Physics—Mechanics | Chapter 1: Units, Physical Quantities, and Vectors | Neptunium. In the fall

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Physics Mechanics: Density, Volume, and Critical Mass of Neptunium

This lesson begins the mechanics portion of the physics crash course series. The objective is not simply obtaining an answer but demonstrating a structured problem solving process that students should eventually learn to perform independently during examinations and professional work. :contentReference[oaicite:0]{index=0}

The Problem

The problem involves a sphere of neptunium with a critical mass of 60 kilograms and a density of 19.5 grams per cubic centimeter.

Students are asked to determine:

The radius of a sphere having the critical mass.

The textbook answer is:

$9.0\text{ cm}$

The lesson focuses on organizing information before performing calculations. :contentReference[oaicite:1]{index=1}

Step 1: Identify the Given Information

The first task is collecting all known quantities from the problem statement.

Given:

$\rho = 19.5\ \frac{g}{cm^3}$

$m = 60\ kg$

Unknown:

$r=?$

The lesson emphasizes reading carefully and identifying every known quantity before attempting any calculations. :contentReference[oaicite:2]{index=2}

Physics problems become much easier when all known values are organized before any algebra begins.

Step 2: Identify the Required Formulas

Two fundamental formulas are needed.

Density:

$\rho=\frac{m}{V}$

Volume of a sphere:

$V=\frac{4}{3}\pi r^3$

These two equations provide all the information required to solve the problem. :contentReference[oaicite:3]{index=3}

Step 3: Solve for Volume

Rearranging the density equation:

$V=\frac{m}{\rho}$

Convert kilograms to grams:

$60\ kg = 60,000\ g$

Substituting:

$V=\frac{60,000}{19.5}$

This produces the volume measured in cubic centimeters. :contentReference[oaicite:4]{index=4}

Step 4: Solve for the Radius

Starting with:

$V=\frac{4}{3}\pi r^3$

Solving for the radius:

$r=\left(\frac{3V}{4\pi}\right)^{1/3}$

Substituting the volume obtained from the density equation:

$r=\left(\frac{3}{4\pi}\cdot\frac{60,000}{19.5}\right)^{1/3}$

Evaluating the expression yields:

$r\approx 9.02283\text{ cm}$

The lesson demonstrates checking this result with computational software to verify the algebra and arithmetic. :contentReference[oaicite:5]{index=5}

Step 5: Apply Significant Figures

The given data contain:

60 kg → 2 significant figures
19.5 g/cm³ → 3 significant figures

The final answer must therefore contain:

$2\text{ significant figures}$

Thus:

$r=9.0\text{ cm}$

This agrees with the textbook answer. :contentReference[oaicite:6]{index=6}

The Importance of Organization

A major theme throughout the lesson is organization.

Rather than immediately plugging numbers into equations, students are encouraged to:

Identify known quantities.
Identify unknown quantities.
Select formulas.
Rearrange equations symbolically.
Substitute values.
Check units.
Apply significant figures.

This structured approach reduces mistakes and creates a clear record of the solution process. :contentReference[oaicite:7]{index=7}

Professional scientists and engineers rarely solve problems by guessing. They solve them through organization and process.

Final Thoughts

This lesson demonstrates how introductory mechanics problems often require only a few formulas but considerable organization. By identifying known quantities, choosing the correct relationships, rearranging equations symbolically, and checking significant figures, students can solve physics problems systematically and with confidence. :contentReference[oaicite:8]{index=8}

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