Sphere floating

Will float a hollow iron ball with an outer diameter d1 = 20cm and an inside diameter d2 = 19cm in the water? The iron density is 7.8 g/cm 3. (Instructions: Calculate the average sphere density and compare with the density of the water. )

Result

x =  0

Solution:

d1=20 cm d2=19 cm ρ=7.8 g/cm3 v=1 g/cm3  r1=d1/2=20/2=10 cm r2=d2/2=19/2=192=9.5 cm  V1=43 π r13=43 3.1416 1034188.7902 cm3 V2=43 π r23=43 3.1416 9.533591.364 cm3   m=ρ (V1V2)=7.8 (4188.79023591.364)4659.9244 g  ρ2=mV1=4659.92444188.79021.1125 g/cm3  ρ2>v  x=0d_{1}=20 \ \text{cm} \ \\ d_{2}=19 \ \text{cm} \ \\ ρ=7.8 \ \text{g/cm}^3 \ \\ v=1 \ \text{g/cm}^3 \ \\ \ \\ r_{1}=d_{1}/2=20/2=10 \ \text{cm} \ \\ r_{2}=d_{2}/2=19/2=\dfrac{ 19 }{ 2 }=9.5 \ \text{cm} \ \\ \ \\ V_{1}=\dfrac{ 4 }{ 3 } \cdot \ \pi \cdot \ r_{1}^3=\dfrac{ 4 }{ 3 } \cdot \ 3.1416 \cdot \ 10^3 \doteq 4188.7902 \ \text{cm}^3 \ \\ V_{2}=\dfrac{ 4 }{ 3 } \cdot \ \pi \cdot \ r_{2}^3=\dfrac{ 4 }{ 3 } \cdot \ 3.1416 \cdot \ 9.5^3 \doteq 3591.364 \ \text{cm}^3 \ \\ \ \\ \ \\ m=ρ \cdot \ (V_{1}-V_{2})=7.8 \cdot \ (4188.7902-3591.364) \doteq 4659.9244 \ \text{g} \ \\ \ \\ ρ_{2}=\dfrac{ m }{ V_{1} }=\dfrac{ 4659.9244 }{ 4188.7902 } \doteq 1.1125 \ \text{g/cm}^3 \ \\ \ \\ ρ_{2} > v \ \\ \ \\ x=0



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