/*! 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 30 Dead Sea About onc-third of the ... [FREE SOLUTION] | 91影视

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Chapter 15: Problem 30

Dead Sea About onc-third of the body of a person floating in the Dead Sea will be above the water line. Assuming that the human body density is \(0.98 \mathrm{~g} / \mathrm{cm}^{3}\), find the density of the water in the Dead Sea. (Why is it so much greater than \(1.0 \mathrm{~g} / \mathrm{cm}^{3} ?\) )

Short Answer

Expert verified
The density of the Dead Sea water is 1.47 g/cm鲁.

Step by step solution

01

Understand the given information

A person floating in the Dead Sea has one-third of their body above the water line. The density of the human body is given as 0.98 g/cm鲁.
02

Analyze the concept of buoyancy

When an object floats, the buoyant force equals the weight of the object. The fraction of the object submerged in the fluid is determined by the ratio of the densities of the object and the fluid.
03

Set up the equation for densities

Let the density of the Dead Sea water be \(\rho_{\text{water}}\). Since one-third of the body is above water, two-thirds is submerged. Thus, the buoyancy principle gives the ratio: \dfrac{\text{density of body}}{\text{density of water}} = \dfrac{2}{3}\.
04

Solve for the density of Dead Sea water

Using the ratio:\[ \dfrac{0.98 \ \text{g/cm}^3}{\rho_{\text{water}}} = \dfrac{2}{3} \]Rearrange to solve for \(\rho_{\text{water}}\):\[ \rho_{\text{water}} = 0.98 \ \text{g/cm}^3 \times \dfrac{3}{2} = 1.47 \ \text{g/cm}^3 \]

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Key Concepts

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

Density
Density is a measure of how much mass is contained in a given volume. It is commonly expressed in units of grams per cubic centimeter (g/cm鲁). To calculate the density, you use the formula: \[\text{Density} = \frac{\text{Mass}}{\text{Volume}} \] This concept is crucial for understanding why certain objects float while others sink. For instance, objects with a density greater than the fluid they are in will sink, while those with a lower density will float. In the exercise, the human body has a density of 0.98 g/cm鲁. This is slightly less dense than fresh water, which typically has a density of 1.0 g/cm鲁. Hence, a person can float in fresh water, but not as prominently as they can in the highly saline Dead Sea.
Archimedes' Principle
Archimedes' Principle states that an object immersed in a fluid experiences a buoyant force equal to the weight of the fluid displaced by the object. This principle helps explain why ships float and how hot air balloons rise. The mathematical expression of Archimedes' Principle is: \[\text{Buoyant Force} = \text{Weight of Displaced Fluid} \] When a person floats in the water, the buoyant force must balance the weight of the person. In this exercise, one-third of the person's body is above water, indicating that the buoyant force is supporting the other two-thirds. This relationship allows us to set up an equation using the densities of the person and the Dead Sea water to find the unknown density of the water.
Fluid Mechanics
Fluid mechanics is the branch of physics concerned with the behavior of liquids and gases at rest and in motion. It encompasses topics like buoyancy, pressure, and flow. It is essential for understanding how forces interact with fluids and is widely applied in fields such as engineering, meteorology, and even medicine. Principles of fluid mechanics help describe why certain fluids, like the highly saline water of the Dead Sea, have higher densities. The high salinity increases the mass per unit volume, thereby increasing the density significantly above that of regular freshwater. This higher density allows for greater buoyancy force, which is why a person floating in the Dead Sea can have approximately one-third of their body above water.

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Most popular questions from this chapter

Block of Wood A block of wood floats in fresh water with twothirds of its volume submerged. In oil the block floats with \(0.90\) of its volume submerged. Find the density of (a) the wood and (b) the oil.

Syringe Find the pressure increase in the fluid in a syringe when a nurse applies a force of \(42 \mathrm{~N}\) to the syringe's circular piston, which has a radius of \(1.1 \mathrm{~cm}\).

Wood with Lead A block of wood has a mass of \(3.67 \mathrm{~kg}\) and a density of \(600 \mathrm{~kg} / \mathrm{m}^{3} .\) It is to be loaded with lead so that it will float in water with \(0.90\) of its volume submerged. What mass of lead is needed (a) if the lead is attached to the top of the wood and (b) if the lead is attached to the bottom of the wood? The density of lead is \(1.13 \times 10^{4} \mathrm{~kg} / \mathrm{m}^{3}\)

Hollow Sphere 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 \(800 \mathrm{~kg} / \mathrm{m}^{3} .\) (a) What is the mass of the sphere? (b) Calculate the density of the material of which the sphere is made.

Venturi Meter A venturi meter is used to measure the flow speed of a fluid in a pipe. The meter is connected between two sections of the pipe (Fig. \(15-48\) ); the cross-sectional area \(A\) of the entrance and exit of the meter matches the pipe's cross-sectional arca At the entrance and cxit, the fluid flows through the pipe with speed \(v_{A}=\left|\vec{v}_{A}\right| .\) But it flows through a narrow "throat" of cross-sectional area \(B\) with speed \(v_{B}=\left|\vec{v}_{B}\right| .\) A manometer connects the wider portion of the meter to the narrower portion. The change in the fluid's speed is accompanied by a change \(\Delta P\) in the fluid's pressure, which causes a height difference \(h\) of the liquid in the two arms of the manometer. (Here \(\Delta P\) means pressure in the throat minus pressure in the pipe.) (a) By applying Bernoulli's equation and the equation of continuity to points 1 and 2 in Fig. \(15-48\), show that $$ \vec{v}_{A}=\sqrt{\frac{2 B^{2} \Delta P}{\rho\left(B^{2}-A^{2}\right)}} $$ where \(\rho\) is the density of the fluid. (b) Suppose that the fluid is fresh water, that the cross-sectional areas are \(64 \mathrm{~cm}^{2}\) in the pipe and \(32 \mathrm{~cm}^{2}\) in the throat, and that the pressure is \(55 \mathrm{kPa}\) in the pipe and \(41 \mathrm{kPa}\) in the throat. What is the rate of water flow in cubic meters per second?
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