In a simple pendulum experiment, the acceleration due to gravity was determined to be $9.67 \\, \\text{ms}^{-2}$. For $T = 2\\pi\\sqrt{\\frac{L}{g}}$, what is the slope of $T^2$ vs $L$?
University of Ilorin PHY107 2021 past question
- 4.08
- 9.67
- 2.42
- 39.5
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