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A cubical block of wood, 10.0 cm on a side, flaats at the inter face between oil and woter with its lower surface 1,75 cm below the interface. The density of oil is 790 kg/m?. (A.) What is the gavge pressure at the upper face of the block? (B.) What is the quage pressure at the lower face of the block? Oil 10.0 Wood cm 10.0 Water cm

A cubical block of wood, 10.0 cm on a side, flaats at the inter face between oil and woter with its lower surface 1,75 cm below the interface. The density of oil is 790 kg/m?. (A.) What is the gavge pressure at the upper face of the block? (B.) What is the quage pressure at the lower face of the block? Oil 10.0 Wood cm 10.0 Water cm

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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### Problem Statement A cubical block of wood, 10.0 cm on a side, floats at the interface between oil and water with its lower surface 1.75 cm below the interface. The density of oil is 790 kg/m³. ##### Questions: (A) What is the gauge pressure at the upper face of the block? (B) What is the gauge pressure at the lower face of the block? ### Diagram Explanation The accompanying diagram shows a container with oil on top and water at the bottom. A cubical block of wood is placed such that it floats, with part of it submerged in oil and a part submerged in water. The block is 10.0 cm tall. The interface between the oil and water is clearly marked. The lower surface of the wood is shown to be 1.75 cm below this oil-water interface. ### Calculation Steps In order to solve the given problem, we need to follow these steps: 1. **Identify the pressurized points**: - The upper face of the wood block - The lower face of the wood block 2. **Calculate the pressure exerted by the oil column on the upper face of the wood block**: - **Height of oil column above the upper face of the block**: This would be the total height the wood is submerged in oil. 3. **Calculate the pressure at the lower face**: - **Gauge pressure at lower face**: Sum of the pressure contributions of both the oil column and the water column. Parameters to use: - Density of oil: \(\rho_{oil} = 790 \; \text{kg/m}^3\) - Gravitational acceleration: \(g \approx 9.81 \; \text{m/s}^2\) - Height or thickness of oil above upper face: \(h = 10.0 \; \text{cm} - 1.75 \; \text{cm} = 8.25 \; \text{cm}\) - Depth of lower face below the interface: \(1.75 \; \text{cm} + 10.0 \; \text{cm} = 11.75 \; \text{cm}\) This approach ensures a step-by-step thorough understanding, aiding students in grasping the concepts related to buoyancy and pressure in fluids.
Transcribed Image Text:### Problem Statement A cubical block of wood, 10.0 cm on a side, floats at the interface between oil and water with its lower surface 1.75 cm below the interface. The density of oil is 790 kg/m³. ##### Questions: (A) What is the gauge pressure at the upper face of the block? (B) What is the gauge pressure at the lower face of the block? ### Diagram Explanation The accompanying diagram shows a container with oil on top and water at the bottom. A cubical block of wood is placed such that it floats, with part of it submerged in oil and a part submerged in water. The block is 10.0 cm tall. The interface between the oil and water is clearly marked. The lower surface of the wood is shown to be 1.75 cm below this oil-water interface. ### Calculation Steps In order to solve the given problem, we need to follow these steps: 1. **Identify the pressurized points**: - The upper face of the wood block - The lower face of the wood block 2. **Calculate the pressure exerted by the oil column on the upper face of the wood block**: - **Height of oil column above the upper face of the block**: This would be the total height the wood is submerged in oil. 3. **Calculate the pressure at the lower face**: - **Gauge pressure at lower face**: Sum of the pressure contributions of both the oil column and the water column. Parameters to use: - Density of oil: \(\rho_{oil} = 790 \; \text{kg/m}^3\) - Gravitational acceleration: \(g \approx 9.81 \; \text{m/s}^2\) - Height or thickness of oil above upper face: \(h = 10.0 \; \text{cm} - 1.75 \; \text{cm} = 8.25 \; \text{cm}\) - Depth of lower face below the interface: \(1.75 \; \text{cm} + 10.0 \; \text{cm} = 11.75 \; \text{cm}\) This approach ensures a step-by-step thorough understanding, aiding students in grasping the concepts related to buoyancy and pressure in fluids.
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