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Nesterboy [21]
3 years ago
7

The period of a simple pendulum in a grandfather clock on another planet is 1.80 s. What is the acceleration due to gravity (in

m/s2) on this planet if the length of the pendulum is 1.00?
Physics
1 answer:
weeeeeb [17]3 years ago
4 0

Answer:

12.17 m/s²

Explanation:

The formula of period of a simple pendulum is given as,

T = 2π√(L/g)........................ Equation 1

Where T = period of the simple pendulum, L = length of the simple pendulum, g = acceleration due to gravity of the planet. π = pie

making g the subject of the equation,

g = 4π²L/T²................... Equation 2

Given: T = 1.8 s, l = 1.00 m

Constant: π = 3.14

Substitute into equation 2

g = (4×3.14²×1)/1.8²

g = 12.17 m/s²

Hence the acceleration due to gravity of the planet = 12.17 m/s²

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3 years ago
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1)  Q ’= 8 Q ,  2)    q ’= 16 q ,  3)   r ’= ¾ r

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For this exercise we will use Coulomb's law

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