3 years ago

How long is a pendulum with a period of 1.0 S on the moon which has 1/6 of the earths gravity

1 answer:

7 0

For small deflections, T = 2*pi*sqrt(L / g) where T is period, L is length and g is gravity. Setting the equations to the same period, 2*pi*sqrt(3.85 / g) = 2*pi*(L / (1/6 * g)) The equation reduces to 3.85 m = 6 * L so L = 0.642 m chrsclrk · 7 years ago

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AlladinOne [14]

By the formula of torque balance we know that

$\tau = r\times F$

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So here we can say

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so output force will increase in this case

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nata0808 [166]

Answer:

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Tasya [4]

uhm i think i don’t understand what it says but i can answer if it was English ! i also tried to translate it but it didn’t help !

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