Answered

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Two students are standing next to one another. One student has a mass of 79.0 kg, and the other has a mass of 93.5 kg. If they are standing so that their centers of mass are a distance of 1.12 m apart, what is the force of the gravitational attraction between them?
A. 4.40 * 10^-7 N
B. 3.93 * 10^-7 N
C. 1.29 * 10^-6 N
D. 8.77 * 10^-6 N

Sagot :

Answer:

[tex]F=3.93\times 10^{-7}\ N[/tex]

Explanation:

Mass of student 1, m₁ = 79 kg

Mass of student 2, m₂ = 93.5 kg

The students are 1.12 m apart, d = 1.12 m

We need to find the force of the gravitational attraction between them. The force of gravitational force is given by :

[tex]F=G\dfrac{m_1m_2}{r^2}\\\\F=6.67\times 10^{-11}\times \dfrac{79\times 93.5}{(1.12)^2}\\\\=3.93\times 10^{-7}\ N[/tex]

So, the force of gravitational attraction between them is [tex]3.93\times 10^{-7}\ N[/tex]. Hence, the correct option is (B).

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