The length of a displaced pendulum ball which passes its lowest point twice every seconds is [g = 10m-2]
Question 1 Report
The length of a displaced pendulum ball which passes its lowest point twice every seconds is [g = 10m-2]
Answer Details
This question is asking for the length of a pendulum that passes its lowest point twice every second, given that the acceleration due to gravity (g) is 10m/s^2. The period of a pendulum is given by the formula T = 2π√(l/g), where T is the period, l is the length of the pendulum, and g is the acceleration due to gravity. In this case, the period is 1/2 seconds (since the pendulum passes its lowest point twice every second). We can rearrange the formula to solve for the length of the pendulum, l = g(T/2π)^2. Plugging in the given values, we get l = 10(1/(2π))^2 ≈ 0.25 meters. Therefore, the answer is 0.25 m.
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