A car starts from rest and moves around a circular track of radius 27.0 m. Its speed increases at the constant rate of 0.420 m/s2. (a) What

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A car starts from rest and moves around a circular track of radius 27.0 m. Its speed increases at the constant rate of 0.420 m/s2. (a) What is the magnitude of its net linear acceleration 13.0 s later

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Ngọc Diệp 4 years 2021-09-02T01:43:39+00:00 1 Answers 6 views 0

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  1. sori
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    2021-09-02T01:45:15+00:00
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    Answer:

    1.18 m/s²

    Explanation:

    The linear speed = 13.0s

    v= v_i + at\\v = 0 + 0.42(13)\\v = 5.46m/s

    where v(i) is the initial linear speed

    the radial acceleration of the car afte (t) = 13s is

    a_r = \frac{v^2}{r} \\= \frac{5.46^{2} }{27} \\= 1.104m/s^2

    the magnitude of the net acceleration is

    a_net = \sqrt{1.104^2 + 0.1764^2} \\= 1.18m/s^2

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