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Time left 1:3 The figure below shows a fluid inside a cubical container of side 1 m. If the pressures inside and outside the container is as shown in the figure, what is the net force on one wall of the container

Time left 1:3 The figure below shows a fluid inside a cubical container of side 1 m. If the pressures inside and outside the container is as shown in the figure, what is the net force on one wall of the container

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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### Pressure and Force in a Cubical Container #### Problem Statement: The figure below shows a fluid inside a cubical container of side 1 m. If the pressures inside and outside the container are as shown in the figure, what is the net force on one wall of the container? #### Diagram Explanation: - The diagram illustrates a transparent cube with fluid inside it. - The side length of the cube is 1 meter. - Inside the cube, the pressure (\( P_{in} \)) is 5 atmospheres (atm). - Outside the cube, the pressure (\( P_{out} \)) is 1 atmosphere (atm). #### Question: Calculate the net force acting on one wall of the container. #### Choices: a) \( 3.039 \times 10^5 \) N b) \( 4.052 \times 10^5 \) N c) \( 5.065 \times 10^5 \) N d) \( 6.078 \times 10^5 \) N ### Detailed Solution: 1. **Conversion of Pressure Units:** - 1 atm = \( 1.013 \times 10^5 \) Pascals (Pa). 2. **Calculate the Pressure Difference:** - Pressure inside (\( P_{in} \)) in Pascals: \( 5 \text{ atm} = 5 \times 1.013 \times 10^5 \text{ Pa} = 5.065 \times 10^5 \text{ Pa} \) - Pressure outside (\( P_{out} \)) in Pascals: \( 1 \text{ atm} = 1.013 \times 10^5 \text{ Pa} \) - Pressure difference (\( \Delta P \)): \( \Delta P = P_{in} - P_{out} \) \( \Delta P = (5.065 \times 10^5 \text{ Pa}) - (1.013 \times 10^5 \text{ Pa}) \) \( \Delta P = 4.052 \times 10^5 \text{ Pa} \) 3. **Calculate the Net Force:** - Force (\( F \)) is given by \( F = \Delta P \times A \), where \( A \)
Transcribed Image Text:### Pressure and Force in a Cubical Container #### Problem Statement: The figure below shows a fluid inside a cubical container of side 1 m. If the pressures inside and outside the container are as shown in the figure, what is the net force on one wall of the container? #### Diagram Explanation: - The diagram illustrates a transparent cube with fluid inside it. - The side length of the cube is 1 meter. - Inside the cube, the pressure (\( P_{in} \)) is 5 atmospheres (atm). - Outside the cube, the pressure (\( P_{out} \)) is 1 atmosphere (atm). #### Question: Calculate the net force acting on one wall of the container. #### Choices: a) \( 3.039 \times 10^5 \) N b) \( 4.052 \times 10^5 \) N c) \( 5.065 \times 10^5 \) N d) \( 6.078 \times 10^5 \) N ### Detailed Solution: 1. **Conversion of Pressure Units:** - 1 atm = \( 1.013 \times 10^5 \) Pascals (Pa). 2. **Calculate the Pressure Difference:** - Pressure inside (\( P_{in} \)) in Pascals: \( 5 \text{ atm} = 5 \times 1.013 \times 10^5 \text{ Pa} = 5.065 \times 10^5 \text{ Pa} \) - Pressure outside (\( P_{out} \)) in Pascals: \( 1 \text{ atm} = 1.013 \times 10^5 \text{ Pa} \) - Pressure difference (\( \Delta P \)): \( \Delta P = P_{in} - P_{out} \) \( \Delta P = (5.065 \times 10^5 \text{ Pa}) - (1.013 \times 10^5 \text{ Pa}) \) \( \Delta P = 4.052 \times 10^5 \text{ Pa} \) 3. **Calculate the Net Force:** - Force (\( F \)) is given by \( F = \Delta P \times A \), where \( A \)
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