Q5 PE - College Physics (Urone)

Question

Verify that a \(2.3 \times {10^{17}}\,{\rm{km}}\) mass of water at normal density would make a cube \(60\,{\rm{km}}\) on a side, as claimed in Example\({\rm{31}}{\rm{.1}}\). (This mass at nuclear density would make a cube \(1.0\,{\rm{m}}\) on a side.)

Step-by-Step Solution

Answer

It is verified that the side of the cube approximately will be: \(60\,{\rm{km}}\).

Step 1: Define Radioactivity

The spontaneous emission of radiation in the form of particles or high-energy photons as a result of a nuclear process is known as radioactivity.

Step 2: Evaluating the side of the cube

The density of water is:

\(p = 1000\,{\rm{kg/}}{{\rm{m}}^{\rm{3}}}\)

The mass of the water is:

\(m = 2.3 \times {10^{17}}\,{\rm{kg}}\)

The volume of water is then obtained as:

\(\begin{align}V &= \frac{m}{p}\\ &= \frac{{2.3 \times {{10}^{17}}\,kg}}{{1000\,kg/{m^3}}}\\ &= {\rm{ }}2.3 \times {10^{14}}\,{m^3}\end{align}\)

The side of cube having this volume is obtained as:

\(\begin{align}l &= \sqrt(3){V}\\ &= \sqrt(3){{2.3 \times {{10}^{14}}\,{m^3}}}\\ &= 6.13 \times {10^4}\,m\\ &= 61\,km\end{align}\)

Therefore, the side of the cube approximately is: \(60\,{\rm{km}}\).