/*! 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 78 A cubical block of wood 0.100 \(... [FREE SOLUTION] | 91Ó°ÊÓ

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A cubical block of wood 0.100 \(\mathrm{m}\) on a side and with a density of 550 \(\mathrm{kg} / \mathrm{m}^{3}\) floats in a jar of water. Oil with a density of 750 \(\mathrm{kg} / \mathrm{m}^{3}\) is poured on the water until the top of the oil layer is 0.035 \(\mathrm{m}\) below the top of the block. (a) How deep is the oil layer? (b) What is the gauge pressure at the block's lower face?

Short Answer

Expert verified
The oil layer is 0.013 m deep, and gauge pressure at the block's lower face is 634.85 Pa.

Step by step solution

01

Understand the Problem

We have a cubical block of wood with side length 0.1 meters and density 550 kg/m³ floating in water. An oil layer with a density of 750 kg/m³ is added such that the oil's surface is 0.035 meters below the top of the block. We need to find the depth of the oil layer and the gauge pressure at the block's lower face.
02

Calculate the Initial Submersion Depth in Water

First, calculate how much of the block is submerged initially just in water. This can be found using Archimedes' principle: the weight of the displaced water equals the weight of the block. Set up the equation:\[\text{Weight of block} = \text{Weight of displaced water}\]The volume submerged, \(V_s\), is given by \(V_s \times \rho_{\text{water}} \times g = V \times \rho_{\text{block}} \times g\), where \(\rho_{\text{water}} = 1000\, \text{kg/m}^3\) and \(\rho_{\text{block}} = 550\, \text{kg/m}^3\). Solve for \(V_s\):\[V_s = \frac{V \times \rho_{\text{block}}}{\rho_{\text{water}}} = \frac{0.001\, \text{m}^3 \times 550\, \text{kg/m}^3}{1000\, \text{kg/m}^3} = 0.00055\, \text{m}^3\]The side that is submerged is then: \(h_s = \frac{0.00055\, \text{m}^3}{0.1\, \text{m}^2} = 0.055\, \text{m}\).
03

Determine the Increase in Submersion Due to Oil

Since 0.035 m of the bloc remains above the oil, the block is now submerged 0.1 m - 0.035 m = 0.065 m into the liquid. Given it was initially 0.055 m submerged in water alone, the additional 0.01 m depth must be due to the oil. The depth of the oil layer can thus be calculated.
04

Calculate the Depth of the Oil Layer

The excess depth (\(\Delta h\)) due to oil, 0.01 meters, relates to the depth of the oil layer (\(d_o\)) using the ratio of densities:\[\Delta h = \frac{\rho_{\text{oil}}}{\rho_{\text{water}}} \times d_o\]We know \(\Delta h = 0.01\, \text{m}\), so solving for \(d_o\):\[0.01 = \frac{750}{1000} \times d_o \d_o = \frac{0.01}{0.75} = 0.01333\, \text{m} \approx 0.013\, \text{m}\]
05

Calculate the Gauge Pressure at the Block's Lower Face

The gauge pressure at the block's lower face is the sum of pressures due to the water and oil layers:\[P = \rho_{\text{water}} \times g \times h_s + \rho_{\text{oil}} \times g \times d_o\]Substitute known values (\(g = 9.8\, \text{m/s}^2\)):\[P = 1000\, \text{kg/m}^3 \times 9.8\, \text{m/s}^2 \times 0.055\, \text{m} + 750\, \text{kg/m}^3 \times 9.8\, \text{m/s}^2 \times 0.013\, \text{m}\]Calculate separately:\[P_{\text{water}} = 539\, \text{Pa} \P_{\text{oil}} = 95.85\, \text{Pa} \P = 539 + 95.85 \approx 634.85\, \text{Pa}\]
06

Conclusion: Final Results

The depth of the oil layer is 0.013 m, and the gauge pressure at the block's lower face is approximately 634.85 Pa.

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Key Concepts

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

Archimedes' Principle
Archimedes' principle is a foundational concept in fluid mechanics and plays a crucial role in understanding how objects behave when submerged in fluids. The principle states that 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. This buoyant force acts in the upward direction, counteracting the gravity pulling the object down. This concept is vital in explaining why objects float or sink when placed in liquids.

For instance, in our exercise, a block of wood floats because the buoyant force generated by the displaced water equals the block's weight. Archimedes' principle allows us to calculate exactly how much of the block is submerged by matching the weight of the water displaced with the block's total weight. This ensures we can determine the submersion depth both initially and after additional layers are added, like oil in our problem.
Buoyancy
Buoyancy is the force that allows objects to float in fluids. It's what keeps ships sailing and balloons rising. This force comes into play when an object is placed in a fluid and can be thought of as the fluid's response to being displaced. The buoyant force is equal to the weight of the fluid displaced by the object, which stems directly from Archimedes' principle.

