/*! 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 63 A piece of wood is 0.600 \(\math... [FREE SOLUTION] | 91Ó°ÊÓ

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A piece of wood is 0.600 \(\mathrm{m}\) long, 0.250 \(\mathrm{m}\) wide, and 0.080 \(\mathrm{m}\) thick. Its density is 600 \(\mathrm{kg} / \mathrm{m}^{3}\) . 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
A volume of 0.000423 m\(^3\) of lead is needed with a mass of 4.8 kg.

Step by step solution

01

Calculate the Volume of the Wood

The volume of a cuboid can be calculated using the formula \( V = ext{length} \times ext{width} \times ext{height} \). Thus, the volume of the wood is given by:\[V = 0.600 \, \text{m} \times 0.250 \, \text{m} \times 0.080 \, \text{m} = 0.012 \, \text{m}^3.\]
02

Calculate the Mass of the Wood

To find the mass of the wood, use the formula \( m = ext{density} \times ext{volume} \). With the density of 600 kg/m\(^3\), the mass of the wood is:\[m = 600 \, \text{kg/m}^3 \times 0.012 \, \text{m}^3 = 7.2 \, \text{kg}.\]
03

Determine the Total Displaced Water Volume Needed

When the wood is floating with its top just even with the water level, the total displaced water volume should equal the volume of the wood and the volume of the lead added. This total volume should also provide a buoyant force equal to the weight of the wood plus lead.
04

Find the Mass of Lead Required for Equilibrium

For the wood and lead to just sink, the mass of the water displaced (volume = 0.012 m\(^3\)) must equal the mass of the wood plus the mass of the lead. The density of water is 1000 kg/m\(^3\), so the water mass is 0.012 m\(^3\) × 1000 kg/m\(^3\) = 12 kg. Hence, mass of lead + mass of wood = 12 kg. Since the mass of the wood is 7.2 kg, the mass of the lead needed is:\[m_{\text{lead}} = 12 \, \text{kg} - 7.2 \, \text{kg} = 4.8 \, \text{kg}.\]
05

Calculate the Volume of Lead Required

We need to find the volume of the lead that corresponds to a mass of 4.8 kg. The density of lead is approximately 11340 kg/m\(^3\), so:\[V_{\text{lead}} = \frac{m_{\text{lead}}}{\text{density of lead}} = \frac{4.8 \, \text{kg}}{11340 \, \text{kg/m}^3} \approx 0.000423 \text{ m}^3.\]

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Key Concepts

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

Density
Density is a fundamental concept in physics that refers to how much mass is contained in a given volume of a substance. It is typically expressed in units of kilograms per cubic meter (kg/m tag). Understanding density helps to compare how much matter is packed into different materials.
  • In the case of the exercise, the wood's density is given as 600 kg/m tag. This means for every cubic meter of wood, it has a mass of 600 kilograms.
  • Density is crucial when determining buoyancy, as a material will float if its density is less than the fluid it's in.
  • For water, the density typically used is 1000 kg/m tag, which serves as a benchmark for determining if an object’s density is higher or lower.
Knowing these values allows you to predict whether the wood will float or need additional mass to fully submerge.
Volume calculation
Volume calculation involves determining the amount of space an object occupies. In mathematical terms, it is usually expressed in cubic meters (m tag). Calculating volume is essential when dealing with objects submerged in a fluid, as it directly relates to the volume of fluid displaced.In the original exercise, the volume of the wood is calculated as:- Using the formula for the volume of a cuboid: \[ V = ext{length} \times \text{width} \times \text{height} \]- For this wood: \[ V = 0.600 \, \text{m} \times 0.250 \, \text{m} \times 0.080 \, \text{m} = 0.012 \, \text{m}^3 \]This calculation informs us on how much water must be displaced for the wood to be submerged. To make an object float just below the surface, the volume of displaced water matches the object's volume. Additional volume translates into additional buoyant force, necessary for balance with the combined weight of wood and any added mass.
Mass calculation
Mass calculation involves using the relationship between density and volume. It determines how much an object weighs—a fundamental step for predicting how an object behaves in a fluid.- **Formula**: The mass of an object can be found using: \[ m = \text{density} \times \text{volume} \]- In the exercise, for the wood with a density of 600 kg/m\( ^3 \) and a volume of 0.012 m\( ^3 \), the mass is: \[ m = 600 \, \text{kg/m}^3 \times 0.012 \, \text{m}^3 = 7.2 \, \text{kg} \]- For the lead, given its density of 11340 kg/m\( ^3 \) and required mass of 4.8 kg: \[ V_{\text{lead}} = \frac{4.8 \, \text{kg}}{11340 \, \text{kg/m}^3} \approx 0.000423 \, \text{m}^3 \]Mass calculation helps determine the necessary conditions for objects to float, sink, or remain in equilibrium in a fluid environment. It also calculates the amount of additional mass required to achieve a specific buoyant state, such as needing a 4.8 kg lead mass to submerge the wood completely.

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Most popular questions from this chapter

A golf course sprinkler system discharges water from a horizontal pipe at the rate of 7200 \(\mathrm{cm}^{3} / \mathrm{s}\) . At one point in the pipe, where the radius is 4.00 \(\mathrm{cm}\) , the water's absolute pressure is \(2.40 \times 10^{5} \mathrm{Pa}\) . At a second point in the pipe, the water passes through a constriction where the radius is \(2.00 \mathrm{cm} .\) What is the water's absolute pressure as it flows through this constriction?

A single ice cube with mass 9.70 g floats in a glass completely full of 420 \(\mathrm{cm}^{3}\) of water. You can ignore the water's surface tension and its variation in density with temperature (as long as it remains a liquid). (a) What volume of water does the ice cube displace? (b) When the ice cube has completely melted, has any water overflowed? If so, how much? If not, explain why this is so. (c) Suppose the water in the glass had been very salty water of density 1050 \(\mathrm{kg} / \mathrm{m}^{3}\) . What volume of salt water would the \(9.70-\mathrm{g}\) ice cube displace? (d) Redo part (b) for the freshwater ice cube in the salty water.

You purchase a rectangular piece of metal that has dimensions \(5.0 \times 15.0 \times 30.0 \mathrm{mm}\) and mass 0.0158 \(\mathrm{kg}\) . The seller tells you that the metal is gold. To check this, you compute the average density of the piece. What value do you get? Were you cheated?

At a certain point in a horizontal pipeline, the water's speed is 2.50 \(\mathrm{m} / \mathrm{s}\) and the gauge pressure is \(1.80 \times 10^{4} \mathrm{Pa}\) . Find the gauge pressure at a second point in the line if the cross-sectional area at the second point is twice that at the first.

A tall cylinder with a cross-sectional area 12.0 \(\mathrm{cm}^{2}\) is partially filled with mercury; the surface of the mercury is 5.00 \(\mathrm{cm}\) above the bottom of the cylinder. Water is slowly poured in on top of the mercury, and the two fluids don't mix. What volume of water must be added to double the gauge pressure at the bottom of the cylinder?

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