Question
A sphere of gold with a radius of 14.9 cm. (density \( =19.3 \frac{\mathrm{~g}}{\mathrm{~cm}^{3}} \) ). Round your answer to 3 significant digits. mass: \( \square \mathrm{g} \)
Ask by Daniel Craig. in the United States
Feb 03,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
The mass of the gold sphere is approximately \(2.67 \times 10^{5} \, \mathrm{g}\).
Solution
To calculate the mass of a gold sphere with a radius of 14.9 cm and a density of \(19.3 \, \frac{\text{g}}{\text{cm}^3}\), follow these steps:
1. **Calculate the Volume of the Sphere:**
The volume \( V \) of a sphere is given by:
\[
V = \frac{4}{3} \pi r^3
\]
Where \( r = 14.9 \, \text{cm} \).
\[
V = \frac{4}{3} \times \pi \times (14.9)^3 \approx 4.18879 \times 3307.949 \approx 13,\!856.3 \, \text{cm}^3
\]
2. **Calculate the Mass:**
Mass \( m \) is the product of density \( \rho \) and volume \( V \):
\[
m = \rho \times V = 19.3 \, \frac{\text{g}}{\text{cm}^3} \times 13,\!856.3 \, \text{cm}^3 \approx 267,\!000 \, \text{g}
\]
3. **Round to Three Significant Digits:**
\[
m \approx 2.67 \times 10^5 \, \text{g}
\]
**Final Answer:**
\[
\text{mass} = 2.67 \times 10^{5} \, \mathrm{g}
\]
Reviewed and approved by the UpStudy tutoring team
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Simplify this solution The Deep Dive
To find the mass of the gold sphere, we first need to calculate its volume using the formula for the volume of a sphere: \[ V = \frac{4}{3} \pi r^3 \] Plugging in the radius (14.9 cm): \[ V = \frac{4}{3} \pi (14.9)^3 \approx 4,143.764 \, \mathrm{cm}^3 \] Next, we use the density to find the mass: \[ \text{mass} = \text{density} \times \text{volume} \] Using the density \( 19.3 \, \mathrm{g/cm}^3 \): \[ \text{mass} \approx 19.3 \times 4,143.764 \approx 79,991.306 \, \mathrm{g} \] Rounding this to three significant digits gives us: mass: \( 8.00 \times 10^4 \, \mathrm{g} \)
