Question

A standard 1.00 kg mass is to be cut from a bar of steel having an equilateral triangular cross section with sides equal to 2.50 in. The density of the steel is 7.70 g/cm3. How many inches long must the section of bar be?

Answer 1

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 = $\frac{\sqrt{3}}{4}a^2$

a = 2.50 inches

Cross sectional area of the steel mass = A

$A=\frac{\sqrt{3}}{4}(2.50 inches)^2=2.71 inches^2$

$V = 2.71 inches^2\times h$

Density of the steel = d =$7.70 g/cm^3$

$1cm^3 = 0.0610237 inches^3$

$d=\frac{m}{v}$

$\frac{7.70 g}{0.0610237 inches^3}=\frac{ 1000 g}{2.71 inches^2\times h}$

$h=\frac{ 1000 g\times 0.0610237 inches^3}{2.71 inches^2\times 7.70 g}$

h = 2.92 inches

The section of the bar is 2.92 inches.