Iron sphere

Iron sphere has weight 100 kg and density ρ = 7600 kg/m3. Calculate the volume, surface and diameter of the sphere.

Correct result:

V =  13.158 dm3
D =  2.929 dm
S =  26.953 dm2

Solution:

m=100 kg h=7600 kg/m3 m=hV V1=mh=1007600≐176≐0.0132 m3 V=V1→dm3=V1⋅ 1000 dm3=0.013157894736842⋅ 1000 dm3=13.158 dm3=13.158 dm3m=100 \ \text{kg} \ \\ h=7600 \ \text{kg/m}^3 \ \\ m=hV \ \\ V_{1}=\dfrac{ m }{ h }=\dfrac{ 100 }{ 7600 } \doteq \dfrac{ 1 }{ 76 } \doteq 0.0132 \ \text{m}_{3} \ \\ V=V_{1} \rightarrow dm^3=V_{1} \cdot \ 1000 \ dm^3=0.013157894736842 \cdot \ 1000 \ dm^3=13.158 \ dm^3=13.158 \ \text{dm}^3
V=43πr3 r=V⋅ 34π3=13.1579⋅ 34⋅ 3.14163≐1.4645 dm D=2⋅ r=2⋅ 1.4645=2.929 dmV=\dfrac{ 4 }{ 3 } \pi r^3 \ \\ r=\sqrt[3]{ V \cdot \ \dfrac{ 3 }{ 4 \pi } }=\sqrt[3]{ 13.1579 \cdot \ \dfrac{ 3 }{ 4 \cdot \ 3.1416 } } \doteq 1.4645 \ \text{dm} \ \\ D=2 \cdot \ r=2 \cdot \ 1.4645=2.929 \ \text{dm}
S=4π⋅ r2=4⋅ 3.1416⋅ 1.46452=26.953 dm2S=4 \pi \cdot \ r^2=4 \cdot \ 3.1416 \cdot \ 1.4645^2=26.953 \ \text{dm}^2

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