Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A piece of wood is 0.600 m long, 0.250 m wide, and 0.080 m thick. Its density is 700 kg/m\(^3\). What volume of lead must be fastened underneath it to sink the wood in calm water so that its top is just even with the water level? What is the mass of this volume of lead?

Short Answer

Expert verified
3.6 kg of lead is needed, and its volume is approximately 0.000317 m³.

Step by step solution

01

Calculate the Volume of the Wood

The volume of the wood can be calculated using the formula for the volume of a rectangular prism:\[ V_{wood} = ext{length} \times ext{width} \times ext{thickness} \]Substituting in the values given:\[ V_{wood} = 0.600 \, \text{m} \times 0.250 \, \text{m} \times 0.080 \, \text{m} = 0.012 \, \text{m}^3 \]
02

Calculate the Mass of the Wood

The mass of the wood can be calculated using its density and volume:\[ m_{wood} = \rho_{wood} \times V_{wood} \]Where the density \( \rho_{wood} \) is 700 kg/m\(^3\) and \( V_{wood} \) is 0.012 m\(^3\):\[ m_{wood} = 700 \, \text{kg/m}^3 \times 0.012 \, \text{m}^3 = 8.4 \, \text{kg} \]
03

Calculate the Buoyant Force Needed

The buoyant force needed to keep the wood completely submerged with its top even with the water level is equal to the weight of the water displaced by the wood:\[ B = \text{density of water} \times \text{gravitational acceleration} \times V_{wood} \]In SI units, the density of water is 1000 kg/m\(^3\) and the gravitational acceleration is approximately 9.8 m/s\(^2\):\[ B = 1000 \, \text{kg/m}^3 \times 9.8 \, \text{m/s}^2 \times 0.012 \, \text{m}^3 = 117.6 \, \text{N} \]
04

Calculate the Required Mass of Lead

The buoyant force must equal the combined weight of the wood and the lead, when the system is just submerged. Using:\[ B = m_{total} \times g \]Where \( m_{total} = m_{wood} + m_{lead} \) and \( g = 9.8 \, \text{m/s}^2 \). Since \( B = 117.6 \, \text{N} \):\[ m_{wood} + m_{lead} = \frac{B}{g} \]\[ 8.4 \, \text{kg} + m_{lead} = \frac{117.6 \, \text{N}}{9.8 \, \text{m/s}^2} = 12 \, \text{kg} \]Solving for \( m_{lead} \):\[ m_{lead} = 12 \, \text{kg} - 8.4 \, \text{kg} = 3.6 \, \text{kg} \]
05

Calculate the Volume of Lead

The volume of lead can be found using its density. Let's denote lead's density as \( \rho_{lead} = 11340 \, \text{kg/m}^3 \):\[ V_{lead} = \frac{m_{lead}}{\rho_{lead}} \]\[ V_{lead} = \frac{3.6 \, \text{kg}}{11340 \, \text{kg/m}^3} \approx 0.000317 \, \text{m}^3 \]

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 Vaia!

Key Concepts

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

Density
Density is a crucial concept in physics and many other scientific areas. It describes how much mass a particular substance contains in a given volume. For example, in the problem of the wood floating on water, the wood's density is given as 700 kg/m³. This means if you have 1 cubic meter of this wood, its mass would be 700 kg.
Understanding density helps in comparing how substances differ in compactness or heaviness per unit volume.
For instance:
  • Higher density indicates that the matter's particles are closely packed together.
  • Lower density suggests the particles are further apart, making the material lighter for its size.
This property is particularly relevant when exploring buoyancy, as it determines whether an object will float or sink when placed in a fluid.
Volume Calculation
Calculating the volume of an object is determining how much space it occupies. It's vital to solve many physics problems, especially those involving buoyancy. The volume of an object is often determined by its shape and dimensions.
In this example, we calculate the volume of wood using the formula for the volume of a rectangular prism, which is:
  • Volume = Length × Width × Thickness
For the wood, with dimensions 0.600 m, 0.250 m, and 0.080 m, the volume comes out to be 0.012 m³.
Accurate volume calculation is key because it directly influences other calculations, such as determining the mass of a material or its buoyant force in a fluid. A mistake in measuring or computing volume can lead to errors in subsequent physics calculations.
Mass Calculation
Mass calculation involves using the density and volume of a substance to determine how much matter is present. From the example:
  • We know the volume of the wood is 0.012 m³
  • The density provided is 700 kg/m³
By multiplying these values, the mass of the wood is found using the formula:
  1. Mass = Density × Volume
This yields a mass of 8.4 kg for the wood.
Understanding how to calculate mass from density and volume is pivotal in physics, as it is a step that often interlinks with calculating forces like buoyant force in various physics problems.
Physics Problem Solving
Solving physics problems involves a structured approach where understanding the problem's physical nature is crucial. In this problem:
  • We started with the goal of finding how much lead needs to be added to sink the wood in water so that its top is just even with the water level.
  • This required calculating the buoyant force necessary, which equaled the combined weight of the wood and lead when just submerged.
  • We calculated the needed volume of lead by first finding the mass of lead using its given density.
Physics problem solving often involves multiple interconnected steps. For buoyancy problems, one often calculates forces, assesses volume and mass, and applies key concepts like Archimedes' principle in determining floating or sinking behavior.

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

A pressure difference of 6.00 \(\times\) 10\(^4\) Pa is required to maintain a volume flow rate of 0.800 m\(^3\)/s for a viscous fluid flowing through a section of cylindrical pipe that has radius 0.210 m. What pressure difference is required to maintain the same volume flow rate if the radius of the pipe is decreased to 0.0700 m?

The densities of air, helium, and hydrogen (at \(p\) \(=\) 1.0 atm and \(T\) \(=\) 20\(^\circ\)C) are 1.20 kg/m\(^3\), 0.166 kg/m\(^3\), and 0.0899 kg/m\(^3\), respectively. (a) What is the volume in cubic meters displaced by a hydrogen- filled airship that has a total "lift" of 90.0 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 barge is in a rectangular lock on a freshwater river. The lock is 60.0 m long and 20.0 m wide, and the steel doors on each end are closed. With the barge floating in the lock, a 2.50 \(\times\) 10\(^6\) N load of scrap metal is put onto the barge. The metal has density 7200 kg/m\(^3\). (a) When the load of scrap metal, initially on the bank, is placed onto the barge, what vertical distance does the water in the lock rise? (b) The scrap metal is now pushed overboard into the water. Does the water level in the lock rise, fall, or remain the same? If it rises or falls, by what vertical distance does it change?

You need to extend a 2.50-inch-diameter pipe, but you have only a 1.00-inch- diameter pipe on hand. You make a fitting to connect these pipes end to end. If the water is flowing at 6.00 cm/s in the wide pipe, how fast will it be flowing through the narrow one?

What gauge pressure is required in the city water mains for a stream from a fire hose connected to the mains to reach a vertical height of 15.0 m? (Assume that the mains have a much larger diameter than the fire hose.)

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.

Sign-up for free