Determine the apparent weight in water

a) Weight density of aluminum in N/m³

Weight density = mass density × g = \( \rho g \)

Since specific gravity = 2.7, \( \rho_{\text{Al}} = 2.7 \times \rho_{\text{water}} \)

Mass density of water, \( \rho_{\text{water}} = 1000 \text{kg/m}^3 \)

So, \( \rho_{\text{Al}} = 2.7 \times 1000 = 2700 \text{kg/m}^3 \)

Weight density of aluminum, \( \rho_{\text{Al}} g = 2700 \times 9.8 = 26460 \text{N/m}^3 \)

Answer: \( \boxed{26460 \text{N/m}^3} \)

b) Weight density of aluminum in N/cm³

Convert \( 1 \text{m}^3 = 10^6 \text{cm}^3 \)

Weight density in N/cm³ = \( \frac{26460}{10^6} = 0.02646 \text{N/cm}^3 \)

Answer: \( \boxed{0.02646 \text{N/cm}^3} \)

c) Weight of the piece of aluminum

Volume \( V = 5 \text{cm}^3 \)

Weight = weight density × volume = \( 0.02646 \text{N/cm}^3 \times 5 \text{cm}^3 = 0.1323 \text{N} \)

Answer: \( \boxed{0.1323 \text{N}} \)

d) Weight density of water in SI units

Mass density of water = \( 1000 \text{kg/m}^3 \)

\( g = 9.8 \text{m/s}^2 \)

Weight density = \( 1000 \times 9.8 = 9800 \text{N/m}^3 \)

Answer: \( \boxed{9800 \text{N/m}^3} \)

e) Weight density of water in N/cm³

\( 1 \text{m}^3 = 10^6 \text{cm}^3 \)

Weight density in N/cm³ = \( \frac{9800}{10^6} = 0.0098 \text{N/cm}^3 \)

Answer: \( \boxed{0.0098 \text{N/cm}^3} \)

f) Buoyant force on 5 cm³ aluminum

Buoyant force = weight of water displaced = weight density of water × volume displaced

Volume displaced = volume of aluminum = \( 5 \text{cm}^3 \)

Buoyant force = \( 0.0098 \text{N/cm}^3 \times 5 \text{cm}^3 = 0.049 \text{N} \)

Answer: \( \boxed{0.049 \text{N}} \)

g) Apparent weight of aluminum in water

Apparent weight = actual weight - buoyant force = \( 0.1323 \text{N} - 0.049 \text{N} = 0.0833 \text{N} \)

Answer: \( \boxed{0.0833 \text{N}} \)