[tex]t = 2\pi \sqrt{ \frac{48}{9.8} } [/tex]
FIND T(TIME PERIOD)
FROM PHYSICS
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[tex]t = 2\pi \sqrt{ \frac{48}{9.8} } [/tex]
FIND T(TIME PERIOD)
FROM PHYSICS
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Answer:
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Explanation:
The formula for the period of a simple pendulum is given by:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where:
- \( T \) is the period of the pendulum,
- \( \pi \) is a mathematical constant (approximately 3.14159),
- \( L \) is the length of the pendulum,
- \( g \) is the acceleration due to gravity.
In your case:
- \( L = 48 \),
- \( g = 9.8 \).
So, substitute these values into the formula:
\[ T = 2\pi \sqrt{\frac{48}{9.8}} \]
Now, calculate the value:
\[ T \approx 6.157 \]
Therefore, the time period \( T \) for a pendulum with a length of 48 meters and in a gravitational field of \( 9.8 \ \text{m/s}^2 \) is approximately \( 6.157 \) seconds.