Answer:
4.6 years
Explanation:
This is solved using Kepler's third law which says:

Where
T = Orbital period of the planet (in seconds)
a = Distance from the star (in meters)
G = Gravitational constant
M = Mass of the parent star (in kg)
From the information given



We put this into Kepler's law and get:

This when converted to years is 4.6 years.
Answer:
Explanation:
As spring season is a yearly phenomenon so, Rita should organize her data on yearly basis. Firstly, she should plan the procedure of her experiment and collect the data according to it. Secondly, identify the attribute of each object of her experiment. Thirdly, she can organize and segregate her data in tabular form, graphical form or diagrammatically.
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The object does not move.