I = 1.9*0.0512 = 4.9419e-3 Acceleration, is given by =3600rpm = 376.991rad/s = 600rpm = 62.83rad/s therefore, =130.90 Thus Torque, T = 130.90*4.9419e-3 since T = I =0.64689471N Question 2 F= where ?…
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Download file to see previous pages 2 =0.5 * 1.5*238.762 =42754.7532 Joules = 42.7547532Kilojoules Question 2 Potential energy, PE = Therefore, PE = 440*9.8*3.5 = 15092N Kinetic energy, KE = = =9295N Total energy = 9295 + 15092 = 24387 joules Assuming all the energy is converted to KE since there is negligible friction, the final velocity can be gotten by = =110.85 Therefore final velocity, = 10.53m/s Task 3 Question 1 When a net force is acting on an object is proportional to the displacement of the object from the equilibrium, a simple harmonic motion occurs. This is as a result of the insignificance of the force acting on the object (in this case it is zero). A sinusoidal curve illustrates, which is presented as a function of time, illustrates the displacement of the object when the force is acting on it (object). Ideologically, this signifies the model of oscillatory motion. This represents ideologically the model of oscillatory motion. ...
Question 2 Angular velocity = 3000rpm = 314.16rad/s Maximum velocity, v =A =0.24*314.16 = 75.4m/s Maximum acceleration is given by 2 =0.24*314.162 = 23685.653376m/s2 =23.683km/s2 Maximum theoretical force F= ma from Newton’s second law 0.28*23685.653376 = 6631.98N Therefore maximum force is ...Download file to see next pagesRead More
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