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KATRIN_1 [288]
3 years ago
10

A large pendulum with a 200-lb gold plated bob 12 inches in diameter is on display in the lobby of the United Nations building.

The pendulum has a length of 75 ft. It is used to show the rotation of the Earth—for this reason it is referred to as a Foucault pendulum.Part A) If the pendulum were to be taken to the Moon, where the acceleration of gravity is g/6, would the period increase, decrease, or stay the same?
Part B) Check your result in part A by calculating the period of the pendulum on the Moon.
Physics
1 answer:
xeze [42]3 years ago
3 0

Answer:

9.59143 s

23.4941 s

Explanation:

L = Length of pendulum = 75 ft

g = Acceleration due to gravity = 9.81 m/s²

Time period is given by

T=2\pi \sqrt{\dfrac{L}{g}}

On Earth

T=2\pi \sqrt{\dfrac{75\times 0.3048}{9.81}}\\\Rightarrow T=9.59143\ s

The time period on Earth is 9.59143 s

On Moon

T=2\pi \sqrt{\dfrac{75\times 0.3048}{\dfrac{9.81}{6}}}\\\Rightarrow T=23.4941\ s

The time period on Moon is 23.4941 s

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Answer:

V = 0.248 L

Explanation:

To do this, use the following equation:

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This equation is used to find a relation between two differents conditions of a same gas, which is this case. From this equation we can solve for V2.

Solving for V2:

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Temperature must be at Kelvin, so, we have to sum the temperature 273 to convert it in K.

Replacing the data we have:

V2 = 1 * 4.91 * (-196+273) / 5.2 * (20+273)

V2 = 378.07 / 1523.6

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