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.
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?
4 answers
thank you
How many inches is the section of the bar
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
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
51 answers