**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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Description: **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 t

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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

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)...

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Concept explainers

Fluid Pressure

The term fluid pressure is coined as, the measurement of the force per unit area of a given surface of a closed container. It is a branch of physics that helps to study the properties of fluid under various conditions of force.

Gauge Pressure

Pressure is the physical force acting per unit area on a body; the applied force is perpendicular to the surface of the object per unit area. The air around us at sea level exerts a pressure (atmospheric pressure) of about 14.7 psi but this doesn’t seem to bother anyone as the bodily fluids are constantly pushing outwards with the same force but if one swims down into the ocean a few feet below the surface one can notice the difference, there is increased pressure on the eardrum, this is due to an increase in hydrostatic pressure.

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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

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