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nignag [31]
2 years ago
13

How is energy transferred through a generator that produces electric current? A.from armature to commutator to brushes to turbin

e
B.from armature to slip rings to brushes to turbine
C.from turbine to armature to slip rings to brushes
D.from turbine to brushes to slip rings to armature
THIS IS SCIENCE
Physics
2 answers:
vaieri [72.5K]2 years ago
7 0
Armature to commutator to brushes to turbine
bagirrra123 [75]2 years ago
6 0
HEY THERE!!

Fully correct answer is A.from armature to commutator to brushes to turbine.

HOPE IT HELPS YOU.
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An electron is projected with an initial speed of 5 3.2 10 / ⋅ m s directly toward a proton that is fixed in place. If the elect
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The distance is d =1.66*10^{-9}m

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From the question we are told that

         The initial speed of the  electron is v_i  = 3.2 *10^5 m/s

         The mass of electron is m = 9.11*10^{-31}kg

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         Let KE_i be the initial kinetic energy of the electron \

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  The energy at the initial position of the electron = The energy at the final position of the electron

      i.e

             KE_i +U_1 = KE_d + U_2

U_1 \ and \ U_2 are the potential energy at the initial  position of the electron and at distance d of the electron to the proton

                Here U_1 = 0

So the equation becomes

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           v_d is the velocity at distance d from the proton = 2v_i

  So the equation becomes

             \frac{1}{2}mv_i^2 = \frac{1}{2} m (2v_i)^2 -\frac{k(q)^2}{d}

            \frac{1}{2} mv_i^2  = 4 [\frac{1}{2}mv_i^2 ]- \frac{k(q)^2}{d}

           3[\frac{1}{2}mv_i^2 ] = \frac{k(q)^2}{d}

Making d the subject of the formula

           d = \frac{2k(q)^2}{3mv_i^2}

              = \frac{2* 8.99*10^9 *(1.602*10^{-19}^2)}{3 * 9.11*10^{-31} *(3.2*10^5)^2}

              =1.66*10^{-9}m

             

           

         

                 

   

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