when a temparature of a coin is 75°C, the coin’s diameter increases. if the original diameter of a coin is 1.8*10^-2 m and its coefficient o

Question

when a temparature of a coin is 75°C, the coin’s diameter increases. if the original diameter of a coin is 1.8*10^-2 m and its coefficient of linear expansion is 1.7*10^5/°C, what is the change in coins diameter?​

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Vân Khánh 3 months 2021-07-19T16:37:15+00:00 1 Answers 2 views 0

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    2021-07-19T16:39:02+00:00

    Answer:

    ΔD = 2.29 10⁻⁵ m

    Explanation:

    This is a problem of thermal expansion, if the temperature changes are not very large we can use the relation

              ΔA = 2α A ΔT

    the area is

             A = π r² = π D² / 4

    we substitute

             ΔA = 2α π D² ΔT/4

    as they do not indicate the initial temperature, we assume that ΔT = 75ºC

        α = 1.7 10⁻⁵ ºC⁻¹

    we calculate

              ΔA = 2 1.7 10⁻⁵ pi (1.8 10⁻²) ² 75/4

              ΔA = 6.49 10⁻⁷ m²

    by definition

               ΔA = A_f- A₀

               A_f = ΔA + A₀

               A_f = 6.49 10⁻⁷ + π (1.8 10⁻²)² / 4

               A_f = 6.49 10⁻⁷ + 2.544 10⁻⁴

               A_f = 2,551 10⁻⁴ m²

    the area is

               A_f = π D_f² / 4

               A_f = \sqrt{4  A_f /\pi }

               D_f = \sqrt{4 \ 2.551 10^{-4} /\pi }

               D_f = 1.80229 10⁻² m

    the change in diameter is

               ΔD = D_f – D₀

               ΔD = (1.80229 – 1.8) 10⁻² m

               ΔD = 0.00229 10⁻² m

               ΔD = 2.29 10⁻⁵ m

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