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Chapter 13: Problem 39

A racquetball with a diameter of \(5.6 \mathrm{~cm}\) and a mass of \(42 \mathrm{~g}\) is cut in half to make a boat for American pennies made after \(1982 .\) The mass and volume of an American penny made after 1982 are \(2.5 \mathrm{~g}\) and \(0.36 \mathrm{~cm}^{3} .\) How many pennies can be placed in the racquetball boat without sinking it?

Short Answer

Expert verified
A maximum of 15 pennies can be placed in the racquetball boat without sinking it, based on the principle of buoyancy and the given mass and diameter of the racquetball boat.

Step by step solution

01

Calculate the volume of the racquetball half sphere

To calculate the volume of the half sphere, we will first calculate the volume of the entire sphere, then divide it by 2. The formula for the volume of a sphere is given by: \(V = \dfrac{4}{3}\pi r^3\) Where V is the volume and r is the radius. Given the diameter of the racquetball as \(5.6~\mathrm{cm}\), the radius will be half of that, which is \(2.8~\mathrm{cm}\). Plugging in the radius into the formula, we find the volume of the entire sphere: \(V = \dfrac{4}{3}\pi (2.8)^3 = 91.95~\mathrm{cm^3}\) Now, divide the volume by 2 to get the volume of the half sphere (boat): \(V_{boat} = \dfrac{91.95}{2} = 45.98~\mathrm{cm^3}\)
02

Calculate the buoyant force of the racquetball boat

The buoyant force (F_b) can be determined using the formula: \(F_b = V_{boat} \rho_{water} g\) Where 蟻 is the density of the water (approximately \(1000~\mathrm{kg/m^3}\) or \(1~\mathrm{g/cm^3}\)) and g is the gravitational acceleration (approximately \(9.81~\mathrm{m/s^2}\) or \(981~\mathrm{cm/s^2}\)). Plugging in the values, we get: \(F_b = (45.98) (1) (981) = 45105.58~\mathrm{dyne}\)
03

Calculate the total weight of the boat and the pennies

Given the mass of the racquetball boat as \(42~\mathrm{g}\), we now convert it to the weight: \(W_{boat} = m_{boat}g = (42)(981) = 41202~\mathrm{dyne}\) Let \(n\) be the number of pennies that can be placed in the racquetball boat without sinking it. The total weight of n pennies is: \(W_{pennies} = nm_{penny}g = (n)(2.5)(981)\)
04

Calculate the number of pennies that can be placed in the racquetball boat without sinking it

Using the principle of buoyancy, the boat will not sink if the buoyant force is equal to or greater than the total weight of the boat and the pennies. We can set up the following inequality: \(F_b \geq W_{boat} + W_{pennies}\) Now, substitute the values we found earlier: \(45105.58 \geq 41202 + (n)(2.5)(981)\) Solving for n, we get: \(n \leq \dfrac{45105.58 - 41202}{(2.5)(981)} \approx 15.78\) Since we cannot have a fraction of a penny, we round down to the nearest whole number: \(n = 15\) Therefore, a maximum of 15 pennies can be placed in the racquetball boat without sinking it.

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Key Concepts

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

Archimedes' Principle
Understanding Archimedes' principle is essential for grasping why objects float or sink. This principle states that any object, wholly or partially submerged in a fluid, is buoyed up by a force that is equal to the weight of the fluid that the object displaces.

When you place an object in water, like a racquetball halved to serve as a boat, it pushes away some water to make room for itself. The 鈥榩ushed away鈥� water has weight, and this weight is exactly what Archimedes' principle says will be the upward force, or the buoyant force, acting on the object.

For the racquetball boat we鈥檙e considering, the displaced water must weigh at least as much as the boat and any pennies it carries, for it to float. If the boat and coins are heavier than the water displaced, the buoyant force is not enough to keep it afloat, and it will sink. This is why calculating the volume of the boat 鈥� which tells us how much water it displaces 鈥� is such a crucial step in figuring out how many pennies it can carry.
Buoyant Force Calculation
Knowing Archimedes' principle, we can calculate the buoyant force to predict whether an object will float. The buoyant force can be expressed numerically using the equation:
\(F_b = V_{displaced} \times \rho_{fluid} \times g\)
Here, \(F_b\) is the buoyant force, \(V_{displaced}\) is the volume of the fluid displaced by the object, \(\rho_{fluid}\) is the density of the fluid, and \(g\) is the acceleration due to gravity.

The solution demonstrates this calculation by first finding the volume of the boat and then multiplying it by the density of water and gravitational acceleration to find the upward force exerted by the water, or the buoyant force. If the buoyant force is at least equal to the weight of the boat with pennies, then it will float; otherwise, it will sink. This is a clear illustration of how the buoyant force is crucial to solving problems related to whether an object will float or sink.
Density and Buoyancy
The concept of density plays a pivotal role in understanding buoyancy. Density is defined as mass per unit volume and is a measure of how much matter is packed into a space. An object that is less dense than the fluid it's in will float because it displaces a volume of fluid whose weight is greater than the object's own weight. This is why the racquetball boat, which is filled with air, is less dense than water and therefore capable of floating.

Density not only applies to the object but also to the fluid in which it is submerged. In the case of our racquetball boat, we use the density of water to calculate the buoyant force. Using the formula for buoyant force, we can infer that if the density of the fluid were to change (say, if the boat was placed in saltwater, which is denser than freshwater), the buoyant force would change as well, potentially altering how many pennies the boat could hold without sinking. This highlights the intricate relationship between density and buoyancy and their roles in predicting the behavior of submerged objects.

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Most popular questions from this chapter

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