Determine 10561

The hollow nickel ball has an outer diameter of 0.4 meters and an inner diameter of 0.3 meters. Determine its weight if the nickel density is 9000 kg/m³.

Correct answer:

m =  174.3584 kg

Step-by-step explanation:

$$\begin{gathered} D_1=0.4\text{m} \\ {D}_2=0.3\text{m} \\ h=9000{\text{kg/m}}^3 \\ {r}_1=D_1/2=0.4/2=\frac{1}{5}=0.2\text{m} \\ {r}_2=D_2/2=0.3/2=\frac{3}{20}=0.15\text{m} \\ {V}_1=\frac{4}{3}\cdot \pi \cdot r_1^3=\frac{4}{3}\cdot 3.1416\cdot 0.2^3\doteq 0.0335{\text{m}}^3 \\ {V}_2=\frac{4}{3}\cdot \pi \cdot r_2^3=\frac{4}{3}\cdot 3.1416\cdot 0.15^3\doteq 0.0141{\text{m}}^3 \\ V=V_1-V_2=0.0335-0.0141\doteq 0.0194{\text{m}}^3 \\ m=h\cdot V=9000\cdot 0.0194=174.3584\text{kg} \end{gathered}$$

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