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Gnesinka [82]
1 year ago
6

How could you increase the force advantage of a lever?Select one:a. Make the effort length longer.b. Make the effort length shor

ter.c. Make the lifting length longer.d. Make the effort force larger.
Physics
1 answer:
11Alexandr11 [23.1K]1 year ago
4 0

The ideal mechanical advantage of a lever (IMA) is given by:

IMA=\frac{Le}{Lr}

Where:

Le = Effort of the arm

Lr = Resistance arm.

Therefore, we can increase the force adventage by increasing the effort arm or reducing the load arm

Answer:

a. Make the effort length longer.

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The Mistuned Piano Strings Two identical piano strings of length 0.800 m are each tuned exactly to 480 Hz. The tension in one of
IRINA_888 [86]

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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