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tresset_1 [31]
3 years ago
8

A pendulum is swinging back and forth with a period of 2.0 seconds here on Earth. This pendulum is then brought to the Moon, whe

re the acceleration due to gravity is much smaller. What will happen to the period of the pendulum, assuming everything else about it (mass, length, initial swing height, etc) remains exactly the same? Explain your answer.
Physics
1 answer:
kompoz [17]3 years ago
7 0

Answer:

A greater period.

Explanation:

The period of a simple pendulum is represented by the following formula:

T = 2\pi\cdot \sqrt{\frac{l}{g} }

Where:

l - Length of the pendulum

g - Gravity constant

Given that gravity constant in the Moon is less than on Earth, period shall be greater.

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Answer:

(a) A = 3.90 \AA

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Solution:

As per the question:

Radius of atom, r = 1.95 \AA = 1.95\times 10^{- 10} m

Now,

(a) For a simple cubic lattice, lattice constant A:

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(b) For body centered cubic lattice:

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A = 2\times \frac{4}{\sqrt{3}}\times 1.95 = 9.02 \AA

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