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jeka94
1 year ago
6

1)What are the three previous manufacturing revolutions Mr. Scalabre mentions? When did these take place?

Engineering
1 answer:
Ostrovityanka [42]1 year ago
8 0

The three previous manufacturing revolutions that Mr. Scalabre mentioned and their year of occurrence are:

  1. The steam engine in the mid-19th Century
  2. The mass-production model in the early 20th Century
  3. The first automation wave in the 1970s

<h3>What is a Manufacturing Revolution?</h3>

This refers to the process of change from a handicraft economy to industry production-based production.

Hence, we can see that Mr. Scalabre believes we are not growing in productivity because there has not been enough automation to perform the tasks needed.

The effect of robotics is making an impact on productivity because a lot of complex, difficult tasks are done by machines.

3D printing has made an impact on productivity because there is a reduction in the pressing cycle and errors due to negligence are reduced.

The role the engineers have to play in the next revolution is that they would have to produce mathematical model that can be used to produce better AIs

Read more about manufacturing revolutions here:

brainly.com/question/14316656

#SPJ1

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A 500-km, 500-kV, 60-Hz, uncompensated three-phase line has a positivesequence series impedance. z = 5 0.03 1 + j 0.35 V/km and
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A) 282.34 - j 12.08 Ω

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C)

A = 0.812 < 1.09° per unit

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C =  2.061 * 10^-3 < 90.32° s

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Explanation:

Given data :

Z ( impedance ) = 0.03 i  + j 0.35 Ω/km

positive sequence shunt admittance ( Y ) = j4.4*10^-6 S/km

A) calculate Zc

Zc = \sqrt{\frac{z}{y} }  =  \sqrt{\frac{0.03 i  + j 0.35}{j4.4*10^-6 } }    

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hence Zc = 282.34 - j 12.08 Ω

B) Calculate  gl

gl = \sqrt{zy} * d  

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 z = 0.03 i  + j 0.35

 y = j4.4*10^-6 S/km

gl =  \sqrt{0.03 i  + j 0.35*  j4.4*10^-6}  * 500

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   = 0.622 < 87.55 °

gl = 0.0266 + j 0.621 / unit

C) exact ABCD parameters for this line

A = cos h (gl) . per unit  =  0.812 < 1.09° per unit ( as calculated )

B = Zc sin h (gl) Ω  = 164.6 < 85.42°Ω  ( as calculated )

C = 1/Zc  sin h (gl) s  =  2.061 * 10^-3 < 90.32° s ( as calculated )

D = cos h (gl) . per unit = 0.812 < 1.09° per unit ( as calculated )

where :  cos h (gl)  = \frac{e^{gl} + e^{-gl}  }{2}

             sin h (gl) = \frac{e^{gl}-e^{-gl}  }{2}

     

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