The period of a pendulum is given by

where L is the pendulum length and g is the gravitational acceleration.
We can write down the ratio between the period of the pendulum on the Moon and on Earth by using this formula, and we find:

where the labels m and e refer to "Moon" and "Earth".
Since the gravitational acceleration on Earth is

while on the Moon is

, the ratio between the period on the Moon and on Earth is
Answer:



Explanation:
Given:
- volume of liquid content in the can,

- mass of filled can,

- weight of empty can,

<u>So, mass of the empty can:</u>



<u>Hence the mass of liquid(soda):</u>



<u>Therefore the density of liquid soda:</u>
(as density is given as mass per unit volume of the substance)


<u>Specific weight of the liquid soda:</u>



Specific gravity is the density of the substance to the density of water:

where:
density of water


Fluorine it has a atomic radius of 42
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