/*! 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 91 The fat-to-muscle ratio of a per... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 91

The fat-to-muscle ratio of a person may be determined from a specific gravity measurement. The measurement is made by immersing the body in a tank of water and measuring the net weight. Develop an expression for the specific gravity of a person in terms of their weight in air, net weight in water, and \(\mathrm{SG}=f(T)\) for water.

Short Answer

Expert verified
The specific gravity of a person is given by the expression: \n \( SG = \frac{\text{Weight in air}}{\text{Weight in air} - \text{Net weight in water}} \times \text{SG of water} \)

Step by step solution

01

Understand Specific Gravity

Specific gravity is a ratio of the density of a substance to the density of a reference substance. It's dimensionless - i.e., it does not have any units. Usually, the reference substance is water, so the formula will be \( SG = \frac{\text{density of the substance}}{\text{density of water}} \)
02

Apply Archimedes Principle

Archimedes' Principle states that the buoyant force on a submerged object is equal to the weight of the fluid that is displaced by the object. From Archimedes principle, we can say \( \text{Weight in air} - \text{Net weight in water} = \text{Weight of the water displaced} \)
03

Deduce New Expression for Specific Gravity

Now we know that the weight of the water displaced equals the volume of the person times the density of water. From this, we can define the density of the person as \( \frac{\text{Weight in air}}{\text{Volume}} \). Substitute the volume from the Archimedes鈥� Principle equation which gives us \n\( \frac{\text{Weight in air}}{\text{Weight in air} - \text{Net weight in water}} \times \text{SG of water} = \text{Specific gravity of the person} \)

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Most popular questions from this chapter

The cross-sectional shape of a canoe is modeled by the curve \(y=a x^{2},\) where \(a=1.2 \mathrm{ft}^{-1}\) and the coordinates are in feet. Assume the width of the canoe is constant at \(w=2 \mathrm{ft}\) over its entire length \(L=18 \mathrm{ft}\). Set up a general algebraic expression relating the total mass of the canoe and its contents to distance \(d\) between the water surface and the gunwale of the floating canoe. Calculate the maximum total mass allowable without swamping the canoe.

Three steel balls (each about half an inch in diameter ) lie at the bottom of a plastic shell floating on the water surface in a partially filled bucket. Someone removes the steel balls from the shell and carefully lets them fall to the bottom of the bucket, leaving the plastic shell to float empty. What happens to the water level in the bucket? Does it rise, go down, or remain unchanged? Explain.

An open tank is filled to the top with water. A steel cylindrical container, wall thickness \(\delta=1 \mathrm{mm},\) outside diameter \(D=100 \mathrm{mm},\) and height \(H=1 \mathrm{m},\) with an open top, is gently placed in the water. What is the volume of water that overflows from the tank? How many 1 kg weights must be placed in the container to make it sink? Neglect surface tension effects.

A door \(1 \mathrm{m}\) wide and \(1.5 \mathrm{m}\) high is located in a plane vertical wall of a water tank. The door is hinged along its upper edge, which is \(1 \mathrm{m}\) below the water surface. Atmospheric pressure acts on the outer surface of the door and at the water surface. (a) Determine the magnitude and line of action of the total resultant force from all fluids acting on the door. (b) If the water surface gage pressure is raised to 0.3 atm, what is the resultant force and where is its line of action? (c) Plot the ratios \(F / F_{9}\) and \(y^{\prime} / y_{c}\) for different values of the surface pressure ratio \(p_{s} / p_{\text {atm }}\). \(\left(F_{0}\right.\) is the resultant force when \(\left.p_{s}=p_{\text {atm }}\right)\).

A reservoir manometer has vertical tubes of diameter \(D=18 \mathrm{mm}\) and \(d=6 \mathrm{mm} .\) The manometer liquid is Meriam red oil. Develop an algebraic expression for liquid deflection \(L\) in the small tube when gage pressure \(\Delta p\) is applied to the reservoir. Evaluate the liquid deflection when the applied pressure is equivalent to \(25 \mathrm{mm}\) of water (gage).
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