/*! 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 43 (a) If the density of an object ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(a) If the density of an object is exactly equal to the density of a fluid, the object will (1) float, (2) sink, (3) stay at any height in the fluid, as long as it is totally immersed. (b) A cube \(8.5 \mathrm{~cm}\) on each side has a mass of \(0.65 \mathrm{~kg}\). Will the cube float or sink in water? Prove your answer.

Short Answer

Expert verified
(a) Stay at any height; (b) Cube will sink.

Step by step solution

01

Understanding Buoyancy

For part (a), we need to apply the principle of buoyancy. When an object is placed in a fluid, it experiences an upward force called the buoyant force. If the object's density is equal to the fluid's density, then the buoyant force is equal to the weight of the object, allowing it to stay at any height as long as it's immersed.
02

Calculation of Density

To solve part (b), we must calculate the density of the cube and compare it with the density of water. The density, \( \rho \), is given by the formula \( \rho = \frac{m}{V} \), where \( m \) is the mass and \( V \) is the volume. We have \( m = 0.65 \mathrm{~kg} \) and need to calculate \( V = (8.5^3) \mathrm{~cm^3} = 614.125 \mathrm{~cm^3} = 0.000614125 \mathrm{~m^3} \).
03

Unit Conversion for Volume

Convert the volume from \( \mathrm{cm^3} \) to \( \mathrm{m^3} \) using the conversion factor \( 1 \mathrm{~cm^3} = 1 \times 10^{-6} \mathrm{~m^3} \). Thus, \( 614.125 \mathrm{~cm^3} = 0.000614125 \mathrm{~m^3} \).
04

Density Calculation

Calculate the density using the formula \( \rho = \frac{m}{V} = \frac{0.65 \mathrm{~kg}}{0.000614125 \mathrm{~m^3}} \approx 1058.24 \mathrm{~kg/m^3} \).
05

Comparison with Water Density

The density of water is about \( 1000 \mathrm{~kg/m^3} \). Since \( 1058.24 \mathrm{~kg/m^3} \) (density of cube) is greater than \( 1000 \mathrm{~kg/m^3} \) (density of water), the cube will sink.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Density
The concept of density is fundamental in understanding why objects float or sink. Density is defined as the mass per unit volume of a substance and is mathematically expressed as \( \rho = \frac{m}{V} \), where \( \rho \) is density, \( m \) is mass, and \( V \) is volume.
For example, if you have two objects of the same size but different masses, the object with the higher mass will have a higher density. This can be visualized when comparing a piece of wood to a metal cube of the same size; typically, the metal is denser.
Understanding density helps in predicting whether an object will float or sink in a fluid, as it directly influences buoyant forces.
Density Calculation
Calculating the density of an object involves determining both its mass and its volume first. Let's take a cube with sides measuring 8.5 cm. To calculate its volume, you use the formula for the volume of a cube: \( V = s^3 \), where \( s \) is the length of a side. In this case, \( V = (8.5 \text{ cm})^3 = 614.125 \text{ cm}^3 \).
To convert this volume into cubic meters (a more commonly used metric unit), apply the conversion \( 1 \text{ cm}^3 = 10^{-6} \text{ m}^3 \). Thus, \( 614.125 \text{ cm}^3 = 0.000614125 \text{ m}^3 \).
With the mass given as 0.65 kg, we use the density formula \( \rho = \frac{m}{V} \) to find that \( \rho = \frac{0.65}{0.000614125} \approx 1058.24 \text{ kg/m}^3 \). This calculation indicates a relatively high density for the cube.
Principle of Buoyancy
Buoyancy refers to the upward force that a fluid exerts on an object placed in it. This force can make objects appear lighter and can even allow them to float. Archimedes' Principle states that the buoyant force on an object is equal to the weight of the fluid displaced by the object.
In practical terms, if a cube is immersed in water, it experiences this upward force. The interplay between the object's weight and the buoyant force determines its ability to float. Specifically:
  • If the object's density is less than the fluid's, it will float.
  • If the object's density is greater, it will sink.
  • If the densities are equal, the object will remain suspended wherever placed in the fluid.
Understanding buoyancy is key to predicting whether the cube from the exercise will float or sink.
Comparison with Water Density
To determine whether the cube will float or sink, its density must be compared with the density of the fluid, typically water. The standard density of water is about \( 1000 \text{ kg/m}^3 \). When we calculated the cube's density as approximately \( 1058.24 \text{ kg/m}^3 \), we found it to be greater than that of water.
This comparison shows that the cube, being denser, will sink when placed in water. Objects only float when their density is less than that of the fluid, allowing the buoyant force to counteract their weight efficiently.
Thus, the exercise helps illustrate how a simple calculation and comparison can predict the behavior of objects in fluids.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In a dramatic lecture demonstration, a physics professor blows hard across the top of a copper penny that is at rest on a level desk. By doing this at the right speed, he can get the penny to accelerate vertically, into the airstream, and then deflect it into a tray, as shown in Fig. 9.42. Assuming the diameter of a penny is \(1.90 \mathrm{~cm}\) and its mass is \(2.50 \mathrm{~g}\), what is the minimum airspeed needed to lift the penny off the tabletop? Assume the air under the penny remains at rest.

The gauge pressure in both tires of a bicycle is \(690 \mathrm{kPa}\). If the bicycle and the rider have a combined mass of \(90.0 \mathrm{~kg}\), what is the area of contact of each tire with the ground? (Assume that each tire supports half the total weight of the bicycle.)

(a) Given a piece of metal with a light string attached, a scale, and a container of water in which the piece of metal can be submersed, how could you find the volume of the piece without using the variation in the water level? (b) An object has a weight of \(0.882 \mathrm{~N}\). It is suspended from a scale, which reads \(0.735 \mathrm{~N}\) when the piece is submerged in water. What are the volume and density of the piece of metal?

In a sample of seawater taken from an oil spill, an oil layer \(4.0 \mathrm{~cm}\) thick floats on \(55 \mathrm{~cm}\) of water. If the density of the oil is \(0.75 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3},\) what is the absolute pressure on the bottom of the container?

An ideal fluid is moving at \(3.0 \mathrm{~m} / \mathrm{s}\) in a section of a pipe of radius \(0.20 \mathrm{~m}\). If the radius in another section is \(0.35 \mathrm{~m},\) what is the flow speed there?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.