Question

A pool is 44.6 m long and 27.6 m wide. If the average depth of water is 8.80 ft, what is the mass (in kg) of water in the pool? Enter your answer in scientific notation.

Answer 1

Answer:

• 3.30 × 10⁶ kg

Explanation:

The form of the pool may be approximated to a reactangular prism with height equal to the average depth of water.

1) Dimensions:

• Length: 44.6 m
• Wide: 27.6 m
• Height: 8.80 ft

2) Volume formula

• V = length × width × height

3) Unit conversions

Convert 8.80 ft to meters:

• 1 ft = 0.3048 m ⇒ 1 = 0.3048 m/ft

• 8.80 ft × 0.3048 m/ft = 2.68 m

4) Volume calculation

Substitute the numbers in the formula:

• V = length × width × height = 44.6 m × 27.6 m × 2.68 m ≈ 3,298.97 m³ ≈ 3,300 m³ (it is rounded to three significant digits)

5) Mass of water:

• Density = mass / volume ⇒ mass = density × mass

• Density of water = 1,000 kg / m³

• mass = 3,300 m³ × 1,000 kg / m³ = 3,300,000 kg

6) Scientific notation

Choose the correct number of significant figures:

• Since the three given factors have 3 significant figures, the correct number of significant figures of the volume and the mass is 3.

The mass must be expressed as a power of ten. The mantissa contains only the significant digits, and the number before the decimal point can only have one digit 1 through 9.

• 3,300,000 kg =  3.30 × 10⁶ kg