Question

A waterbed filled with water has the dimensions 8ft by 7 ft by .75 ft taking the density of water to be 1.00/gm cubed how many kilgrams of water are required to fill the waterbd

Answer 1

By definition, the density is given by:

$D = \frac{m}{V}$

Where,

m: mass

V: volume

Clearing the mass we have:

$m = DV$

The volume is given by:

$V = (8) * (7) * (0.75)$

$V = 42ft ^ 3$

Then, we have the following conversion:

$1foot = 0.3048m$

Applying the conversion we have:

$V = 42 * (0.3048) ^ 3$

$V = 1.19m ^ 3$

On the other hand we have the following conversions:

$1m = 100cm$

$1Kg = 1000g$

Applying the conversions for the density we have:

$D = (1\frac{g}{cm^3})((\frac{100}{1})^3\frac{cm^3}{1m^3})(\frac{1}{1000}\frac{Kg}{g})=1000\frac{Kg}{m^3}$

Then, the mass of the water is:

$m = (1000) * (1.19)$

$m = 1190$

Answer:

1190 kilgrams of water are required to fill the waterbed