1. a 6.3 kg bowling ball and a 7.1 kg bowling ball rest on a rack 0.85 m apart. (a) what is the force of…

Explanation:

Step1: Identify the formula

The gravitational - force formula is $F = G\frac{m_1m_2}{r^2}$, where $G = 6.67\times10^{- 11}\ N\cdot m^2/kg^2$, $m_1$ and $m_2$ are the masses of the two objects, and $r$ is the distance between them.

Step2: Calculate the force in part (a)

Given $m_1 = 6.3\ kg$, $m_2 = 7.1\ kg$, and $r = 0.85\ m$. Substitute the values into the formula: $F=G\frac{m_1m_2}{r^2}=6.67\times 10^{-11}\frac{6.3\times7.1}{(0.85)^2}$ $F = 6.67\times10^{-11}\frac{44.73}{0.7225}$ $F=6.67\times10^{-11}\times61.91$ $F = 4.13\times10^{-9}\ N$

Step3: Solve for $r$ in part (b)

We know $F = 2.0\times10^{-9}\ N$, $m_1 = 6.3\ kg$, $m_2 = 7.1\ kg$, and $G = 6.67\times10^{-11}\ N\cdot m^2/kg^2$. From $F = G\frac{m_1m_2}{r^2}$, we can re - arrange it to get $r^2=G\frac{m_1m_2}{F}$. Substitute the values: $r^2 = 6.67\times10^{-11}\frac{6.3\times7.1}{2.0\times10^{-9}}$ $r^2=6.67\times10^{-11}\frac{44.73}{2.0\times10^{-9}}$ $r^2=6.67\times10^{-11}\times22365$ $r^2 = 0.149$ $r=\sqrt{0.149}\approx0.386\ m$

Answer:

(a) $4.13\times10^{-9}\ N$ (b) $0.386\ m$