A student is studying the potential energy change of a 50 kg object raised 110 km above Earth’s surface. What will be the percentage error i

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A student is studying the potential energy change of a 50 kg object raised 110 km above Earth’s surface. What will be the percentage error if he simply used the approximate relation ΔU = mgΔy? Hint

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Ngọc Diệp 5 years 2021-08-04T12:35:34+00:00 1 Answers 108 views 0

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  1. kimchi2
    0
    2021-08-04T12:37:04+00:00
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    Answer:

    The percentage error is given by 99.9 %

    Explanation:

    Given:

    Mass of object m = 50 kg

    Height h = 110 km

    From the formula of potential energy,

       U = mgh

    Where g = 9.8 \frac{m}{s}

       U = 50\times 9.8 \times 110000

    Here true value of potential energy,

       U = 50 \times 9.8 \times 11 \times 10^{4}

    Approximate value of student,

       U = 50 \times 9.8 \times 110

    Absolute error is given by

        = true value – approximate value

        = 50 \times 9.8 \times 11 \times 10^{4} - 50 \times 9.8 \times 110

        = 53846100

    Hence percentage error,

        = \frac{53846100}{50 \times 9.8 \times 11 \times 10^{4} }

        = 0.999 \times 100 %

        = 99.9 %

    Therefore, the percentage error is given by 99.9 %

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