Answered

Calculate the following conversion:

[tex]\[
\frac{2350 \, \text{kg}}{1 \, \text{m}^3} \times \frac{2.2 \, \text{lb}}{1 \, \text{kg}} \times \frac{1 \, \text{m}^3}{35.3 \, \text{ft}^3} = \square \frac{\text{lb}}{\text{ft}^3}
\][/tex]



Answer :

Sure, let's go through this step by step.

1. Identify the given values and conversion factors:
- Density: [tex]\( \frac{2350 \text{ kg}}{1 \text{ m}^3} \)[/tex]
- Conversion factor from kilograms to pounds: [tex]\( 2.2 \frac{\text{lb}}{\text{kg}} \)[/tex]
- Conversion factor from cubic meters to cubic feet: [tex]\( \frac{1 \text{ m}^3}{35.3 \text{ ft}^3} \)[/tex]

2. Set up the equation using the given conversions:
[tex]\[ \frac{2350 \text{ kg}}{1 \text{ m}^3} \times \frac{2.2 \text{ lb }}{1 \text{ kg}} \times \frac{1 \text{ m}^3}{35.3 \text{ ft}^3} \][/tex]

3. Perform the multiplications and cancellations:
- First, multiply the density in kg per m³ by the conversion factor to pounds:
[tex]\[ 2350 \text{ kg/m}^3 \times 2.2 \text{ lb/kg} = 5170 \text{ lb/m}^3 \][/tex]

- Next, divide by the conversion factor to cubic feet:
[tex]\[ 5170 \text{ lb/m}^3 \times \frac{1 \text{ m}^3}{35.3 \text{ ft}^3} = \frac{5170 \text{ lb}}{35.3 \text{ ft}^3} \][/tex]

4. Calculate the final result:
- Perform the division:
[tex]\[ \frac{5170 \text{ lb}}{35.3 \text{ ft}^3} \approx 146.4589235127479 \text{ lb/ft}^3 \][/tex]

Thus, the density [tex]\( \frac{2350 \text{ kg}}{1 \text{ m}^3} \)[/tex] is approximately equal to [tex]\( 146.4589235127479 \text{ lb/ft}^3 \)[/tex].