Steel tube

The steel tube has an inner diameter of 4 cm and an outer diameter of 4.8 cm. The density of the steel is 7800 kg/m3. Calculate its length if it weighs 15 kg.

Result

l =  3.478 m

Solution:

D1=4 cm=4/100 m=0.04 m D2=4.8 cm=4.8/100 m=0.048 m  r1=D1/2=0.04/2=150=0.02 m r2=D2/2=0.048/2=3125=0.024 m  S=π r22π r12=3.1416 0.02423.1416 0.0220.0006 m2  m=15 kg h=7800 kg/m3  V=m/h=15/780015200.0019 m3  l=V/S=0.0019/0.00063.4783.478 mD_{1}=4 \ cm=4 / 100 \ m=0.04 \ m \ \\ D_{2}=4.8 \ cm=4.8 / 100 \ m=0.048 \ m \ \\ \ \\ r_{1}=D_{1}/2=0.04/2=\dfrac{ 1 }{ 50 }=0.02 \ \text{m} \ \\ r_{2}=D_{2}/2=0.048/2=\dfrac{ 3 }{ 125 }=0.024 \ \text{m} \ \\ \ \\ S=\pi \cdot \ r_{2}^2 - \pi \cdot \ r_{1}^2=3.1416 \cdot \ 0.024^2 - 3.1416 \cdot \ 0.02^2 \doteq 0.0006 \ \text{m}^2 \ \\ \ \\ m=15 \ \text{kg} \ \\ h=7800 \ \text{kg/m}^3 \ \\ \ \\ V=m/h=15/7800 \doteq \dfrac{ 1 }{ 520 } \doteq 0.0019 \ \text{m}^3 \ \\ \ \\ l=V/S=0.0019/0.0006 \doteq 3.478 \doteq 3.478 \ \text{m}



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