Respuesta :
[tex]\bf \textit{volume of a sphere}\\\\
V=\cfrac{4\pi r^3}{3}\qquad
\begin{cases}
r=radius\\
-----\\
r_1=5.10\\
r_2=5
\end{cases}\implies \cfrac{V_1}{V_2}\implies \cfrac{\frac{4\pi \cdot 5.10^3}{3}}{\frac{4\pi \cdot 5^3}{3}}
\\\\\\
\cfrac{\underline{4\pi }\cdot 5.10^3}{\underline{3}}\cdot \cfrac{\underline{3}}{\underline{4\pi }\cdot 5^3}\implies \cfrac{5.10^3}{5^3}\implies \cfrac{132.651}{125}[/tex]
By definition, the volume of a sphere is given by:
[tex] V = (\frac{4}{3}) * \pi * r ^ 3
[/tex]
Where,
r: sphere radio
Therefore, the difference between the volume of the two sphere is given by:
[tex] V1 - V2 = (\frac{4}{3}) * \pi * r1 ^ 3 - (\frac{4}{3}) * \pi * r2 ^ 3
[/tex]
Rewriting we have:
[tex] V1 - V2 = (\frac{4}{3}) * \pi * (r1 ^ 3 - r2 ^ 3)
[/tex]
Substituting values we have:
[tex] V1 - V2 = (\frac{4}{3}) * 3.1416 * (5.10 ^ 3 - 5 ^ 3)
V1 - V2 = 32.05 cm ^ 3
[/tex]
Answer:
The difference in volume (in cubic centimeters) between the two spheres is:
[tex] V1 - V2 = 32.05 cm ^ 3 [/tex]