Understanding buoyancy explained why our wooden block, despite being heavier than water per unit volume, maintains its position on the water's surface. The density of the wooden block determines how much of it stays above water, as buoyancy balances the gravitational pull. In the exercise, the introduction of an oil layer modifies the total fluid in which the wood is submerged, affecting the buoyancy and leading to a change in the submersion depth, showcasing how different fluid layers interact to affect buoyant forces.
Density
Density is a measure of how much mass is contained in a given volume of a substance. It is usually expressed in kilograms per cubic meter (kg/m³). Density is an essential property in fluid mechanics as it helps determine how substances interact with each other in terms of buoyancy and pressure. It governs how substances stack when layered together as seen in the exercise with the wooden block, water, and oil.

In our scenario, the wood has a density of 550 kg/m³, which is less than that of water (1000 kg/m³), allowing it to float. The oil, with a higher density than the wood (750 kg/m³), but less than water, forms a distinct layer above the water but supports the wooden block differently from water. Calculating the densities allows us to understand how much of the oil versus the water the block displaces, thus altering the block's buoyancy and submersion behavior accordingly.
Gauge Pressure
Gauge pressure is the pressure of a system above the ambient atmospheric pressure. It's often used in practical scenarios because it represents the actual pressure exerted by a fluid in a closed environment, such as within pipes or under a submerged block in fluid mechanics problems.

In the exercise, gauge pressure is determined at the block's lower face—right where the interface with water occurs. Since the oil layer is added, both the oil and water contribute to the overall pressure exerted on the block's submerged face. This can be calculated by summing the pressures due to each fluid layer separately. Using the densities and depths known for water and oil, along with gravitational acceleration, students can calculate the downward force exerted by each layer and thus find the gauge pressure at the block's interface with water, crucial for understanding the net force and overall stability of the system.

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Most popular questions from this chapter

Black Smokers. Black smokers are hot volcanic vents that emit smoke deep in the ocean floor. Many of them teem with exotic creatures, and some biologists think that life on earth may have begun around such vents. The vents range in depth from about 1500 \(\mathrm{m}\) to 3200 \(\mathrm{m}\) below the surface. What is the gauge pressure at a \(3200-\mathrm{m}\) deep vent, assuming the density of water does not vary? Express your answer in pascals and atmospheres.

The densities of air, helium, and hydrogen (at \(p=1.0\) atm and \(T=20^{\circ} \mathrm{C} )\) are \(1.20 \mathrm{kg} / \mathrm{m}^{3}, 0.166 \mathrm{kg} / \mathrm{m}^{3},\) and 0.0899 \(\mathrm{kg} / \mathrm{m}^{3}\) , respectively. (a) What is the volume in cubic meters displaced by a hydrogen-filled airship that has a total "Iift" of 120 \(\mathrm{kN}\) ? (The "lift" is the amount by which the buoyant force exceeds the weight of the gas that fills the airship. \((b)\) What would be the "lift" if helium were used instead of hydrogen? In view of your answer, why is helium used in modern airships like advertising blimps?

A single ice cube with mass 9.70 g floats in a glass completely full of 420 \(\mathrm{cm}^{3}\) of water. You can ignore the water's surface tension and its variation in density with temperature (as long as it remains a liquid). (a) What volume of water does the ice cube displace? (b) When the ice cube has completely melted, has any water overflowed? If so, how much? If not, explain why this is so. (c) Suppose the water in the glass had been very salty water of density 1050 \(\mathrm{kg} / \mathrm{m}^{3}\) . What volume of salt water would the \(9.70-\mathrm{g}\) ice cube displace? (d) Redo part (b) for the freshwater ice cube in the salty water.

A rock with mass \(m=3.00 \mathrm{kg}\) is suspended from the roof of an elevator by a light cord. The rock is totally immersed in a bucket of water that sits on the floor of the elevator, but the rock doesn't touch the bottom or sides of the bucket. (a) When the elevator is at rest, the tension in the cord is 21.0 \(\mathrm{N}\) . Calculate the volume of the rock. ( b) Derive an expression for the tension in the cord when the elevator is accelerating upward with an acceleration of magnitude a. Calculate the tension when \(a=2.50 \mathrm{m} / \mathrm{s}^{2}\) upward. (c) Derive an expression for the tension in the cord when the clevator is accelerating downwand with an acceleration of magnitude \(a\) . Calculate the tension when \(a=2.50 \mathrm{m} / \mathrm{s}^{2}\) downward. (d) What is the tension when the elevator is in free fall with a downward acceleration equal to \(g ?\)

A closed container is partially filled with water. Initially, the air above the water is at atmospheric pressure \(\left(1.01 \times 10^{5} \mathrm{Pa}\right)\) and the gauge pressure at the bottom of the water is 2500 Pa. Then additional air is pumped in, increasing the pressure of the air above the water by 1500 \(\mathrm{Pa}\) . (a) What is the gauge pressure at the bottom of the water? (b) By how much must the water level in the container be reduced, by drawing some water out through a valve at the bottom of the container, to return the gauge pressure at the bottom of the water to its original value of 2500 Pa? The pressure of the air above the water is maintained at 1500 Pa above atmospheric pressure.

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