You are standing on a bathroom scale in an elevator in a tall building. Your mass is 64 kg. The elevator starts from rest and travels upward

Question

You are standing on a bathroom scale in an elevator in a tall building. Your mass is 64 kg. The elevator starts from rest and travels upward with a speed that varies with time according to v(t) = (3.0 m/s2)t + (0.20 m/s3)t2. When t = 4.0 s, what is the reading on the bathroom scale?

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Nem 1 year 2021-09-05T01:00:16+00:00 1 Answers 2 views 0

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    2021-09-05T01:01:41+00:00

    Answer:

    = 922N

    Explanation:

    m = 64kg

    v(t) = (3.0 m/s²)t + (0.20 m/s³)t².

    t = 4.0s

    differentiation of v(t) so the acceleration is given by

    [tex]a(t) = \frac{ dv(t)}{dt} \\ = 3.0 + 0.4t\\ t = 4s\\a(4) = 3.0 + 0.4(4)\\= 4.6ms^-^2[/tex]

    using Newton’s second law

    F (net) = ΣFg = ma

    ⇒R – w = ma

    R = m(a + g)

     = 64 (9.81 + 4.6)

     = 922N

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