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
where area=sqrt(s(s-a)^3) s=half perimeter=(3in*2.54cm/in)*3/2
- Chem Help!! - Lisa , Saturday, February 2, 2013 at 9:33pm
- 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
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
Converting to inches
5.06cm*(1in/2.54)= 2 inches/side
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