Answer The Following Problems: Pistons Are Fitted To Two Cyl

Answer The Following Problems1 Pistons Are Fitted To Two Cylindrical

Answer the following problems. 1. Pistons are fitted to two cylindrical chambers connected through a horizontal tube to form a hydraulic system. The piston chambers and the connecting tube are filled with an incompressible fluid. The cross-sectional areas of piston 1 and piston 2 are A₁ and A₂, respectively. A force F₁ is exerted on piston 1. Rank the resultant force F₂ on piston 2 that results from the combinations of F₁, A₁, and A₂ given from greatest to smallest. If any of the combinations yield the same force, give them the same ranking. (Use only ">" or "=" symbols. Do not include any parentheses around the letters or symbols.) F₁ = 6.0 N; A₁ = 1.1 m²; and A₂ = 2.2 m². F₁ = 3.0 N; A₁ = 1.1 m²; and A₂ = 0.55 m². F₁ = 3.0 N; A₁ = 2.2 m²; and A₂ = 4.4 m². F₁ = 6.0 N; A₁ = 0.55 m²; and A₂ = 2.2 m². F₁ = 6.0 N; A₁ = 0.55 m²; and A₂ = 1.1 m². F₁ = 3.0 N; A₁ = 2.2 m²; and A₂ = 1.1 m. A bicycle tire pump has a piston with area 0.49 in². If a person exerts a force of 24 lb on the piston while inflating a tire, what pressure does this produce on the air in the pump? psi=

Paper For Above instruction

This collection of physics problems covers key principles related to fluid mechanics, including Pascal's principle, pressure calculations, buoyant force, and properties of gases. The solutions demonstrate how to analyze various scenarios involving hydraulic systems, pressures at different depths, and the behavior of gases under specific temperature and pressure conditions, providing a comprehensive understanding of fundamental physics concepts relevant to engineering and everyday phenomena.

Solution to Problems

Problem 1: Hydraulic System Force Analysis

In a hydraulic system comprising two pistons interconnected through an incompressible fluid, the force transmitted is governed by Pascal's principle: the pressure applied to the fluid is transmitted equally throughout. The force on piston 2 (F₂) is related through the areas and the applied force F₁ by the relation:

F₂ = (A₂/A₁) * F₁

Using the data provided:

• F₁ = 6.0 N; A₁ = 1.1 m²; A₂ = 2.2 m²
• F₁ = 3.0 N; A₁ = 1.1 m²; A₂ = 0.55 m²
• F₁ = 3.0 N; A₁ = 2.2 m²; A₂ = 4.4 m²
• F₁ = 6.0 N; A₁ = 0.55 m²; A₂ = 2.2 m²
• F₁ = 6.0 N; A₁ = 0.55 m²; A₂ = 1.1 m²
• F₁ = 3.0 N; A₁ = 2.2 m²; A₂ = 1.1 m

Calculating each case:

1. F₂ = (2.2/1.1) 6.0 = 2 6.0 = 12.0 N

2. F₂ = (0.55/1.1) 3.0 = 0.5 3.0 = 1.5 N

3. F₂ = (4.4/2.2) 3.0 = 2 3.0 = 6.0 N

4. F₂ = (2.2/0.55) 6.0 = 4 6.0 = 24.0 N

5. F₂ = (1.1/0.55) 6.0 = 2 6.0 = 12.0 N

6. F₂ = (1.1/2.2) 3.0 = 0.5 3.0 = 1.5 N

Ranking from greatest to smallest:

F₂ = 24.0 N > 12.0 N = 12.0 N > 6.0 N > 1.5 N = 1.5 N

Problem 2: Pressure in a Bicycle Pump

The pressure (P) exerted by a force (F) on an area (A) is P = F/A.

Given:

• F = 24 lb
• A = 0.49 in²

Calculate:

P = 24 lb / 0.49 in² ≈ 48.98 psi

Problem 3: Force on Large Truck Tire Sidewall

Given the gauge pressure P_g= 82 psi, and the area A = 1,330 in², the net outward force (F) is pressure times area:

F = P_g A = 82 psi 1,330 in² ≈ 108,860 lb

The force magnitude is approximately 108,860 lb.

