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

To fill the waterbed, the volume of the waterbed is calculated in cubic meters and then multiplied by the density of water, resulting in a mass of 1186.5 kilograms of water that is required.

Step-by-step explanation:

To determine how many kilograms of water are required to fill the waterbed, we must first calculate its volume in cubic meters and then use the density of water to convert this volume to mass.

Step 1: Calculate the volume of the waterbed

Convert the dimensions from feet to meters (since 1 ft = 0.3048 m):

• 8 ft = 8 × 0.3048 m = 2.4384 m
• 7 ft = 7 × 0.3048 m = 2.1336 m
• 0.75 ft = 0.75 × 0.3048 m = 0.2286 m

Now, multiply these dimensions to get the volume in cubic meters:
Volume = 2.4384 m × 2.1336 m × 0.2286 m = 1.1865 m³

Step 2: Convert volume to mass using the density

The density of water is 1,000 kg/m³. Therefore, the mass of water required to fill the waterbed is:
Mass = Volume × Density = 1.1865 m³ × 1000 kg/m³ = 1186.5 kg

Answer 2

By definition, the density is given by:

`D = (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(g)/(cm^3))(((100)/(1))^3(cm^3)/(1m^3))((1)/(1000)(Kg)/(g))=1000(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