# Conversions Again

• September 10th 2008, 06:07 PM
>_<SHY_GUY>_<
Conversions Again
I Am So Lost...I Try Getting To Convert One Unit To The One I Need. The Math Is Good....Yet I Come With The Wrong Answer :(

A Block Of Of Ice Measures 14.5 cm. by 6.4 in. by 1.05 x 10^-4 km.
How Much Would The Cube Weight If The Density Of Ice Is 0.98 g. Per Cubic Centimeters.
• September 10th 2008, 06:14 PM
11rdc11
1st convert everything to cm and then multiply the 3 measurements to get $cm^3$

$\bigg(\frac{6.4in}{1}\bigg)\bigg(\frac{2.54cm}{1in }\bigg) =$

$\bigg(\frac{1.05 X 10^{-4}km}{1}\bigg)\bigg(\frac{100000cm}{1km}\bigg) =$
• September 10th 2008, 06:17 PM
Shyam
Quote:

Originally Posted by >_<SHY_GUY>_<
I Am So Lost...I Try Getting To Convert One Unit To The One I Need. The Math Is Good....Yet I Come With The Wrong Answer :(

A Block Of Of Ice Measures 14.5 cm. by 6.4 in. by 1.05 x 10^-4 km.
How Much Would The Cube Weight If The Density Of Ice Is 0.98 g. Per Cubic Centimeters.

1 inch = 2.54 cm

$1 km = 10^5 cm$ convert all units to cm. {because density is given in $g/cm^3$}

volume of cube $= 14.5 \times (6.4 \times 2.54) \times [(1.05\times 10^{-4})\times (10^5)]$

$
= 14.5 \times 16.256 \times 10.5$

$=2474.976 cm^3$

mass = volume \times density

$
= 2474.976 \times 0.98$

= 2425.48 grams

= 2.425 kg