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mars1129 [50]
1 year ago
6

Phase unbalance causes three-phase motors to operate at temperatures ? and, therefore, the motor cannot deliver its rated horsep

ower.
Physics
1 answer:
anastassius [24]1 year ago
7 0

Phase unbalance causes three-phase motors to operate at temperatures higher than nameplate ratings and, therefore, the motor cannot deliver its rated horsepower.

<h3>What is an electric vehicle (EV)?</h3>

An EV stands for an electric vehicle. Electric vehicles (EVs) are autos that are powered totally or partially by electricity.

Electric cars are extremely cost-effective to run because they have fewer moving parts to maintain and use little to no fossil fuels (petrol or diesel).

Hence, three-phase motors operating with phase imbalance run at temperatures over nameplate limits, which prevents the motor from producing its rated horsepower.

.

To learn more about electric vehicles here,

brainly.com/question/27434548

#SPJ4

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An interference pattern is produced by light with a wavelength 550 nm from a distant source incident on two identical parallel s
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Answer:

a

 \theta  =  0.0022 rad

b

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Explanation:

From the question we are told that  

   The  wavelength of the light is \lambda  = 550 \ nm  =  550 *10^{-9} \ m

    The  distance of the slit separation is  d = 0.500 \ mm = 5.0 *10^{-4} \ m

 

Generally the condition for two slit interference  is  

     dsin \theta =  m \lambda

Where m is the order which is given from the question as  m = 2

=>    \theta  =  sin ^{-1} [\frac{m \lambda}{d} ]

 substituting values  

      \theta  =  0.0022 rad

Now on the second question  

   The distance of separation of the slit is  

       d =  0.300 \ mm  =  3.0 *10^{-4} \ m

The  intensity at the  the angular position in part "a" is mathematically evaluated as

      I  =  I_o  [\frac{sin \beta}{\beta} ]^2

Where  \beta is mathematically evaluated as

       \beta  =  \frac{\pi *  d  *  sin(\theta )}{\lambda }

  substituting values

     \beta  =  \frac{3.142  *  3*10^{-4}  *  sin(0.0022 )}{550 *10^{-9} }

    \beta  = 0.06581

So the intensity is  

    I  =  I_o  [\frac{sin (0.06581)}{0.06581} ]^2

   I  =  0.000304 I_o

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