Posted by **Lisa ** on Saturday, February 2, 2013 at 7:43pm.

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?

- Chem Help!! -
**bobpursley**, Saturday, February 2, 2013 at 8:43pm
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.

- Chem Help!! -
**Lisa **, Saturday, February 2, 2013 at 9:33pm
thank you

- Chem Help!! -
**Greg **, Saturday, February 2, 2013 at 9:46pm
How many inches is the section of the bar

- Chem Help!! -
**Devron**, Saturday, February 2, 2013 at 10:33pm
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

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