/*! 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 38 A beaker contains oil of density... [FREE SOLUTION] | 91Ó°ÊÓ

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A beaker contains oil of density \(0.80 \mathrm{~g} / \mathrm{cm}^{3}\). A \(1.6-\mathrm{cm}\) cube of aluminum \(\left(\rho=2.70 \mathrm{~g} / \mathrm{cm}^{3}\right)\) hanging ertically on a thread is submerged in the oil. Find the tension in the thread.

Short Answer

Expert verified
The tension in the thread is approximately 0.7622 N.

Step by step solution

01

Calculate the Volume of the Aluminum Cube

The cube has sides of length \(1.6\) cm. The volume \(V\) of a cube is given by \(V = ext{side}^3\). Calculate the volume: \[ V = (1.6 ext{ cm})^3 = 4.096 ext{ cm}^3 \]
02

Calculate the Weight of the Aluminum Cube

The weight \(W\) of the cube is the gravitational force acting on it, which can be calculated using the formula \(W = V \times \rho \times g\), where \(\rho\) is the density of aluminum \(2.70 \, \text{g/cm}^3\), and \(g\) is the acceleration due to gravity \(9.8 \, \text{m/s}^2\) or equivalently \(980 \, \text{cm/s}^2\) since volume was in \(\text{cm}^3\).Calculate the weight:\[ W = 4.096 \, \text{cm}^3 \times 2.70 \, \text{g/cm}^3 \times 980 \, \text{cm/s}^2 = 10836.864 \, ext{dynes} \]
03

Calculate the Buoyant Force

The buoyant force \(F_b\) is the force exerted by a fluid that opposes the weight of an object immersed in it. It can be calculated using Archimedes' principle: \(F_b = V \times \rho_{fluid} \times g\), where \(\rho_{fluid}\) is the density of the oil \(0.80 \, \text{g/cm}^3\).Calculate the buoyant force:\[ F_b = 4.096 \, \text{cm}^3 \times 0.80 \, \text{g/cm}^3 \times 980 \, \text{cm/s}^2 = 3214.976 \, ext{dynes} \]
04

Calculate the Tension in the Thread

The tension \(T\) in the thread is the net force required to keep the aluminum cube submerged. This is given by: \(T = W - F_b\).Subtract to find the tension:\[ T = 10836.864 \, ext{dynes} - 3214.976 \, ext{dynes} = 7621.888 \, ext{dynes} \]
05

Convert the Tension to SI Units (Newtons)

Since dynes are non-SI units, convert the tension to newtons, where \(1 \, ext{dyne} = 10^{-5} \, ext{N}\).Convert the tension:\[ T = 7621.888 \, ext{dynes} \times 10^{-5} = 0.7622 \, ext{N} \]

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Key Concepts

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

Density
Density is a fundamental concept in physics, providing a measure of how much mass is contained within a given volume. Imagine squeezing a big cloud into a tight ball - the mass stays the same, but the volume changes, making the ball denser. Similar to this, density is basically mass divided by volume \[ \rho = \frac{m}{V} \] where \( \rho \) is the density, \( m \) is the mass, and \( V \) is the volume. For the exercise mentioned, we use the density of aluminum, which is \(2.70 \, \text{g/cm}^3\), and calculate how much stuff (mass) is packed into a space (volume) to determine how heavy the aluminum cube is. When calculating the density of fluids like oil, this same formula helps us understand how well a fluid can support or oppose an object submerged in it.
Buoyant Force
Buoyant force is the uplifting force exerted by a fluid that allows objects to float or sink with less effort. Think of it as the resistance you feel when pressing down on an inflatable ball under water. This force acts upwards and counteracts the gravity pulling the object down. The buoyant force can be calculated using the formula: \[ F_b = V \times \rho_{fluid} \times g \] where \( F_b \) is the buoyant force, \( V \) represents the volume of the submerged portion of the object, \( \rho_{fluid} \) is the density of the fluid, and \( g \) is the acceleration due to gravity. In our example, the aluminum cube submerged in oil experiences a buoyant force which is different from air due to the oil's density (\(0.80 \, \text{g/cm}^3\)). This force makes the cube seem lighter than its actual weight, just like how a submerged stone feels lighter underwater.
Archimedes' Principle
Archimedes' Principle is a pivotal rule in understanding buoyant forces. It states that any object immersed in a fluid experiences a buoyant force equal to the weight of the fluid displaced by the object. This principle explains why ships float and why some objects sink or remain suspended. To grasp this, imagine placing a solid wood block in a tub of water. The block displaces a certain volume of water, and the weight of that water determines the upward buoyant force acting on the block. When considering the submerged aluminum cube in oil, Archimedes' Principle helps explain that the cube displaces a specific volume of oil. The weight of this volume of oil equals the buoyant force pushing upwards on the cube. Consequently, the tension in the thread holding the cube is not just the weight of the cube but the difference between its weight and the buoyant force. This principle is what sailors rely on to ensure that cargo ships remain buoyant and stable, no matter how loaded they are with goods.

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Most popular questions from this chapter

A reservoir dam holds an \(8.00-\mathrm{km}^{2}\) lake behind it. Just behind the dam, the lake is \(12.0 \mathrm{~m}\) deep. What is the water pressure \((a)\) at the base of the dam and \((b)\) at a point \(3.0 \mathrm{~m}\) down from the lake's surface? The area of the lake behind the dam has no effect on the pressure against the dam. At any point, \(P=\rho_{w} g h\). (a) \(P=\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(12.0 \mathrm{~m})=118 \mathrm{kPa}\) (b) \(P=\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(3.0 \mathrm{~m})=29 \mathrm{kPa}\)

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?

In a hydraulic press such as the one shown in Fig. \(13-3\), the large piston has cross-sectional area \(A_{1}=200 \mathrm{~cm}^{2}\) and the small piston has cross-sectional area \(A_{2}=5.0 \mathrm{~cm}^{2} .\) If a force of \(250 \mathrm{~N}\) is applied to the small piston, find the force \(F_{1}\) on the large piston. By Pascal's principle, Pressure under large piston \(=\) Pressure under small piston \(\quad\) or \(\quad \frac{F_{1}}{A_{1}}=\frac{F_{2}}{A_{2}}\) so that $$F_{1}=\frac{A_{1}}{A_{2}} F_{2}=\frac{200}{5.0} 250 \mathrm{~N}=10 \mathrm{kN}$$ Note that atmospheric pressure acting on both pistons cancels out of the calculation.

A vertical test tube has \(2.0 \mathrm{~cm}\) of oil \(\left(\rho=0.80 \mathrm{~g} / \mathrm{cm}^{3}\right)\) floating on \(8.0 \mathrm{~cm}\) of water. What is the pressure at the bottom of the tube due to the liquid in it? \(\begin{aligned} P &=\rho_{1} g h_{1}+\rho_{2} g h_{2}=\left(800 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(0.020 \mathrm{~m})+\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(0.080 \mathrm{~m}) \\ &=0.94 \mathrm{kPa} \end{aligned}\)

A glass tube is bent into the form of a U. A \(50.0-\mathrm{cm}\) height of olive oil in one arm is found to balance \(46.0 \mathrm{~cm}\) of water in the other. What is the density of the olive oil?

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