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 5.8.0 x 104 Pa The density of the oil is 531kg/m³. Give your answer in meters with two decimal places. (numerical answer only, no units.) oil water 3.0 m

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)...
icon
Related questions
icon
Concept explainers
Topic Video
Question
**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
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
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps with 2 images

Blurred answer
Knowledge Booster
Fluid Pressure
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
Recommended textbooks for you
College Physics
College Physics
Physics
ISBN:
9781305952300
Author:
Raymond A. Serway, Chris Vuille
Publisher:
Cengage Learning
University Physics (14th Edition)
University Physics (14th Edition)
Physics
ISBN:
9780133969290
Author:
Hugh D. Young, Roger A. Freedman
Publisher:
PEARSON
Introduction To Quantum Mechanics
Introduction To Quantum Mechanics
Physics
ISBN:
9781107189638
Author:
Griffiths, David J., Schroeter, Darrell F.
Publisher:
Cambridge University Press
Physics for Scientists and Engineers
Physics for Scientists and Engineers
Physics
ISBN:
9781337553278
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning
Lecture- Tutorials for Introductory Astronomy
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:
9780321820464
Author:
Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:
Addison-Wesley
College Physics: A Strategic Approach (4th Editio…
College Physics: A Strategic Approach (4th Editio…
Physics
ISBN:
9780134609034
Author:
Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:
PEARSON