Four rods that obey Hooke’s law are each put under tension. (a) A rod 50.0 cm 50.0 cm long with cross-sectional area 1.00 mm 2 1.00 mm2 and

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Four rods that obey Hooke’s law are each put under tension. (a) A rod 50.0 cm 50.0 cm long with cross-sectional area 1.00 mm 2 1.00 mm2 and with a 200 N 200 N force applied on each end. (b) A rod 25.0 cm 25.0 cm long with cross-sectional area 1.00 mm 2 1.00 mm2 and with a 200 N 200 N force applied on each end. (c) A rod 20.0 cm 20.0 cm long with cross-sectional area 2.00 mm 2 2.00 mm2 and with a 100 N 100 N force applied on each end.

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Kim Chi 5 days 2021-07-22T19:43:18+00:00 1 Answers 0 views 0

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    2021-07-22T19:44:45+00:00

    Answer:

    Four rods that obey Hooke’s law are each put under tension. (a) A rod 50.0 cm long with cross-sectional area 1.00 mm2 and with a 200 N force applied on each end. A rod 25.0 cm long with cross-sectional area 1.00 mm2 and with a 200 N force applied on each end. .00 mm2 and with a 100 N on each end. (c) A rod 20.0 cm long with cross-sectional area 2 force applied Order the rods according to the tensile stress on each rod, from smallest to largest.

    Explanation:

    Tensile stress is given as =

    σ = Force/Area

    A. Given that

    Length of rod = 50cm

    Cross-sectional Area A = 1mm²

    A = 1 × 10^-6m²

    Force applied F = 200N

    Then, σ = F/A

    σ = 200/1×10^-6

    σ = 200 × 10^6 N/m²

    σ = 200 Mpa

    B. Given that,

    Length of rod = 25cm.

    Cross-sectional Area A = 1mm²

    A = 1 × 10^-6m²

    Force applied F = 200N

    Then, σ = F/A

    σ = 200/1×10^-6

    σ = 200 × 10^6 N/m²

    σ = 200 Mpa

    C. Given that,

    Length of rod = 20cm.

    Cross-sectional Area A = 2mm²

    A = 2 × 10^-6m²

    Force applied F = 100N

    Then, σ = F/A

    σ = 100/2×10^-6

    σ = 50 × 10^6 N/m²

    σ = 50 Mpa

    It is noticed that

    50Mpa < 20Mpa =200Mpa

    Then, C < B = A

    The tensile strength of rod A and B are equal and their tensile strength is greater than that of Rod C

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