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6 days ago

5

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 k

ilgrams of water are required to fill the waterbd

1 answer:

Svet_ta [4.3K]6 days ago

7 0

Density can be defined as:

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

Where:

m is mass

V stands for volume

To isolate the mass, we derive:

$m = DV$

Volume can be expressed as:

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

$V = 42ft ^ 3$

We then use the following conversion:

$1foot = 0.3048m$

After applying the conversion, we arrive at:

$V = 42 * (0.3048) ^ 3$

$V = 1.19m ^ 3$

Additionally, we have these conversions:

$1m = 100cm$

$1Kg = 1000g$

When we apply these conversions for density, we conclude:

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

Finally, the mass of the water required is:

$m = (1000) * (1.19)$

$m = 1190$

Answer:

A total of 1190 kilograms of water is needed to fill the waterbed.

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