Domanda del libro di testo
The deepest point in the ocean is 11 km below sea level, deeper than Mt. Everest is tall. What is the pressure in atmospheres at this depth?
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Ch 14: Fluids and Elasticity
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796
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TOC.tableOfContents
- Ch 01: Concepts of Motion35
- Ch 02: Kinematics in One Dimension75
- Ch 03: Vectors and Coordinate Systems58
- Ch 04: Kinematics in Two Dimensions60
- Ch 05: Force and Motion23
- Ch 06: Dynamics I: Motion Along a Line53
- Ch 07: Newton's Third Law27
- Ch 08: Dynamics II: Motion in a Plane52
- Ch 09: Work and Kinetic Energy56
- Ch 10: Interactions and Potential Energy50
- Ch 11: Impulse and Momentum56
- Ch 12: Rotation of a Rigid Body71
- Ch 13: Newton's Theory of Gravity52
- Ch 14: Fluids and Elasticity44
- Ch 15: Oscillations55
- Ch 16: Traveling Waves37
- Ch 17: Superposition39
- Ch 18: A Macroscopic Description of Matter53
- Ch 19: Work, Heat, and the First Law of Thermodynamics60
- Ch 20: The Micro/Macro Connection53
- Ch 21: Heat Engines and Refrigerators34
- Ch 22: Electric Charges and Forces40
- Ch 23: The Electric Field39
- Ch 24: Gauss' Law32
- Ch 25: The Electric Potential63
- Ch 26: Potential and Field57
- Ch 27: Current and Resistance42
- Ch 28: Fundamentals of Circuits39
- Ch 29: The Magnetic Field47
- Ch 30: Electromagnetic Induction60
- Ch 31: Electromagnetic Fields and Waves32
- Ch 32: AC Circuits63
- Ch 33: Wave Optics32
- Ch 34: Ray Optics57
- Ch 35: Optical Instruments30
- Ch 36: Special Relativity38
- Ch 37: The Foundations of Modern Physics25
- Ch 38: Quantization49
- Ch 39: Wave Functions and Uncertainty31
- Ch 40: One-Dimensional Quantum Mechanics35
- Ch 41: Atomic Physics18
- Ch 42: Nuclear Physics27
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Capitolo 14, Problema 11a
The container shown in FIGURE EX14.11 is filled with oil. It is open to the atmosphere on the left. What is the pressure at point 1?
Guida verificata passo dopo passo1
Step 1: Identify the key concepts involved in the problem. This is a fluid mechanics problem, and the pressure at a given point in a fluid can be determined using the hydrostatic pressure formula. The formula is: , where is atmospheric pressure, is the density of the fluid, is the acceleration due to gravity, and is the depth of the fluid below the surface.
Step 2: Determine the atmospheric pressure. Since the container is open to the atmosphere on the left, the pressure at the surface of the oil is equal to the atmospheric pressure, . This value is typically given in the problem or can be approximated as 101,325 Pa (standard atmospheric pressure).
Step 3: Identify the depth of point 1 below the surface of the oil. Measure or calculate the vertical distance from the surface of the oil to point 1. This depth is crucial for calculating the hydrostatic pressure contribution.
Step 4: Use the density of the oil. The problem may provide the density of the oil, , or you may need to look it up. Ensure the units are consistent (e.g., kg/m³).
Step 5: Substitute the values into the hydrostatic pressure formula. Combine the atmospheric pressure, the density of the oil, the acceleration due to gravity (, approximately 9.8 m/s²), and the depth to calculate the pressure at point 1 using the formula: .
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Hydrostatic Pressure
Hydrostatic pressure is the pressure exerted by a fluid at rest due to the weight of the fluid above it. It increases with depth in a fluid and is calculated using the formula P = ρgh, where P is the pressure, ρ is the fluid density, g is the acceleration due to gravity, and h is the height of the fluid column above the point in question.
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Percorso guidato
17:04
Pressure and Atmospheric Pressure
Atmospheric Pressure
Atmospheric pressure is the pressure exerted by the weight of the atmosphere above a given point. At sea level, this pressure is approximately 101,325 Pa (Pascals). In open containers, the pressure at the surface of the fluid is equal to the atmospheric pressure, which influences the pressure at various points within the fluid.
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17:04Pressure and Atmospheric Pressure
Pascal's Principle
Pascal's Principle states that when pressure is applied to a confined fluid, the pressure change is transmitted undiminished throughout the fluid. This principle is crucial for understanding how pressure is distributed in a fluid, allowing us to determine the pressure at different points within the fluid, such as point 1 in the given scenario.
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14:47Diffraction with Huygen's Principle
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