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Compute the torque developed by an industrial motor whose output is 175 kw at an angular speed of 4300 rev/min.

Sagot :

The torque developed by an industrial motor whose output is 175 kw at an angular speed of 4300 rev/min will be 389 Nm.

Given

  • Power = 175kw = 17500 W
  • Angular Speed(ω) = 4300 rev/min = 450 rad/s

Let Torque = т

P = т * ω

т = P/ ω

т = 17500 / 450 = 388.88 Nm (approx. 389 Nm)

Hence , the torque developed by an industrial motor whose output is 175 kw at an angular speed of 4300 rev/min will be 389 Nm.

In physics and mechanics, torque is the rotational equivalent of linear force. It is also referred to as the moment, moment of force, rotational force or turning effect, depending on the field of study. It represents the capability of a force to produce change in the rotational motion of the body.

We can define power as the rate of doing work, it is the work done in unit time. The SI unit of power is Watt (W) which is joules per second (J/s). Sometimes the power of motor vehicles and other machines is given in terms of Horsepower (hp), which is approximately equal to 745.7 watts.

Angular speed is the speed of the object in rotational motion.

To learn more about torque : https://brainly.com/question/20691242

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