College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Question
![**Problem Statement:**
**3.** What is the mass of an iron sphere whose diameter is 58 cm? The density of iron is \(7.9 \times 10^3 \, \text{kg/m}^3\)?
**Solution:**
To solve this problem, we need to calculate the mass of the sphere using the formula for the volume of a sphere and the relationship between mass, density, and volume.
1. **Calculate the Radius:**
The diameter of the sphere is given as 58 cm. Therefore, the radius \( r \) is half of the diameter.
\[
r = \frac{58}{2} = 29 \, \text{cm}
\]
Convert the radius to meters since the density is in kg/m³.
\[
r = 29 \, \text{cm} = 0.29 \, \text{m}
\]
2. **Calculate the Volume of the Sphere:**
The formula for the volume \( V \) of a sphere is:
\[
V = \frac{4}{3} \pi r^3
\]
Substituting the radius:
\[
V = \frac{4}{3} \pi (0.29)^3
\]
\[
V \approx \frac{4}{3} \times 3.1416 \times (0.024389)
\]
\[
V \approx 0.102 \, \text{m}^3
\]
3. **Calculate the Mass of the Sphere:**
Using the formula \( \text{mass} = \text{density} \times \text{volume} \), and given that the density \( \rho \) of iron is \( 7.9 \times 10^3 \, \text{kg/m}^3 \):
\[
\text{mass} = 7.9 \times 10^3 \, \text{kg/m}^3 \times 0.102 \, \text{m}^3
\]
\[
\text{mass} \approx 805.8 \, \text{kg}
\]
Therefore, the mass of the iron sphere is approximately 805.8 kg.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F883e3e80-624e-4e0d-88a1-3570ac65b326%2Fd98dc12e-b2d2-4a99-b21a-7b413ca3e860%2Feblfcsxh_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
**3.** What is the mass of an iron sphere whose diameter is 58 cm? The density of iron is \(7.9 \times 10^3 \, \text{kg/m}^3\)?
**Solution:**
To solve this problem, we need to calculate the mass of the sphere using the formula for the volume of a sphere and the relationship between mass, density, and volume.
1. **Calculate the Radius:**
The diameter of the sphere is given as 58 cm. Therefore, the radius \( r \) is half of the diameter.
\[
r = \frac{58}{2} = 29 \, \text{cm}
\]
Convert the radius to meters since the density is in kg/m³.
\[
r = 29 \, \text{cm} = 0.29 \, \text{m}
\]
2. **Calculate the Volume of the Sphere:**
The formula for the volume \( V \) of a sphere is:
\[
V = \frac{4}{3} \pi r^3
\]
Substituting the radius:
\[
V = \frac{4}{3} \pi (0.29)^3
\]
\[
V \approx \frac{4}{3} \times 3.1416 \times (0.024389)
\]
\[
V \approx 0.102 \, \text{m}^3
\]
3. **Calculate the Mass of the Sphere:**
Using the formula \( \text{mass} = \text{density} \times \text{volume} \), and given that the density \( \rho \) of iron is \( 7.9 \times 10^3 \, \text{kg/m}^3 \):
\[
\text{mass} = 7.9 \times 10^3 \, \text{kg/m}^3 \times 0.102 \, \text{m}^3
\]
\[
\text{mass} \approx 805.8 \, \text{kg}
\]
Therefore, the mass of the iron sphere is approximately 805.8 kg.
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