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Chapter 12: Problem 24

Estimate the fraction of your body's total volume that is above the surface of the water when you float in seawater (density \(1030 \mathrm{~kg} / \mathrm{m}^{3}\) ). (a) Use this estimate and your weight to calculate the total volume of your body. (b) What is your average density? How does your average density compare to the density of seawater?

Short Answer

Expert verified
Body volume is estimated using the principle of buoyancy, and density by mass over volume relationship. Comparison of average body density with seawater density can reveal if one floats or sinks in the sea. The actual values would depend on the individual's weight and estimated volume fraction above the water.

Step by step solution

01

Identify Known Quantities

Firstly, identify the known quantities from the exercise. Here, the density of seawater \(1030 \, \mathrm{kg/m^3}\) is given. Also, one can consider about half the body protrudes above the surface of water when floating in it. So, estimate a fraction (approximately 0.5) of body's total volume that is above the water.
02

Calculate Body's Total Volume

Now, calculate the total volume of the body using the principle of buoyancy. To float, the weight of the water displaced must equal the weight of the body. Thus, one can write: \[ \text{Weight} = \text{Density_seawater} \times \text{Volume_displaced} \] Solving for Volume_displaced: \[ \text{Volume_displaced} = \frac{\text{Weight}}{\text{Density_seawater}} \] To find the total volume of the body, calculate: \[ \text{Volume_body} = \frac{\text{Volume_displaced}}{1 - \text{Fraction_above_water}} \]
03

Calculate Average Body Density

The density of an object is its mass divided by its volume. Therefore, the mass of the body would be equal to its weight divided by the acceleration due to gravity. So body density can be calculated as: \[ \text{Density_body} = \frac{\text{Mass}}{\text{Volume_body}} = \frac{\text{Weight}}{\text{g}} \times \frac{1}{\text{Volume_body}} \] where \(g\) is the acceleration due to gravity.
04

Comparison with Seawater Density

Lastly, compare body density with the density of seawater to understand how buoyancy depends on the densities of both body and displaced fluid.

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Key Concepts

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

Density of Seawater
Understanding the density of seawater is essential when exploring the principles of buoyancy. Density is defined as the mass per unit volume of a substance, represented by the formula \( \text{density} = \frac{\text{mass}}{\text{volume}} \). Seawater has a density of about \( 1030 \text{ kg/m}^3 \), which varies slightly due to temperature and salinity differences. Because of its density, objects that are less dense can float in seawater, as they displace an amount of water equal in weight to themselves.

When it comes to floating in seawater, the density difference between the human body and the seawater dictates whether a person will float or sink. If the person's average body density is less than that of seawater, they will float, and if it's more, they will tend to sink. The interplay of these densities is governed by a fundamental principle in physics, commonly known as Archimedes' principle.
Archimedes' Principle
Archimedes' principle is a law of physics fundamental to fluid mechanics. It states that any object, wholly or partially submerged in a fluid, is buoyed up by a force equal to the weight of the fluid displaced by the object. This principle is the reason why ships made of steel float and how hot air balloons rise.

In application to our scenario, when a person is floating in seawater, the water that is pushed away or 'displaced' by their body weighs as much as the person does. This is because the buoyant force (upthrust) provided by the displaced seawater must balance the weight of the person for them to float. A practical way to use this principle is in calculating the volume of an object through the displacement of fluid, which ties in with the next concept of volume displacement.
Average Body Density
Average body density is a measure that can be more variable than many students think. It's the overall density of a person's body, considering all its various components - bones, muscles, fat, and other tissues. On average, human body density is slightly less than water at \( 985 \text{ kg/m}^3 \), allowing humans to float more easily in seawater.

To compute average body density using the principles of buoyancy, you can follow the method outlined in the exercise, which uses weight and volume displacement. Knowing your body's density is vital since it helps explain why your body floats or sinks in water. This average density affects the proportion of a person's body that remains above the water's surface while floating.
Volume Displacement
Volume displacement is a method to determine the volume of an object by the amount of fluid it displaces. According to Archimedes' principle mentioned earlier, the volume of fluid displaced is directly related to the volume of the object submerged.

