/*! 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 28 A flat-bottomed barge, loaded wi... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 28

A flat-bottomed barge, loaded with coal, has a mass of $3.0 \times 10^{5} \mathrm{kg} .\( The barge is \)20.0 \mathrm{m}\( long and \)10.0 \mathrm{m}$ wide. It floats in fresh water. What is the depth of the barge below the waterline? (W) tutorial: boat)

Short Answer

Expert verified
Answer: The depth of the barge below the waterline in fresh water is 1.5 meters.

Step by step solution

01

Calculate the weight of the barge

Remember that the weight of an object is given by its mass times the acceleration due to gravity (g = 9.81 m/s虏). So, the weight of the barge is given by: Weight = Mass 脳 g Weight = \( 3.0\times10^{5}\,\text{kg} \times 9.81\,\text{m/s}^2\) Weight = \(2.94\times10^6\,\text{N}\)
02

Calculate the buoyant force on the barge

According to Archimedes' principle, the buoyant force acting on an object is equal to the weight of the fluid displaced by the object. Since the barge is floating, the buoyant force must be equal to the weight of the barge. Buoyant force = Weight Buoyant force = \(2.94\times10^6\,\text{N}\)
03

Calculate the volume of water displaced by the barge

Now we can use the buoyant force to find the volume of water displaced by the barge. Recall that the buoyant force is equal to the weight of the fluid displaced. Therefore, we can use the formula: Volume = Buoyant force / (Density of water 脳 g) The density of fresh water is \(1000\,\text{kg/m}^3\). So, we get: Volume = \(\frac{2.94\times10^6\,\text{N}}{1000\,\text{kg/m}^3 \times 9.81\,\text{m/s}^2}\) Volume = \(300\,\text{m}^3\)
04

Calculate the depth below the waterline

Finally, we can find the depth of the barge below the waterline by dividing the volume of the displaced water by the area of the barge. The area can be calculated as the product of the length and width of the barge. Area = Length 脳 Width Area = \(20.0\,\text{m} \times 10.0\,\text{m}\) Area = \(200\,\text{m}^2\) Now we can calculate the depth: Depth = Volume / Area Depth = \(\frac{300\,\text{m}^3}{200\,\text{m}^2}\) Depth = \(1.5\,\text{m}\) So the depth of the barge below the waterline in fresh water is \(1.5\,\text{m}\).

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Most popular questions from this chapter

A cube that is \(4.00 \mathrm{cm}\) on a side and of density $8.00 \times 10^{2} \mathrm{kg} / \mathrm{m}^{3}$ is attached to one end of a spring. The other end of the spring is attached to the base of a beaker. When the beaker is filled with water until the entire cube is submerged, the spring is stretched by \(1.00 \mathrm{cm} .\) What is the spring constant?
A garden hose of inner radius \(1.0 \mathrm{cm}\) carries water at $2.0 \mathrm{m} / \mathrm{s} .\( The nozzle at the end has radius \)0.20 \mathrm{cm} .$ How fast does the water move through the nozzle?
A horizontal segment of pipe tapers from a cross sectional area of $50.0 \mathrm{cm}^{2}\( to \)0.500 \mathrm{cm}^{2} .$ The pressure at the larger end of the pipe is \(1.20 \times 10^{5} \mathrm{Pa}\) and the speed is $0.040 \mathrm{m} / \mathrm{s} .$ What is the pressure at the narrow end of the segment?
Are evenly spaced specific gravity markings on the cylinder of a hydrometer equal distances apart? In other words, is the depth \(d\) to which the cylinder is submerged linearly related to the density \(\rho\) of the fluid? To answer this question, assume that the cylinder has radius \(r\) and mass \(m .\) Find an expression for \(d\) in terms of \(\rho, r,\) and \(m\) and see if \(d\) is a linear function of \(\rho\).
A shallow well usually has the pump at the top of the well. (a) What is the deepest possible well for which a surface pump will work? [Hint: A pump maintains a pressure difference, keeping the outflow pressure higher than the intake pressure.] (b) Why is there not the same depth restriction on wells with the pump at the bottom?
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