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=15/7800=15200.0019 m3  l=V/S=0.0019/0.00063.478=3.478  m D_{ 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 \ m \ \\ r_{ 2 } = D_{ 2 }/2 = 0.048/2 = \dfrac{ 3 }{ 125 } = 0.024 \ 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 \ m^2 \ \\ \ \\ m = 15 \ kg \ \\ h = 7800 \ kg/m^3 \ \\ \ \\ V = m/h = 15/7800 = \dfrac{ 1 }{ 520 } \doteq 0.0019 \ m^3 \ \\ \ \\ l = V/S = 0.0019/0.0006 \doteq 3.478 = 3.478 \ \text { m }







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