Problem 4: Force on Aquarium Window

The pressure exerted by water at depth h is P = ρgh, but here we are given the gauge pressure P = 7 psi. The total force is:

F = P A = 7 psi (59 in * 69 in)

Area A = 59 in * 69 in = 4,071 in²

Since 1 psi = 1 lb/in², the force:

F = 7 lb/in² * 4,071 in² ≈ 28,497 lb

Problem 5: Volume of Hydrogen Gas in Hindenburg

Using the ideal gas law:

PV = nRT

Where:

• P = 1 atm = 101,325 Pa
• V = ?
• n = mass / molar mass = 18,000 kg / 2.016 kg/mol ≈ 8,928 mol
• R = 8.314 J/(mol·K)
• T = 0°C = 273 K

V = nRT / P = (8928 mol)(8.314 J/(mol·K))(273 K) / 101,325 Pa ≈ 2, that is, approximately 2, or more accurately:

V ≈ 632,388.3 m³ (after calculation)

Problem 6: Mass of Hydrogen in Sample

Given volume V = 590 m³, density of hydrogen gas at STP: 0.0899 kg/m³.

Mass = density × volume = 0.0899 kg/m³ * 590 m³ ≈ 53.0 kg

Problem 7: Gauge Pressure at Pool Depth

The hydrostatic pressure: P = ρgh

g = 32.2 ft/s², ρ (water) ≈ 62.4 lb/ft³

P = 62.4 lb/ft³ * 18.6 ft ≈ 1,161.8 lb/ft²

Convert to psi: 1,161.8 lb/ft² / 144 ≈ 8.07 psi

Problem 8: Pressure in Mariana Trench

Pressure at depth h: P = ρgh

Using same density and g:

P = 62.4 lb/ft³ * 36,198 ft ≈ 2,259,490 lb/ft²

Convert to psi: 2,259,490 / 144 ≈ 15,673 psi gauge pressure

Problem 9: Buoyant Force on Ebony Log

Buoyant force: F_b = ρ_water V g

Density of water: 62.4 lb/ft³, volume V = 15.0 ft³

F_b = 62.4 lb/ft³ * 15.0 ft³ ≈ 936 lb

Problem 10: Buoyant Force on Empty Tank

The tank volume: 9,490 ft³. Using the same density:

F_b = 62.4 lb/ft³ * 9,490 ft³ ≈ 592,176 lb

Problem 11: Maximum Payload of Zeppelins Helium Volume

Using density of helium: about 0.1786 kg/m³ at STP.

Mass of helium: m= ρ V = 0.1786 kg/m³ 9,770 m³ ≈ 1,743 kg.

Maximum payload: the difference between the weight of displaced air and the weight of helium.

Displaced air weight: 1,743 kg (helium) displaced by the same volume of air, which weighs:

→ 1,743 kg * 9.81 m/s² ≈ 17,122 N

Since the question asks for maximum payload, roughly about 17,122 N converted to mass:

Mass = 17,122 N / 9.81 ≈ 1,745 kg.

Problem 12: Submersion Depth of Boat

Total weight: 6,400 N.

Area of bottom: 5 m².

The weight supported by buoyancy: F_b = ρ_water g V_displaced.

Pressure at depth h: P = ρgh, and force: F = P * A = ρghA.

Set F = ρghA:

h = F / (ρ g A) = 6,400 N / (1000 kg/m³ 9.81 m/s² 5 m²) ≈ 0.13 m

Problem 13: Scale Reading with Submerged Iron

When submerged, the apparent weight = actual weight - buoyant force.

Buoyant force = ρ_water V g.

Mass of iron: m = 378 N / 9.81 ≈ 38.5 kg.

Volume of iron: V = m / 7,860 kg/m³ ≈ 0.0049 m³.

Buoyant force = ρ_water V g = 1000 kg/m³ 0.0049 m³ 9.81 m/s² ≈ 48.07 N.

Scale reading when submerged: 378 N - 48.07 N ≈ 330 N.

Problem 14: Hydraulic System Area Ratio

The pressure exerted by force: P = F/A. Since the pressure is transmitted equally:

F_input/A_input = F_output/A_output, with F_output = weight of chair = 2000 N.

And F_input = 44 N.

Assuming A_output is the area of the plunger:

A_plunger / A_piston = F_plunger / F_piston = 44 N / 2000 N ≈ 0.022

Problem 15: Pressure Difference on Airplane Wing

Lift force: F = ΔP * A

ΔP = F / A = 91,000 N / 13 m² ≈ 7,000 N/m²

Converted to psi: 7,000 / 6894.76 ≈ 1.016 psi.

Problem 16: Air Flow Speed into Room

Continuity equation: A₁ v₁ = A₂ v₂

A₁ = (9 in)² = (0.75 ft)² = 0.5625 ft²

A₂ = (15 in)² = (1.25 ft)² = 1.5625 ft²

v₂ = (A₁/A₂) v₁ = (0.5625 / 1.5625) 4.1 ≈ 1.48 ft/s

Problem 17: Displacement of Olive Oil by Floating Bowl

Weight of the bowl: 1.45 N.

The buoyant force equals the weight, so volume displaced V_disp satisfies: