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Aloiza [94]
3 years ago
15

A simple pendulum is 1.15 meters long and has a period of 6.29 seconds. The pendulum is on an unknown planet. What is the accele

ration of gravity of the Unknown Planet
Physics
1 answer:
Nastasia [14]3 years ago
7 0

Answer:

The value of the acceleration of gravity of the Unknown Planet = 1.14 \frac{m}{s^{2} }

Explanation:

length of the pendulum (L)= 1.15 m

Time period (T)= 6.29 seconds

We know that time period  of a simple pendulum is given by

⇒ T = 2\pi × \sqrt\frac{L}{g}

put the values in the above formula we get

⇒ T = 2\pi × \sqrt \frac{1.15}{g}

⇒ 6.29 = 2\pi × \sqrt \frac{1.15}{g}

By solving the above equation we get

⇒ g = 1.14 \frac{m}{s^{2} }

This is the value of the acceleration of gravity of the Unknown Planet.

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A rectangular block floats in pure water with 0.400 in. above the surface and 1.60 in. below the surface. When placed in an aque
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Answer:

specific gravity = 0.8

specific gravity of  solution  = 2

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given data

rectangular block above water  = 0.400 in

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solution

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put here value we get

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3 years ago
A student attaches a block to a vertical spring of unknown spring constant so that the block-spring system will oscillate if the
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A mother and daughter are on a seesaw in the park. How far from the center must the 160.9lb mother sit in order to balance the 6
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Answer:

The mother has to sit 2.17 ft from the center on the other side of the seesaw.

Explanation:

We are trying to find the sum of torques given by the weights of mother and daughter to be zero.

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we need that the absolute value of the torque exerted by the mom (160.9 lb) to be the same in magnitude (and of course opposite direction). So we assume that "d" is the distance at which the mother locates to make this torque equal in magnitude to her daughter's torque:

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d = 2.17 ft

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3 years ago
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