**Question:**Two masses are attracted by a gravitational force of 9.3 N. What will the force of attraction be if the distance between the two masses is quadrupled? (Round the final answer to four decimal places.)**Incorrect Response Given:**The force of attraction will be \(148.8\) N.**Explanation:**The gravitational force between two masses is governed by Newton's Law of Universal Gravitation, which is mathematically expressed as:\[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \]Where:- \( F \) is the gravitational force between the masses,- \( G \) is the gravitational constant,- \( m_1 \) and \( m_2 \) are the masses,- \( r \) is the distance between the centers of the two masses.When the distance \( r \) is quadrupled, the force \( F \) becomes:\[ F' = \frac{G \cdot m_1 \cdot m_2}{(4r)^2} = \frac{G \cdot m_1 \cdot m_2}{16r^2} = \frac{1}{16} \cdot F \]So the new force \( F' \) is \( \frac{1}{16} \) of the original force.Given the original force \( F = 9.3 \) N, the new force is:\[ F' = \frac{9.3}{16} \approx 0.5813 \, \text{N} \]

Gravitational force = Gm1 m2 /r2

Answered: d by a gravitational force of 9.3 N.…

d by a gravitational force of 9.3 N. What will the force of attraction be if the distance between the two masses e final answer to four decimal places.) ill be 148.8 ON.

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Two masses are attracted by a gravitational force of 9.3 N. What will the force of attraction be if the distance between the two masses is quadrupled? The force of attraction will be ? N.
the answer should be in expanded form