A cube of unknown material is 0.200 m on each side and has a mass of 17.6 kg. It is suspended in a vat of honey (density h = 1.42 x 103 kg/m³) so that its top surface is at a depth of 2.00 m below the surface of the honey (see figure-not to scale). What is the size of the honey force pressing on the top side of the cube?
A cube of unknown material is 0.200 m on each side and has a mass of 17.6 kg. It is suspended in a vat of honey (density h = 1.42 x 103 kg/m³) so that its top surface is at a depth of 2.00 m below the surface of the honey (see figure-not to scale). What is the size of the honey force pressing on the top side of the cube?
College Physics
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Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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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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![### Question on Buoyancy and Fluid Forces
**Problem Statement:**
A cube of unknown material is 0.200 m on each side and has a mass of 17.6 kg. It is suspended in a vat of honey (density \( \rho_h = 1.42 \times 10^3 \) kg/m³) so that its top surface is at a depth of 2.00 m below the surface of the honey (see figure—not to scale). What is the size of the honey force pressing on the top side of the cube?
**Diagram Explanation:**
The diagram illustrates a vat containing honey, and within this fluid, a cube is submerged. The cube is shown as being suspended, with its top face 2.00 meters below the surface of the honey. The depth from the top surface of the honey to the top face of the cube is labeled as \( h = 2.0 \,m \), and the level representing the top surface of the honey is labeled as \( h = 0 \, m \).
**Options:**
- ( ) 1110 N
- ( ) 354 N
- ( ) 784 N
- ( ) 1730 N
**Explanation of Calculations:**
To find the force exerted by the honey on the top side of the cube, we use the formula:
\[ F = P \cdot A \]
where:
- \( F \) is the force exerted by the fluid (honey) on the surface of the cube.
- \( P \) is the pressure exerted by the honey at the depth of the top face of the cube.
- \( A \) is the area of the top face of the cube.
The pressure at depth \( h \) in a fluid of density \( \rho \) is given by:
\[ P = \rho g h \]
where:
- \( \rho \) is the density of the honey.
- \( g \) is the acceleration due to gravity (approximately \( 9.8 \) m/s²).
- \( h \) is the depth of the fluid above the surface in question.
For this particular problem:
- \( \rho = 1.42 \times 10^3 \, \text{kg/m}^3 \)
- \( h = 2.0 \, \text{m} \)
- \( A = \text{area](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3fc0f41c-1d8b-41a4-aee9-3a82e03a5afd%2F8ce9217d-d89e-487e-869a-2645e5c58955%2Fms2alzr.png&w=3840&q=75)
Transcribed Image Text:### Question on Buoyancy and Fluid Forces
**Problem Statement:**
A cube of unknown material is 0.200 m on each side and has a mass of 17.6 kg. It is suspended in a vat of honey (density \( \rho_h = 1.42 \times 10^3 \) kg/m³) so that its top surface is at a depth of 2.00 m below the surface of the honey (see figure—not to scale). What is the size of the honey force pressing on the top side of the cube?
**Diagram Explanation:**
The diagram illustrates a vat containing honey, and within this fluid, a cube is submerged. The cube is shown as being suspended, with its top face 2.00 meters below the surface of the honey. The depth from the top surface of the honey to the top face of the cube is labeled as \( h = 2.0 \,m \), and the level representing the top surface of the honey is labeled as \( h = 0 \, m \).
**Options:**
- ( ) 1110 N
- ( ) 354 N
- ( ) 784 N
- ( ) 1730 N
**Explanation of Calculations:**
To find the force exerted by the honey on the top side of the cube, we use the formula:
\[ F = P \cdot A \]
where:
- \( F \) is the force exerted by the fluid (honey) on the surface of the cube.
- \( P \) is the pressure exerted by the honey at the depth of the top face of the cube.
- \( A \) is the area of the top face of the cube.
The pressure at depth \( h \) in a fluid of density \( \rho \) is given by:
\[ P = \rho g h \]
where:
- \( \rho \) is the density of the honey.
- \( g \) is the acceleration due to gravity (approximately \( 9.8 \) m/s²).
- \( h \) is the depth of the fluid above the surface in question.
For this particular problem:
- \( \rho = 1.42 \times 10^3 \, \text{kg/m}^3 \)
- \( h = 2.0 \, \text{m} \)
- \( A = \text{area
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