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1. The distance \(D\) between them is given by: [tex]r = D^2[/tex]

2. Translate each variation equation into words using the given variable names.

a) [tex]F = k A v^2[/tex]: Force \(F\) is equal to the constant \(k\) multiplied by the acceleration \(A\) and the square of velocity \(v\).

b) [tex]P V = k[/tex]: The product of pressure \(P\) and volume \(V\) is a constant.

3. In sonography, sound waves travel through human tissues in the body to create an optimal image. The wavelength \(W\) of the ultrasound sound wave in tissue varies inversely with the frequency \(F\) of the sound waves.

a) Write a general formula relating the given variables \(W\) and \(F\). Use \(k\) for the constant of variation.
[tex]
W = \frac{k}{F}
[/tex]

b) Given that a sonographer uses a frequency of 3 million cycles per second \((F = 3,000,000)\) and a wavelength of 0.5133 millimeters \((W = 0.5133)\), find \(k\).
[tex]
W \times F = 0.5133 \times 3,000,000 = 1,539,900
[/tex]

c) Determine the wavelength for a superficial structure like a carotid artery where the sonographer uses a frequency of 7 million cycles per second.
[tex]
W = \frac{1,539,900}{7,000,000} = 0.21999 \text{ millimeters}
[/tex]

4. In respiratory therapy, cardiac output is the volume of blood the heart pumps per minute (mL/min). Cardiac output \(Q\) is directly related to oxygen consumption \(V\) (mL/min) and inversely related to the oxygen content difference \(C\) between arterial and venous blood (mL/100 mL).

a) Write a general formula relating the given variables \(Q, V\), and \(C\). Use \(k\) for the constant of variation.
[tex]
Q = \frac{kV}{C}
[/tex]

b) Given that a patient with severe cardiac impairment has a cardiac output of 500 mL/min \((Q = 500)\), an oxygen consumption of 250 mL/min \((V = 250)\), and an arteriovenous oxygen content difference of 5 mL/100 mL \((C = 5)\), find \(k\).

c) Determine the cardiac output (in mL/min) for a healthy adult at rest with an oxygen consumption of 200 mL/min and an arteriovenous oxygen content difference of 0.5 mL/100 mL.

Answer :

Sure! Let's go through the problems step by step.

### Part 1: Translating Variation Equations

a) The equation [tex]\( F = kAv^2 \)[/tex] is read as "The force [tex]\( F \)[/tex] is equal to the constant [tex]\( k \)[/tex] times the acceleration [tex]\( A \)[/tex] multiplied by the square of the velocity [tex]\( v \)[/tex]."

b) The equation [tex]\( PV = k \)[/tex] is read as "The product of pressure [tex]\( P \)[/tex] and volume [tex]\( V \)[/tex] is a constant."

### Part 2: Sonography Problem

a) We start with the inverse variation relationship between wavelength [tex]\( W \)[/tex] and frequency [tex]\( F \)[/tex], expressed as:
[tex]\[ W = \frac{k}{F} \][/tex]
where [tex]\( k \)[/tex] is the constant of variation.

b) To find [tex]\( k \)[/tex], we use the given values for the kidney examination:
- Frequency [tex]\( F = 3{,}000{,}000 \)[/tex] cycles per second
- Wavelength [tex]\( W = 0.5133 \)[/tex] millimeters

Plug these into the equation:
[tex]\[ W \times F = k \][/tex]
[tex]\[ 0.5133 \times 3{,}000{,}000 = k \][/tex]
[tex]\[ k = 1{,}539{,}900 \][/tex]

c) For the carotid artery, with a frequency of [tex]\( 7{,}000{,}000 \)[/tex] cycles per second:
[tex]\[ W = \frac{k}{F} = \frac{1{,}539{,}900}{7{,}000{,}000} \][/tex]
[tex]\[ W \approx 0.22 \text{ millimeters} \][/tex]

### Part 3: Respiratory Therapy Problem

a) The relationship in the context of cardiac output, oxygen consumption, and oxygen content difference is expressed as:
[tex]\[ Q = \frac{k \times V}{C} \][/tex]
where:
- [tex]\( Q \)[/tex] is cardiac output
- [tex]\( V \)[/tex] is oxygen consumption
- [tex]\( C \)[/tex] is the oxygen content difference
- [tex]\( k \)[/tex] is the constant of variation

b) For a patient with severe cardiac impairment:
- [tex]\( Q = 500 \)[/tex] mL/min
- [tex]\( V = 250 \)[/tex] mL/min
- [tex]\( C = 5 \)[/tex] mL/100 mL

To find [tex]\( k \)[/tex]:
[tex]\[ Q = \frac{k \times V}{C} \][/tex]
[tex]\[ 500 = \frac{k \times 250}{5} \][/tex]
[tex]\[ k = 10 \][/tex]

c) For a healthy adult at rest:
- [tex]\( V = 200 \)[/tex] mL/min
- [tex]\( C = 0.5 \)[/tex] mL/100 mL

Calculate the cardiac output:
[tex]\[ Q = \frac{k \times V}{C} \][/tex]
[tex]\[ Q = \frac{10 \times 200}{0.5} \][/tex]
[tex]\[ Q = 4000 \text{ mL/min} \][/tex]

And that's the detailed solution for each part of the problem!

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