/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 34 A metal cube, \(2.00 \mathrm{~cm... [FREE SOLUTION] | 91Ó°ÊÓ

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A metal cube, \(2.00 \mathrm{~cm}\) on each side, has a density of \(6600 \mathrm{~kg} / \mathrm{m}^{3}\). Find its apparent mass when it is totally submerged in water.

Short Answer

Expert verified
The apparent mass of the cube when submerged in water is approximately 0.0457 kg.

Step by step solution

01

Calculate the Volume of the Cube

To find the volume of the cube, use the formula for the volume of a cube, which is side length cubed: \[ V = 2.00 \text{ cm} \times 2.00 \text{ cm} \times 2.00 \text{ cm} = 8.00\text{ cm}^3 \] Convert the volume to cubic meters (since 1 cm = 0.01 m): \[ V = 8.00 \text{ cm}^3 = 8.00 \times 10^{-6} \text{ m}^3 \]
02

Calculate the Mass of the Metal Cube

The mass of the cube can be found using the formula: \[ \text{mass} = \text{density} \times \text{volume} \] Where the density is given as \(6600\, \text{kg/m}^3\). Thus, the mass is: \[ \text{mass} = 6600\, \text{kg/m}^3 \times 8.00 \times 10^{-6}\, \text{m}^3 \] \[ \text{mass} = 0.0528\, \text{kg} \]
03

Calculate the Buoyant Force Acting on the Cube

The buoyant force can be calculated using Archimedes’ principle, which states that the buoyant force is equal to the weight of the displaced fluid. For water, with the density \(1000\, \text{kg/m}^3\): \[ \text{buoyant force} = \text{density}_\text{water} \times V \times g \]\[ = 1000\, \text{kg/m}^3 \times 8.00 \times 10^{-6}\, \text{m}^3 \times 9.81 \, \text{m/s}^2 \] \[ = 0.07848\, \text{N} \]
04

Calculate the Apparent Mass of the Cube

The apparent mass is the actual mass minus the buoyant force divided by gravitational acceleration: \[ \text{apparent mass} = (\text{actual mass} - F_b) / g \]\[ = (0.0528\, \text{kg} \times 9.81 \text{ m/s}^2 - 0.07848\, \text{N}) / 9.81 \text{ m/s}^2 \]\[ = 0.0457\, \text{kg} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Density Calculation
Density is a fundamental property that relates the mass of an object to its volume. To calculate the density of a substance, you use the formula:
\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]
In this exercise, the density of the metal cube is given as \(6600 \, \text{kg/m}^3\), and this value helps us understand how much mass exists per unit volume within the cube.
Understanding how to calculate density is crucial because it allows us to predict how materials will interact with different environments, especially fluids.
Knowing the density also helps in calculating other vital aspects, such as buoyancy when the object is submerged in a fluid like water.
Volume of a Cube
The volume of a cube can be determined using a simple formula because all sides of a cube are equal in length. The volume formula is:
\[ V = \text{side length}^3 \]
In our example, the side length is given as \(2.00 \text{ cm}\). So, the calculated volume in cubic centimeters is:
- \( V = 2.00 \times 2.00 \times 2.00 \)- \( V = 8.00 \text{ cm}^3 \)
It's often helpful to convert this into cubic meters for consistency with other units. Thus,
\[ 1 \text{ cm}^3 = 1 \times 10^{-6} \text{ m}^3 \]
So in cubic meters, it becomes \(8.00 \times 10^{-6} \text{ m}^3\). Knowing the volume is crucial for determining the space an object occupies, which is fundamental in deciding the amount of fluid displaced when submerged.
Buoyant Force
Buoyant force is the upward force exerted by a fluid on an object submerged in it, as explained by Archimedes' Principle.
According to this principle, the magnitude of the buoyant force is equal to the weight of the fluid displaced by the object. The formula to calculate the buoyant force is:
\[ F_b = \text{density}_\text{fluid} \times V \times g \]
Where:
  • \(F_b\) is the buoyant force
  • \(\text{density}_\text{fluid}\) is the density of the liquid
  • \(V\) is the volume of the object immersed
  • \(g\) is the acceleration due to gravity \(9.81 \text{ m/s}^2\)
For our metal cube fully submerged in water, which has a density of \(1000 \text{ kg/m}^3\), the buoyant force computes to \(0.07848 \text{ N}\).
This force acts to support the submerged object against gravity, explaining why things feel lighter in water.
Apparent Mass in Fluids
The apparent mass of an object in a fluid is the mass it seems to have while submerged. It is different from its actual mass due to the buoyant force acting upon it.
We calculate apparent mass by considering:
\[ \text{Apparent Mass} = \frac{\text{Actual Mass} - F_b}{g} \]
Where:
  • \(\text{Actual Mass}\) is the mass of the object in air
  • \(F_b\) is the buoyant force
  • \(g\) is the acceleration due to gravity
In our example, the metal cube's actual mass is \(0.0528 \text{ kg}\). However, once in water, its apparent mass becomes \(0.0457 \text{ kg}\).
This concept is critical in understanding how objects behave differently in fluids, affecting engineering designs and providing insight into fluid dynamics.

