Question

A standard 1.000kg- mass is to be cut from a bar of steel having an equilateral triangular cross section with sides equal to 3.00in. The density of the steel is 7.70g/cm^3.


How many inches long must the section of bar be?

Answer 1

mass=density(volume)=density*area*length

where area=sqrt(s(s-a)^3) s=half perimeter=(3in*2.54cm/in)*3/2
area=sqrt(9*1.27(3*2.54*2))

check that.

Answer 2

thank you

Answer 3

How many inches is the section of the bar

Answer 4

I'm not sure that I understand what you did, but this is what I did.


1kg=1 x10^3 g

Density=Mass/volume, so

Volume =mass/density= (1 x10^3 g/7.70g/cm^3)= 130cm^3

Since in a equilateral triangle all sides are equal and since volume = cm*cm*cm or l*w*h

(130cm^3)^(1/3)=5.06cm

Converting to inches
5.06cm*(1in/2.54)= 2 inches/side