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Mariulka [41]
4 years ago
14

Rotor windings in wound rotor motors are connected to ? on the shaft of the motor. Brushes allow those windings to be connected

to external resistors to control the speed of the motor.
Physics
1 answer:
Mashutka [201]4 years ago
6 0

Answer:

Slip Rings

Explanation:

The wound rotor motor has a three-phase winding with each one connected to seperate slip rings. These slip rings contain brushes which form a secondary circuit where resistance can be inserted and this will allow for the rotor current to run more in phase with the stator current which will result in increased torque that is created

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The answer is the 3rd option!

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Which of the following statements are true of an object in orbit around Earth? (Select all that apply.) The gravity force on the
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Answer:

All these statement are true

Explanation:

Gravity will be acting like a centripetal force for the circular motion of object around earth, which makes it perpendicular to the velocity vector. In the case of elliptical motion, gravity can still be divided into 2 vectors, one parallel and the other perpendicular to the velocity. At the nearest point in elliptical motion, gravity is directly perpendicular to velocity just like in circular motion. At the farthest point, the potential energy is minimized and has been converted into kinetic energy. Therefore at this point the speed is greatest.

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4 years ago
A single slit of width 0.3 mm is illuminated by a mercury light of wavelength 254 nm. Find the intensity at an 11° angle to the
MArishka [77]

Answer:

The  the intensity at an 11° angle to the axis in terms of the intensity of the central maximum is  

   I_c = \frac{I}{I_o}  =8.48 *10^{-8}

Explanation:

From the question we are told that

   The  width of the slit is  D  =  0.3 \ mm =  0.3 *10^{-3} \ m

    The  wavelength is  \lambda =  254 \ nm =  254 *10^{-9} \ m

     The angle is  \theta  =  11^o

The intensity of at 11^o to the axis in terms of the intensity of the central maximum. is mathematically represented as

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

Where \beta is mathematically represented as

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

substituting values

      \beta  =  \frac{0.3 *10^{-3} sin (11 ) * 3.142}{254 *10^{-9} }

     \beta  =  708.1 \ rad

So

  I_c = \frac{I}{I_o}  = [ \frac{sin (708.1)  }{(708.1)}] ^2

   I_c = \frac{I}{I_o}  =8.48 *10^{-8}

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