After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 50.0 cm. She finds that the pendulum makes 10

Question

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 50.0 cm. She finds that the pendulum makes 100 complete swings in 136 s. What is the value of g on this planet?

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Eirian 2 months 2021-07-31T14:38:20+00:00 2 Answers 5 views 0

Answers ( )

    0
    2021-07-31T14:39:45+00:00

    Answer: 10.67 m/s²

    Explanation:

    Given, l = 50 cm = 0.5 m

    T = 2π √(L / g) , where

    L = length of spring

    g = acceleration due to gravity

    T = period

    T can also be gotten using the formula,

    T = total number of oscillations / time taken, so that

    T = 136 / 100

    T = 1.36

    T = 2π √(L / g)

    1.36 = 2 * π * √(0.5 / g)

    1.36 = 6.284 * √(0.5 / g)

    1.36² = 6.284² * (0.5 / g)

    1.85 = 39.49 * 0.5 / g

    g = 19.745 / 1.85

    g = 10.67 m/s²

    Thus, we can infer that the value of g on this planet is 10.67 m/s²

    0
    2021-07-31T14:40:10+00:00

    Answer:

    g = 10.67 m/s²

    Explanation:

    The period of a simple pendulum is given as;

    T = 2π√(L / g)

    Where;

    T = period

    L = length of spring

    g = acceleration due to gravity

    Now, in the question we want to find g and we are given L but not given the period.

    We are given

    So let’s find the period (T)

    We are given total number of oscillations and time taken to complete those oscillations. Thus, we can express period as;

    T = time taken/total number of oscillations

    So, T = 136/100 = 1.36

    Thus,

    1.36 = 2π√(L / g)

    1.36/2π = √(L / g)

    Taking the square of both sides, we have;

    (1.36/2π)² = L/g

    0.04685 = L/g

    We are given that; L = 50cm = 0.5m

    So,

    0.04685 = 0.5/g

    g = 0.5/0.04685

    g = 10.67 m/s²

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