Answer:
D. two positively charged objects

Explanation:
The acceleration due to gravity g is defined as

and solving for R, we find that

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration
experienced by the satellite is equal to the gravitational force
or

The orbital velocity <em>v</em> is the velocity of the satellite around the planet defined as

where <em>r</em><em> </em>is the radius of the satellite's orbit in meters and <em>T</em> is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

Solving for <em>M</em>, we get

Putting this expression back into Eqn(1), we get




Hi! The answer is ‘B’! Because the nucleus is found at the center and contains protons (positive charge) and neutrons (no charge)
I think the answer would be 70
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<h3>What is a survey?</h3>
A survey is a useful tool based on population samples, which is used to make statistical analyses in a given investigation.
Differences in surveys are generally due to small sample sizes, which may lead to errors in the analysis of data.
In conclusion, the article is not found here but surveys are important because they are representative samples of a population.
Learn more about surveys here:
brainly.com/question/13624055
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