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Chapter 13: Problem 88

A ship, carrying fresh water to a desert island in the Caribbean, has a horizontal cross-sectional area of \(2240 \mathrm{~m}^{2}\) at the waterline. When unloaded, the ship rises \(8.50 \mathrm{~m}\) higher in the sea. How many cubic meters of water was delivered?

Short Answer

Expert verified
The ship delivered 19,040 cubic meters of water.

Step by step solution

01

Understand the Relationship of Volume and Height

The volume of water delivered is related to the cross-sectional area of the ship at the waterline and the change in height of the ship. The ship rises by this height as the load is removed.
02

Recognize the Formula for Volume Calculation

Since the ship rises 8.50 m when the water is unloaded and the cross-sectional area at the waterline remains constant, the volume of water delivered can be found using the formula: \( \text{Volume} = \text{Area} \times \text{Height} \).
03

Calculate the Volume of Water Delivered

Here, \( \text{Area} = 2240 \ m^2 \) and \( \text{Height} = 8.50 \ m \). Substitute these values into the formula: \( \text{Volume} = 2240 \ m^2 \times 8.50 \ m \).
04

Perform the Multiplication

Calculate the volume: \[ \text{Volume} = 2240 \times 8.50 = 19040 \ m^3 \] Thus, the ship delivered 19,040 cubic meters of water.

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Key Concepts

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

Understanding Buoyancy
Buoyancy is an essential concept to grasp when studying how objects float or sink in a fluid. It explains why certain objects are able to float while others sink, based on the interaction between the object and the fluid it is in.
The principle behind this phenomenon is known as Archimedes' principle, which states that an object is buoyed up by a force equal to the weight of the fluid it displaces.
For a ship carrying water, like in the exercise, this principle enables it to stay afloat. When the ship is loaded with water, it displaces an additional volume of sea water equivalent to the weight of the freshwater it carries.
This causes the ship to sink lower in the water. Conversely, when the freshwater is unloaded, the ship rises as less sea water is displaced. Overall, understanding buoyancy helps us analyze the ship's behavior and how it interacts with the water around it.
Concept of Displacement
Displacement is closely tied to buoyancy and is crucial for determining how much fluid an object displaces when it is placed in the fluid.
In the context of our exercise with the ship, displacement refers to the volume of sea water that the ship has pushed aside when it is carrying its load of fresh water.
As the load is removed, the ship rises by a given height because the displaced sea water moves back into the space the ship vacates.
To calculate the volume of water the ship was carrying, we use the relationship between cross-sectional area and height.
Since displacement involves the space the object occupies in the water, we measure this by multiplying the cross-sectional area of the ship at the waterline by the vertical rise observed when the load is removed.
Significance of Cross-Sectional Area
Cross-sectional area plays a vital role in calculating both buoyancy and displacement. It refers to the area of an object's slice that is parallel to the top surface of the liquid, like the deck of the ship at the waterline.
In the given exercise, the cross-sectional area is given as 2240 square meters.
This measurement remains unchanged regardless of the ship's load and impacts how much a ship will rise or fall in the water when the load is changed.
  • The larger the cross-sectional area, the more water is displaced when the ship is loaded or unloaded.
  • This influences the amount of buoyant force acting on the vessel.
Therefore, knowing the cross-sectional area is fundamental for accurately calculating the volume of water that the ship was carrying.

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Most popular questions from this chapter

(I) A 15 -cm-radius air duct is used to replenish the air of a room 8.2 \(\mathrm{m} \times 5.0 \mathrm{m} \times 3.5 \mathrm{m}\) every 12 \(\mathrm{min.}\) How fast does the air flow in the duct?

(II) Archimedes' principle can be used not only to determine the specific gravity of a solid using a known liquid (Example \(13-10\) ); the reverse can be done as well. ( \(a\) ) As an example, a \(3.80-\mathrm{kg}\) aluminum ball has an apparent mass of \(2.10 \mathrm{~kg}\) when submerged in a particular liquid: calculate the density of the liquid. (b) Derive a formula for determining the density of a liquid using this procedure.

(III) A cylindrical bucket of liquid (density \(\rho\) ) is rotated about its symmetry axis, which is vertical. If the angular velocity is \(\omega\), show that the pressure at a distance \(r\) from the rotation axis is $$ P=P_{0}+\frac{1}{2} \rho \omega^{2} r^{2} $$ where \(P_{0}\) is the pressure at \(r=0\)

(II) An open-tube mercury manometer is used to measure the pressure in an oxygen tank. When the atmospheric pressure is \(1040 \mathrm{mbar}\), what is the absolute pressure (in \(\mathrm{Pa}\) ) in the tank if the height of the mercury in the open tube is (a) \(21.0 \mathrm{~cm}\) higher, \((b) 5.2 \mathrm{~cm}\) lower, than the mercury in the tube connected to the tank?

(III) Water stands at a height \(h\) behind a vertical dam of uniform width \(b .(a)\) Use integration to show that the total force of the water on the dam is \(F=\frac{1}{2} \rho g h^{2} b .\) (b) Show that the torque about the base of the dam due to this force can be considered to act with a lever arm equal to \(h / 3\). (c) For a freestanding concrete dam of uniform thickness \(t\) and height \(h,\) what minimum thickness is needed to prevent overturning? Do you need to add in atmospheric pressure for this last part? Explain.
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