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

The percentage error is given by 99.9 %

Explanation:

Given:

Mass of object kg

Height km

From the formula of potential energy,

Where

Here true value of potential energy,

Approximate value of student,

Absolute error is given by

    = true value - approximate value

    =

    =

Hence percentage error,

    %

    %

Therefore, the percentage error is given by 99.9 %

answer: strength i guess

explanation:

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