College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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![**Problem Statement:**
In the figure, an open tank contains a layer of oil floating on top of a layer of water (of density 1000 kg/m³) that is 3.0 m thick, as shown. What must be the thickness of the oil layer if the gauge pressure at the bottom of the tank is to be 5.8 x 10⁴ Pa? The density of the oil is 531 kg/m³. Give your answer in meters with two decimal places. (Numerical answer only, no units.)
**Diagram Explanation:**
The diagram accompanying the problem shows a vertical cross-section of the open tank. The tank has two distinct layers of liquids:
- The bottom layer is water with a thickness of 3.0 meters.
- The top layer is oil with an unknown thickness, denoted by '?'.
**Solution Approach:**
We will use the hydrostatic pressure formula to solve for the thickness of the oil layer:
\[ P = \rho g h \]
where:
- \( P \) is the pressure,
- \( \rho \) is the density of the liquid,
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)),
- \( h \) is the height (or thickness) of the liquid.
Since the gauge pressure at the bottom of the tank is the sum of the pressures exerted by both the water and the oil layers, we set up the following equation:
\[ P_{total} = P_{water} + P_{oil} \]
Given:
- \( P_{total} = 5.8 \times 10^4 \, \text{Pa} \)
- Density of water \( \rho_{water} = 1000 \, \text{kg/m}^3 \)
- Thickness of water \( h_{water} = 3.0 \, \text{m} \)
- Density of oil \( \rho_{oil} = 531 \, \text{kg/m}^3 \)
Let's solve for the thickness of the oil layer \( h_{oil} \):
\[ 5.8 \times 10^4 = (1000 \times 9.81 \times 3.0) + (531 \times 9.81 \times h_{oil}) \]
\[ 5.8 \times 10^4](https://content.bartleby.com/qna-images/question/feee2f86-f729-492a-b5ec-fb1726086ce7/5e318ca0-d76f-4089-8f78-4fe0bff7d85c/hjns67_processed.jpeg)
Transcribed Image Text:**Problem Statement:**
In the figure, an open tank contains a layer of oil floating on top of a layer of water (of density 1000 kg/m³) that is 3.0 m thick, as shown. What must be the thickness of the oil layer if the gauge pressure at the bottom of the tank is to be 5.8 x 10⁴ Pa? The density of the oil is 531 kg/m³. Give your answer in meters with two decimal places. (Numerical answer only, no units.)
**Diagram Explanation:**
The diagram accompanying the problem shows a vertical cross-section of the open tank. The tank has two distinct layers of liquids:
- The bottom layer is water with a thickness of 3.0 meters.
- The top layer is oil with an unknown thickness, denoted by '?'.
**Solution Approach:**
We will use the hydrostatic pressure formula to solve for the thickness of the oil layer:
\[ P = \rho g h \]
where:
- \( P \) is the pressure,
- \( \rho \) is the density of the liquid,
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)),
- \( h \) is the height (or thickness) of the liquid.
Since the gauge pressure at the bottom of the tank is the sum of the pressures exerted by both the water and the oil layers, we set up the following equation:
\[ P_{total} = P_{water} + P_{oil} \]
Given:
- \( P_{total} = 5.8 \times 10^4 \, \text{Pa} \)
- Density of water \( \rho_{water} = 1000 \, \text{kg/m}^3 \)
- Thickness of water \( h_{water} = 3.0 \, \text{m} \)
- Density of oil \( \rho_{oil} = 531 \, \text{kg/m}^3 \)
Let's solve for the thickness of the oil layer \( h_{oil} \):
\[ 5.8 \times 10^4 = (1000 \times 9.81 \times 3.0) + (531 \times 9.81 \times h_{oil}) \]
\[ 5.8 \times 10^4
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