The period of a pendulum (for very small swings) can be estimated as ...T = 2 pi (L/G)0.5... so, plugging in 0.024 m for L, and 9.81 m s-2 for G, we get L = 0.31 seconds.
Use the formula T = 2Pi * Square root (L)/ Square root (g) Set T to .75; L is length of string and g is gravity (9.8 m/s)
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S: actually, there is no S in the roman's counting system. D: 500 L: 50 M: 1000 X: 10 V: 5
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I would solve this problem by starting from the equation for a pendulum's period:T ≈ 2π√( L / g )Rearranging for g, and substituting the length in Cambridge:g ≈ ( 4π² * L ) / T². . = ( 4π * 0.9942 m ) / ( 1.000 s + 1.000 s )². . = 9.812 m/s²And for Tokyo:g ≈ ( 4π² * L ) / T². . = ( 4π * 0.9927 m ) / ( 1.000 s + 1.000 s )². . = 9.798 m/s²
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The period of a pendulum (for very small swings) can be estimated as ...T = 2 pi (L/G)0.5... so, plugging in 0.024 m for L, and 9.81 m s-2 for G, we get L = 0.31 seconds.
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