Question

The volume of a soap bubble is 2,756.9 mm^3 cubed. Find the radius and diameter of the soap bubble. Use 3.14 for π.

Answer 1

To find the radius and diameter of a soap bubble with a volume of 2,756.9 mm³, we can use the formula for the volume of a sphere:

\[
V = \frac{4}{3} \pi r^3
\]

Given:
- \( V = 2,756.9 \, \text{mm}^3 \)
- \( \pi \approx 3.14 \)

We can rearrange the formula to solve for the radius \( r \):

\[
r^3 = \frac{3V}{4\pi}
\]

Now, substituting the values we have:

\[
r^3 = \frac{3 \times 2,756.9}{4 \times 3.14}
\]

Calculating the right side:

1. Calculate \( 3 \times 2,756.9 \):
\[
3 \times 2,756.9 = 8,270.7
\]

2. Calculate \( 4 \times 3.14 \):
\[
4 \times 3.14 = 12.56
\]

3. Divide \( 8,270.7 \) by \( 12.56 \):
\[
r^3 = \frac{8,270.7}{12.56} \approx 658.4
\]

Now, take the cube root to find \( r \):

\[
r \approx \sqrt[3]{658.4}
\]

Using a calculator, we find:

\[
r \approx 8.66 \, \text{mm}
\]

Now, to find the diameter \( d \):

\[
d = 2r \approx 2 \times 8.66 \approx 17.32 \, \text{mm}
\]

Thus, the radius and diameter of the soap bubble are approximately:

- **Radius**: \( 8.66 \, \text{mm} \)
- **Diameter**: \( 17.32 \, \text{mm} \)