The volume of the larger ball bearing is 8 times the volume of the
smaller one
Step-by-step explanation:
All the spheres are similar, and there is a ratio between their radii [tex]\frac{r_{1}}{r_{2}}[/tex]
So the ratio between their:
1. Diameters⇒ [tex]\frac{d_{1}}{d_{2}}[/tex] = [tex]\frac{r_{1}}{r_{2}}[/tex]
2. Areas ⇒ [tex]\frac{A_{1}}{A_{2}}[/tex] = [tex](\frac{r_{1}}{r_{2}})^{2}[/tex]
3. Volumes ⇒ [tex]\frac{V_{1}}{V_{2}}[/tex] = [tex](\frac{r_{1}}{r_{2}})^{3}[/tex]
∵ The radius of the larger ball bearing is twice the radius of the
smaller one
- That means the radius of the larger sphere is equal to the radius
of the smaller sphere multiplied by 2
∴ [tex]\frac{r_{1}}{r_{2}}[/tex] = [tex]\frac{2}{1}[/tex]
∵ [tex]\frac{V_{1}}{V_{2}}[/tex] = [tex](\frac{r_{1}}{r_{2}})^{3}[/tex]
∴ [tex]\frac{V_{1}}{V_{2}}[/tex] = [tex](\frac{2}{1})^{3}[/tex]
∴ [tex]\frac{V_{1}}{V_{2}}[/tex] = [tex]\frac{8}{1}[/tex]
The volume of the larger ball bearing is 8 times the volume of the
smaller one
Learn more:
You can learn more about ratios in brainly.com/question/4302397
brainly.com/question/4713715
#LearnwithBrainly
