8) newton’s law of gravitation states that any two bodies attract each other with a force that is inversely…
8) newton’s law of gravitation states that any two bodies attract each other with a force that is inversely proportional to the square of the distance between them. the force of attraction between two bodies is 450×10^(-12) n when they are 5 cm apart. what is the force on the bodies when they are 23 cm apart? express the answer in engineering notation.
Answer
Explanation:
Step1: Write the inverse - square law formula
According to Newton's law of gravitation, $F_1d_1^{2}=F_2d_2^{2}$, where $F_1$ and $d_1$ are the initial force and distance, and $F_2$ and $d_2$ are the final force and distance.
Step2: Identify the given values
We have $F_1 = 450\times10^{- 12}\text{ N}$, $d_1 = 5\text{ cm}=0.05\text{ m}$, and $d_2 = 23\text{ cm}=0.23\text{ m}$.
Step3: Solve for $F_2$
From $F_1d_1^{2}=F_2d_2^{2}$, we can express $F_2$ as $F_2=\frac{F_1d_1^{2}}{d_2^{2}}$. Substitute the values: [ \begin{align*} F_2&=\frac{450\times 10^{-12}\times(0.05)^{2}}{(0.23)^{2}}\ &=\frac{450\times10^{-12}\times0.0025}{0.0529}\ &=\frac{1.125\times10^{-12}}{0.0529}\ &\approx21.3\times10^{-12}\text{ N}\ & = 2.13\times10^{-11}\text{ N} \end{align*} ]
Answer:
$2.13\times 10^{-11}\text{ N}$