In our floating body example, part of the body volume is submerged to displace water equal to the body's weight. From this, we can calculate the body's total volume, considering the proportion of it above water (fraction above water). The formula derived from these relationships allows us to solve for the body's volume using body weight and the known density of seawater. Hence, determining the volume displacement plays a crucial role in calculating not just the total volume of the body but also contribute to identifying the average body density.

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Most popular questions from this chapter

On the afternoon of January \(15,1919,\) an unusually warm day in Boston, a 17.7 -m-high, 27.4-m-diameter cylindrical metal tank used for storing molasses ruptured. Molasses flooded into the streets in a 5-m-deep stream, killing pedestrians and horses and knocking down buildings. The molasses had a density of \(1600 \mathrm{~kg} / \mathrm{m}^{3}\). If the tank was full before the accident, what was the total outward force the molasses exerted on its sides? (Hint: Consider the outward force on a circular ring of the tank wall of width \(d y\) and at a depth \(y\) below the surface. Integrate to find the total outward force. Assume that before the tank ruptured, the pressure at the surface of the molasses was equal to the air pressure outside the tank.)

You are designing a diving bell to withstand the pressure of seawater at a depth of \(250 \mathrm{~m}\). (a) What is the gauge pressure at this depth? (You can ignore changes in the density of the water with depth.) (b) At this depth, what is the net force due to the water outside and the air inside the bell on a circular glass window \(30.0 \mathrm{~cm}\) in diameter if the pressure inside the diving bell equals the pressure at the surface of the water? (Ignore the small variation of pressure over the surface of the window.)

A closed and elevated vertical cylindrical tank with diameter \(2.00 \mathrm{~m}\) contains water to a depth of \(0.800 \mathrm{~m}\). A worker accidently pokes a circular hole with diameter \(0.0200 \mathrm{~m}\) in the bottom of the tank. As the water drains from the tank, compressed air above the water in the tank maintains a gauge pressure of \(5.00 \times 10^{3} \mathrm{~Pa}\) at the surface of the water. Ignore any effects of viscosity. (a) Just after the hole is made, what is the speed of the water as it emerges from the hole? What is the ratio of this speed to the efflux speed if the top of the tank is open to the air? (b) How much time does it take for all the water to drain from the tank? What is the ratio of this time to the time it takes for the tank to drain if the top of the tank is open to the air?

If the elephant were to snorkel in salt water, which is more dense than freshwater, would the maximum depth at which it could snorkel be different from that in freshwater? (a) Yes - that depth would increase, because the pressure would be lower at a given depth in salt water than in freshwater; (b) yes - that depth would decrease, because the pressure would be higher at a given depth in salt water than in freshwater; (c) no, because pressure differences within the submerged elephant depend on only the density of air, not the density of the water; (d) no, because the buoyant force on the elephant would be the same in both cases.

The Environmental Protection Agency is investigating an abandoned chemical plant. A large, closed cylindrical tank contains an unknown liquid. You must determine the liquid's density and the height of the liquid in the tank (the vertical distance from the surface of the liquid to the bottom of the tank). To maintain various values of the gauge pressure in the air that is above the liquid in the tank, you can use compressed air. You make a small hole at the bottom of the side of the tank, which is on a concrete platform - so the hole is \(50.0 \mathrm{~cm}\) above the ground. The table gives your measurements of the horizontal distance \(R\) that the initially horizontal stream of liquid pouring out of the tank travels before it strikes the ground and the gauge pressure \(p_{\mathrm{g}}\) of the air in the tank. $$\begin{array}{l|lllll}p_{\mathrm{g}}(\mathrm{atm}) & 0.50 & 1.00 & 2.00 & 3.00 & 4.00 \\\\\hline \boldsymbol{R}(\mathrm{m}) & 5.4 & 6.5 & 8.2 & 9.7 & 10.9\end{array}$$ (a) Graph \(R^{2}\) as a function of \(p_{\mathrm{g}}\). Explain why the data points fall close to a straight line. Find the slope and intercept of that line. (b) Use the slope and intercept found in part (a) to calculate the height \(h\) (in meters) of the liquid in the tank and the density of the liquid (in \(\mathrm{kg} / \mathrm{m}^{3}\) ). Use \(g=9.80 \mathrm{~m} / \mathrm{s}^{2} .\) Assume that the liquid is nonviscous and that the hole is small enough compared to the tank's diameter so that the change in \(h\) during the measurements is very small.
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