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Most popular questions from this chapter

What fraction of the volume of a piece of quartz \(\left(\rho=2.65 \mathrm{~g} / \mathrm{cm}^{3}\right)\) will be submerged when it is floating in a container of mercury ( \(\rho=13.6 \mathrm{~g} / \mathrm{cm}^{3}\) )

A metal object "weighs" \(26.0 \mathrm{~g}\) in air and \(21.48 \mathrm{~g}\) when totally immersed in water. What is the volume of the object? Its mass density?

A foam plastic \(\left(\rho_{p}=0.58 \mathrm{~g} / \mathrm{cm}^{3}\right)\) is to be used as a life preserver. What volume of plastic must be used if it is to keep 20 percent (by volume) of an 80 -kg man above water in a lake? The average density of the man is \(1.04 \mathrm{~g} / \mathrm{cm}^{3}\). Keep in mind that a density of \(1 \mathrm{~g} / \mathrm{cm}^{3}\) equals \(1000 \mathrm{~kg} / \mathrm{m}^{3}\). At equilibrium $$\begin{array}{l} F_{B} \text { on man }+F B \text { on plastic }=\text { Weight of man }+\text { Weight of plastic } \\ \quad\left(\rho_{w}\right)\left(0.80 V_{m}\right) g+\rho_{w} V_{p} g=\rho_{m} V_{m} g+\rho_{p} V_{p} g \end{array}$$ or $$\left(\rho_{w}-\rho_{p}\right) V_{p}=\left(\rho_{m}-0.80 \rho_{w}\right) V_{m}$$ where subscripts \(m, w\), and \(p\) refer to man, water, and plastic, respectively. But \(\rho_{m} V_{m}=80 \mathrm{~kg}\) and so \(V_{m}=(80 / 1040) \mathrm{m}^{3}\). Substitution gives $$\left[(1000-580) \mathrm{kg} / \mathrm{m}^{3}\right] V_{p}=\left[(1040-800) \mathrm{kg} / \mathrm{m}^{3}\right]\left[(80 / 1040) \mathrm{m}^{3}\right]$$ from which \(V_{p}=0.044 \mathrm{~m}^{3}\).

A \(60-\mathrm{kg}\) rectangular box, open at the top, has base dimensions of \(1.0 \mathrm{~m}\) by \(0.80 \mathrm{~m}\) and a depth of \(0.50 \mathrm{~m} .(a)\) How deep will it sink in fresh water? \((b)\) What weight \(F_{W b}\) of ballast will cause it to sink to a depth of \(30 \mathrm{~cm}\) ? (a) Assuming that the box floats, $$\begin{array}{c} F_{B}=\text { Weight of displaced water }=\text { Weight of box } \\ \left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(1.0 \mathrm{~m} \times 0.80 \mathrm{~m} \times y)=(60 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right) \end{array}$$ where \(y\) is the depth the box sinks. Solving yields \(y=0.075 \mathrm{~m}\). Because this is smaller than \(0.50 \mathrm{~m}\), our assumption is shown to be correct. (b) \(F_{B}=\) weight of box \(+\) weight of ballast But the \(F_{B}\) is equal to the weight of the displaced water. Therefore, the above equation becomes $$\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(1.0 \mathrm{~m} \times 0.80 \mathrm{~m} \times 0.30 \mathrm{~m})=(60)(9.81) \mathrm{N}+F_{W b}$$ from which \(F_{W b}=1760 \mathrm{~N}=1.8 \mathrm{kN}\). The ballast must have a mass of \((1760 / 9.81) \mathrm{kg}=180 \mathrm{~kg}\).

A solid piece of aluminum \(\left(\rho=2.70 \mathrm{~g} / \mathrm{cm}^{3}\right)\) has a mass of \(8.35 \mathrm{~g}\) when measured in air. If it is hung from a thread and submerged in a vat of oil \(\left(\rho=0.75 \mathrm{~g} / \mathrm{cm}^{3}\right)\), what will be the tension in the thread?

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