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nordsb [41]
2 years ago
14

An electrician reads a resistance of 65 ohms and when energize the circuit runs 1.77 amps. what voltage to the system run on?

Physics
1 answer:
Stels [109]2 years ago
3 0

Answer:

115.05 Volts

Explanation:

We can use Ohm's Law!

V=IR

V= 1.77 * 65

V=115.05 Volts

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Complete Question

The complete question is shown on the first uploaded image

Answer:

The answer is

     T_2 = 1.008 % higher than T_1

    T_2 = 0.99 % lower than T_1

Explanation:

   From the question we are told that

         The first string has a frequency of   f_1 = 230 Hz

          The period of the beat is  t_{beat}  = 0.99s

Generally the frequency of the beat is

             f_{beat} = \frac{1}{t_{beat}}

  Substituting values

            f_{beat} = \frac{1}{0.99}

                   = 1.01 Hz

From the question

        f_2 - f_1 = f_{beat}   for  f_2  having a  higher tension

So

       f_2 - 230 = 1.01

               f_2 = 231.01Hz

 From the question

            \frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1} }

         \frac{T_2}{T_1}  = \frac{f_2^2}{f_1^2}

Substituting values

         \frac{T_2}{T_1}  = \frac{(231.01)^2}{(230)^2}

      T_2 = 1.008 % higher than T_1

    For f_2 having a lower tension

           f_1 - f_2 = f_{beat}

  So

       230 - f_2 = 1.01

            f_2 = 230 -1.01

                  = 228.99

  From the question

            \frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1} }

         \frac{T_2}{T_1}  = \frac{f_2^2}{f_1^2}

    Substituting values

         \frac{T_2}{T_1}  = \frac{(228.99)^2}{(230)^2}

      T_2 = 0.99 % lower than T_1        

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