/*! 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 63 A block of wood floats in fresh ... [FREE SOLUTION] | 91Ó°ÊÓ

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A block of wood floats in fresh water with two-thirds of its volume \(V\) submerged and in oil with \(0.92 V\) submerged. Find the density of (a) the wood and (b) the oil.

Short Answer

Expert verified
(a) The density of the wood is \(\frac{2}{3}\, \text{g/cm}^3\). (b) The density of the oil is approximately \(0.724\, \text{g/cm}^3\).

Step by step solution

01

Understanding the Problem

A block of wood is floating in two different liquids, and we need to determine the density of the wood and the oil. We know that in static floating conditions, the weight of the displaced liquid is equal to the weight of the object.
02

Apply Archimedes' Principle for Fresh Water

According to Archimedes' principle, the weight of the liquid displaced by the block is equal to the weight of the block. Let the density of water be \( \rho_{water} = 1\, \text{g/cm}^3 \). Since two-thirds of the wood's volume \( V \) is submerged, the buoyant force (which equals the weight of the wood) can be expressed as: \[ \rho_{wood} \cdot V \cdot g = \rho_{water} \cdot \left(\frac{2}{3}V\right) \cdot g \] Simplifying this gives: \[ \rho_{wood} = \frac{2}{3} \cdot \rho_{water} = \frac{2}{3} \text{g/cm}^3 \]
03

Apply Archimedes' Principle for Oil

Using the same principle for oil, where 0.92 of the wood's volume is submerged, we express the buoyant force as: \[ \rho_{wood} \cdot V \cdot g = \rho_{oil} \cdot (0.92V) \cdot g \] We know \( \rho_{wood} = \frac{2}{3} \text{g/cm}^3 \). Solving for \( \rho_{oil} \): \[ \frac{2}{3} = \rho_{oil} \cdot 0.92 \] \[ \rho_{oil} = \frac{\frac{2}{3}}{0.92} \] Simplifying this gives: \[ \rho_{oil} \approx 0.724 \text{ g/cm}^3 \]
04

Concluding the Results

The calculated densities are consistent with the floating conditions described. We confirmed that the wood's density is less than both the water and the oil, which is why it floats in both fluids. The density of the oil is less than that of water, which explains why the block displaces more volume when floating in oil.

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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 concept in physics, especially when understanding buoyancy and fluid displacement. The formula for density is:- Density, \( \rho \) = mass/volumeIn simpler terms, it tells us how much mass an object has in a specific volume. When comparing different materials, density helps explain how "heavy" an object feels relative to its size.During our exercise, the density of the wood and oil were calculated using Archimedes' principle, which involved comparing the volume of the wood submerged in water versus oil. By knowing that the density of water is typically \( 1 \, \text{g/cm}^3 \), we could determine the wood's density. Then, substituting in the known values when the wood is submerged in oil, further calculations gave us the oil's density.
Buoyancy
Buoyancy is the force that allows objects to float in fluids. It can be understood as the upward push on an object submerged in a liquid, counteracting the weight of the object. The concept of buoyancy can be observed when floating a block of wood in fresh water and oil. Key considerations:
  • The block of wood displaces a certain volume of water equal to its own weight to float.
  • The less dense the object relative to the fluid, the greater the percentage of its volume will rise above the fluid.
As observed in our task, the part of the wood submerged changes when placed in water versus oil, revealing insights into both the density of the fluid and the buoyant force involved.
Fluid Displacement
Fluid displacement relates to how much fluid is pushed aside by an object immersed in it. According to Archimedes' principle, "any object, wholly or partially immersed in a fluid, is buoyed up by a force equal to the weight of the fluid displaced by the object." For the block of wood:
  • In water, two-thirds of its volume causes water to be displaced, equal to its weight.
  • In oil, a greater portion of the block's volume displaces oil, indicating a difference in fluid density.
By observing the change in submerged volume between oil and water, we calculated each fluid's density, thereby solving the problem logically.
Physics Problem Solving
Solving physics problems can be straightforward with systematic approaches. Here's how one might solve a fluid-based density problem:
  • Understand the scenario: Identify what is known and what needs to be found. In our task, the densities of the wood and oil needed to be discovered.
  • Apply appropriate principles: Using Archimedes' principle, equate forces acting on the wood in water and oil.
  • Calculate step-by-step: Substitute known values, solve using logic, and simplify the equations.
When learned with persistence, these methods can help demystify complex physics problems, making them approachable and solvable, much like the way we untangled the densities of the wood and oil here.

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Most popular questions from this chapter

A hollow sphere of inner radius \(8.0 \mathrm{~cm}\) and outer radius \(9.0 \mathrm{~cm}\) floats half-submerged in a liquid of density \(820 \mathrm{~kg} / \mathrm{m}^{3}\). (a) What is the mass of the sphere? (b) Calculate the density of the material of which the sphere is made.

Suppose that you release a small ball from rest at a depth of \(0.400 \mathrm{~m}\) below the surface in a pool of water. If the density of the ball is \(0.450\) that of water and if the drag force on the ball from the water is negligible, how high above the water surface will the ball shoot as it emerges from the water? (Neglect any transfer of energy to the splashing and waves produced by the emerging ball.)

You inflate the front tires on your car to 32 psi. Later, you measure your blood pressure, obtaining a reading of \(135 / 70\), the readings being in \(\mathrm{mm} \mathrm{Hg}\). In metric countries (which is to say, most of the world), these pressures are customarily reported in kilopascals (kPa). In kilopascals, what are (a) your tire pressure and (b) your blood pressure?

Water is pumped steadily out of a flooded basement at \(4.5 \mathrm{~m} / \mathrm{s}\) through a hose of radius \(1.0 \mathrm{~cm}\), passing through a window \(3.5 \mathrm{~m}\) above the waterline. What is the pump's power?

An inflatable device with a density of \(1.20 \mathrm{~g} / \mathrm{cm}^{3}\) is fully submerged in fresh water and then inflated enough for the device's density to match that of the water. What then is the ratio of the expanded volume to the full volume of the device?

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