A waterbed filled with water has the dimensions 8.0 ft ×7.0 ft ×0.75 ft. Taking the density of w

step 1 given dimension of water wet is 8 feet into 7 feet into 0 .75 feet and density of water is 1 gra

A waterbed filled with water has the dimensions 8.0 ft ×7.0...

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?

Key Concepts

Volume Calculation

This concept involves determining the total space occupied by an object by multiplying its dimensions. In the context of a prism, like a waterbed, the volume is obtained by multiplying its length, width, and height. Understanding this is fundamental for converting physical dimensions into a measurable quantity such as volume.

Unit Conversion

Unit conversion is key when measurements are given in different systems (such as feet and centimeters) or need to be changed to be consistent with the units used in formulas. It involves the systematic transformation of one set of units into another, ensuring that the computed values are accurate for subsequent calculations.

Density

Density is a physical property that relates an object's mass to its volume. It is defined as mass per unit volume and is used to determine how much mass is contained in a given volume. In problems involving substances like water, knowing the density allows us to link the calculated volume to the mass of the substance.

Mass Calculation

Mass calculation involves using the relationship between mass, density, and volume. By multiplying the density of a material by its volume, you obtain the mass. This concept is crucial when converting a calculated volume part into a mass measurement, particularly in problems where unit consistency (such as converting grams to kilograms) is important.

Transcript

Please give Ace some feedback