## Related questions with answers

A waterbed filled with water has the dimensions $8.0 \mathrm{ft} \times 7.0 \mathrm{ft} \times 0.75 \mathrm{ft}$. Taking the density of water to be $1.00 \mathrm{~g} / \mathrm{cm}^3$, how many kilograms of water are required to fill the waterbed?

Solutions

VerifiedFor this problem, we are asked to calculate the **kilograms of water required to fill the waterbed** given the following information:

Dimensions of the waterbed: 8.0 ft $\times$ 7.0 ft $\times$ 0.75 ft Density of water: 1.00 g/cm$^3$

To get the kilograms of water required to fill the water bed, we have to get first the volume of the waterbed. It is easier to convert the volume of the waterbed from ft$^3$ to mL. **Hence, we use the density of water that is still equal to 1.00 g/mL.**

Then, we get the grams of water and convert it again into kg.

The goal of this exercise is to determine the mass of water ($\rho$ = 1.00 g/cm$^{3}$) in kilograms that is enough to fill a waterbed with the dimensions 8.0 ft $\times$ 7.0 ft $\times$ 0.75 ft.

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