Question 9bdaa

Sep 8, 2015

Volume: ${\text{2750 in}}^{3}$

Explanation:

You're going to have to play around with the units given to you to get them to match those required for the volume of the silver bar.

Notice that silver's density is given to you in grams per cubic centimeter, but that the mass of silver is given in pounds.

Your first conversion will take you from pounds of silver to grams of silver

3.89color(red)(cancel(color(black)("lbs"))) * "453.5937 g"/(1color(red)(cancel(color(black)("lbs")))) = "1764.5 g"

Now you can use silver's density to find the volume of the bar in cubic centimeters

1764.5color(red)(cancel(color(black)("g"))) * ("1 cm"""^3)/(10.5color(red)(cancel(color(black)("g")))) = "168.05 cm"""^3

Now all you have to do is convert the volume from cubic centimeters to cubinc inches. You know that $\text{1 inch" = "2.54 cm}$, which means that

${\text{1 cm" xx "1 cm" xx "1cm" = "1 cm}}^{3}$

and

${\text{2.54 in" xx "2.54 in" xx "2.54 in" = "16.387064 in}}^{3}$

Therefore, ${\text{1 cm"""^3 = "16.387064 in}}^{3}$. The volume of the silver bar will be

168.05color(red)(cancel(color(black)("cm"""^3))) * ("16.387064 in"""^3)/(1color(red)(cancel(color(black)("cm"""^3)))) = "2753.8 in"""^3#

Rounded to three sig figs, the answer will be

$V = \textcolor{g r e e n}{{\text{2750 in}}^{3}}$