Given:

• Side length: $L = 5.0$ cm $= 0.050$ m
• Density: $\rho_{Al} = 2700$ kg/m³

Find: Mass $m$

Solution:

Step 1: Calculate volume of cube. $V = L^3 = (0.050)^3 = 1.25 \times 10^{-4} \text{ m}^3$

Step 2: Use density formula. $m = \rho V = (2700)(1.25 \times 10^{-4}) = 0.338 \text{ kg}$

Answer: 338 g

Given:

• Depth: $h = 200$ m
• Seawater density: $\rho = 1025$ kg/m³
• $g = 9.8$ m/s²
• $P_{atm} = 1.01 \times 10^5$ Pa

Find: Absolute and gauge pressure

Solution:

Step 1: Calculate gauge pressure (water only). $P_{gauge} = \rho g h = (1025)(9.8)(200)$ $$

Step 2: Calculate absolute pressure (total). $P_{absolute} = P_{atm} + P_{gauge}$

Step 3: Convert to atmospheres. $P_{absolute} = \frac{2.11 \times 10^6}{1.01 \times 10^5} \approx 20.9 \text{ atm}$

Answer:

• Gauge: 2.01 MPa
• Absolute: 2.11 MPa or 20.9 atm

Given:

• Depth: $h = 200$ m
• Seawater density: $\rho = 1025$ kg/m³
• $g = 9.8$ m/s²
• $P_{atm} = 1.01 \times 10^5$ Pa

Find: Absolute and gauge pressure

Solution:

Step 1: Calculate gauge pressure (water only). $P_{gauge} = \rho g h = (1025)(9.8)(200)$ $$

Step 2: Calculate absolute pressure (total). $P_{absolute} = P_{atm} + P_{gauge}$

Step 3: Convert to atmospheres. $P_{absolute} = \frac{2.11 \times 10^6}{1.01 \times 10^5} \approx 20.9 \text{ atm}$

Answer:

• Gauge: 2.01 MPa
• Absolute: 2.11 MPa or 20.9 atm

Solution:

Given: side = 5.0 cm = 0.050 m, m = 1.5 kg

(a) Density: Volume = (0.050)³ = 1.25 × 10⁻⁴ m³ ρ = m/V = 1.5/(1.25 × 10⁻⁴) = 12,000 kg/m³

(b) Will it float? ρ_metal = 12,000 kg/m³ > ρ_water = 1000 kg/m³ No, it will sink (denser than water)

(c) Pressure at bottom: Weight: W = mg = 1.5 × 10 = 15 N Area: A = (0.050)² = 2.5 × 10⁻³ m² P = F/A = 15/(2.5 × 10⁻³) = 6,000 Pa or 6.0 kPa

Given:

• $d_1 = 5.0$ cm $= 0.050$ m, $d_2 = 30$ cm $= 0.30$ m
• Car mass: $m = 2000$ kg

Solution:

Part (a): Find input force

Step 1: Calculate areas. $A_1 = \pi r_1^2 = \pi (0.025)^2 = 1.96 \times 10^{-3} \text{ m}^2$

Step 2: Find output force (car weight). $F_2 = mg = (2000)(9.8) = 19,600 \text{ N}$

Step 3: Apply Pascal's principle. $F_1 = F_2 \frac{A_1}{A_2} = (19,600) \frac{1.96 \times 10^{-3}}{7.07 \times 10^{-2}}$

Part (b): Find output displacement

Step 4: Use volume conservation. $A_1 d_1 = A_2 d_2$

Answer:

• (a) Input force: 543 N
• (b) Car rises: 5.5 mm

Mechanical advantage = 36, but distance disadvantage = 36. Energy conserved!

Solution:

Given: h = 2.5 m, ρ = 1000 kg/m³, P_atm = 1.01 × 10⁵ Pa

(a) Absolute pressure: P = P_atm + ρgh P = 1.01 × 10⁵ + (1000)(10)(2.5) P = 1.01 × 10⁵ + 2.5 × 10⁴ P = 1.26 × 10⁵ Pa or 126 kPa

(b) Gauge pressure: P_gauge = ρgh = (1000)(10)(2.5) = 2.5 × 10⁴ Pa or 25 kPa

(c) Force on bottom: Area = 2.0 × 1.0 = 2.0 m² F = PA = (1.26 × 10⁵)(2.0) F = 2.52 × 10⁵ N or 252 kN

Note: If using gauge pressure, F = (2.5 × 10⁴)(2.0) = 5.0 × 10⁴ N (force due to water only)

Given:

• $d_1 = 5.0$ cm $= 0.050$ m, $d_2 = 30$ cm $= 0.30$ m
• Car mass: $m = 2000$ kg

Solution:

Part (a): Find input force

Step 1: Calculate areas. $A_1 = \pi r_1^2 = \pi (0.025)^2 = 1.96 \times 10^{-3} \text{ m}^2$

Step 2: Find output force (car weight). $F_2 = mg = (2000)(9.8) = 19,600 \text{ N}$

Step 3: Apply Pascal's principle. $F_1 = F_2 \frac{A_1}{A_2} = (19,600) \frac{1.96 \times 10^{-3}}{7.07 \times 10^{-2}}$

Part (b): Find output displacement

Step 4: Use volume conservation. $A_1 d_1 = A_2 d_2$

Answer:

• (a) Input force: 543 N
• (b) Car rises: 5.5 mm

Mechanical advantage = 36, but distance disadvantage = 36. Energy conserved!