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?

Short Answer

Expert verified
Answer: Approximately 1,194.77 kg.

Step by step solution

01

Convert dimensions to centimeters

Since the dimensions of the waterbed are given in feet, we must first convert them to centimeters, since the density of water is given in grams per cubic centimeter. 1 foot = 30.48 centimeters. So, the dimensions in centimeters are: 8.0 ft × 30.48 cm/ft = 243.84 cm 7.0 ft × 30.48 cm/ft = 213.36 cm 0.75 ft × 30.48 cm/ft = 22.86 cm
02

Calculate the volume of the waterbed

Now, let's calculate the volume of the waterbed by multiplying its length, width, and height. Volume = length × width × height = 243.84 cm × 213.36 cm × 22.86 cm
03

Find the weight of the water

Now we can find the weight of the water needed to fill the waterbed using the given density of water (1.00 g/cm³). Weight = Volume × Density = 243.84 cm × 213.36 cm × 22.86 cm × 1.00 g/cm³
04

Convert weight to kilograms

Finally, we need to convert the weight of the water from grams to kilograms. 1 kilogram = 1000 grams Weight in kilograms = Weight in grams / 1000 g/kg Now, let's perform the calculations: Volume = 243.84 cm × 213.36 cm × 22.86 cm = 1,194,772.97 cm³ (approximately) Weight = 1,194,772.97 cm³ × 1.00 g/cm³ = 1,194,772.97 g Weight in kilograms = 1,194,772.97 g / 1000 g/kg = 1,194.77 kg (approximately) So, approximately 1,194.77 kilograms of water are required to fill the waterbed.

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