Answer:
The section of the bar is 2.92 inches.
Explanation:
Mass of the steel cut ,m = 1.00 kg = 1000 g
Volume of the steel bar = V = Area × height
Height of the of the section of bar = h
Area of Equilateral triangular = [tex]\frac{\sqrt{3}}{4}a^2[/tex]
a = 2.50 inches
Cross sectional area of the steel mass = A
[tex]A=\frac{\sqrt{3}}{4}(2.50 inches)^2=2.71 inches^2[/tex]
[tex]V = 2.71 inches^2\times h[/tex]
Density of the steel = d =[tex]7.70 g/cm^3[/tex]
[tex]1cm^3 = 0.0610237 inches^3[/tex]
[tex]d=\frac{m}{v}[/tex]
[tex]\frac{7.70 g}{0.0610237 inches^3}=\frac{ 1000 g}{2.71 inches^2\times h}[/tex]
[tex]h=\frac{ 1000 g\times 0.0610237 inches^3}{2.71 inches^2\times 7.70 g}[/tex]
h = 2.92 inches
The section of the bar is 2.92 inches.
