After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 55.0 cm. The explorer finds that the pendulum

Question

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 55.0 cm. The explorer finds that the pendulum completes 93.0 full swing cycles in a time of 129 s.

What is the value of the acceleration of gravity on this planet?

in progress 0
Thái Dương 5 years 2021-08-10T17:44:44+00:00 1 Answers 17 views 0

Answers ( )

  1. RuslanHeatt
    0
    2021-08-10T17:45:50+00:00
    Reply

    Answer:

    The value of the acceleration of gravity on this planet is 11.4\ m/s^2.  

    Explanation:

    Given that,

    Length of the simple pendulum, l = 55 cm = 0.55 m

    The explorer finds that the pendulum completes 93.0 full swing cycles in a time of 129 s. The time period is given by :

    T=\dfrac{129}{93}=1.38\ s

    We need to find the value of the acceleration of gravity on this planet. The time period in case of simple pendulum is given by :

    T=2\pi \sqrt{\dfrac{l}{g}}\\\\g=\dfrac{4\pi ^2l}{T^2}

    g=\dfrac{4\pi ^2\times 0.55}{(1.38)^2}\\\\g=11.4\ m/s^2

    So, the value of the acceleration of gravity on this planet is 11.4\ m/s^2. Hence, this is the required solution.

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )