3. A certain wire, 3 m long, stretches by 1.2 mm when under tension of 200 N. By how much does an equally thick wire 6 m long, made of

Question

3. A certain wire, 3 m long, stretches by 1.2 mm when under tension of 200 N. By how much does
an equally thick wire 6 m long, made of the same material and under the same tension, stretch?
​

in progress 0
Nho 3 years 2021-07-14T01:27:44+00:00 1 Answers 33 views 0

Answers ( )

  1. Ben
    0
    2021-07-14T01:29:36+00:00
    Reply

    Answer:

    The extension of the second wire is   e_2 = 0.0024 \ m = 2.4 mm

    Explanation:

    From the question we are told that

        The length of the wire is L = 3 \ m

         The elongation of the wire is  e = 1.2mm = \frac{1.2}{1000} = 0.0012 m

            The tension is F = 200 \ N

           The length of the second wire is  L_2 = 6 \ m

         

    Generally the Young’s modulus(Y) of this material is  

            Y = \frac{stress}{strain }

    Where stress = \frac{F}{A}

        Where A is the area which is evaluated as  

               A = \pi r^2

      and   strain = \frac{extention}{length} = \frac{e}{L}

       So

            Y = \frac{\frac{F}{\pi r^2 } }{ \frac{e}{L} }

    Since the wire are of the same material Young’s modulus(Y)  is constant

    So we have  

                  \frac{F * L }{r^2 e} = \pi * Y = constant

                  F * L = constant * r^2 e

    Now the ration between the first and the second wire is

             \frac{F_1}{F_2} * \frac{L_1}{L_2} = \frac{r*2_1}{r^2} * \frac{e_1}{e_2}

    Since tension , radius are constant

       We have

               \frac{L_1}{L_2} = \frac{e_1}{e_2}

    substituting values

              \frac{3}{6} = \frac{0.0012}{e_2}

              0.5 e_2 = 0.0012

             e_2 = \frac{ 0.0012 }{0.5}

              e_2 = 0.0024 \ m = 2.4 mm

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )