/*! 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 85 If the aorta (diameter of \(d_{a... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

If the aorta (diameter of \(d_{a} )\) branches into two equal-sized arteries with a combined area equal to the aorta, what is the diameter of one of the branches? A. \(\sqrt{d_{a}}\) B. \(d_{a} / \sqrt{2}\) C. 2\(d_{a}\) D. \(d_{a} / 2\)

Short Answer

Expert verified
The diameter of one branch is \( \frac{d_a}{\sqrt{2}} \), so the answer is B.

Step by step solution

01

Understanding the Problem

We need to find the diameter of one of the arteries when the aorta divides into two equal-sized arteries, and their combined cross-sectional area is equal to the area of the aorta.
02

Express Given Components Mathematically

Let the diameter of the aorta be denoted as \( d_a \). The cross-sectional area of the aorta, assuming it is circular, is \( \pi \left(\frac{d_a}{2}\right)^2 = \frac{\pi d_a^2}{4} \). Each artery has a diameter of \( d_b \), and since there are two arteries with combined area equal to the aorta, each artery also has an area of \( \frac{\pi d_b^2}{4} \).
03

Set-up the Equation Relating Areas

We know that the combined area of the two arteries is the area of the aorta: \[ \frac{\pi d_a^2}{4} = 2 \times \frac{\pi d_b^2}{4} \]. Simplifying, we have \( \frac{\pi d_a^2}{4} = \frac{\pi d_b^2}{2} \).
04

Solve for the Diameter of the Branches

By canceling \( \pi \) and simplifying the equation \( \frac{d_a^2}{4} = \frac{d_b^2}{2} \), we multiply both sides by 4 to get \( d_a^2 = 2d_b^2 \). Solving for \( d_b \) gives \( d_b^2 = \frac{d_a^2}{2} \), and taking the square root, \( d_b = \frac{d_a}{\sqrt{2}} \).
05

Verify Against Answer Choices

From the options given, option B, \( \frac{d_a}{\sqrt{2}} \), matches our calculated diameter \( d_b \). Therefore, the correct answer is B.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Equation setup
Setting up the right equation is a crucial part of solving any physics problem, especially when it involves geometric shapes like circles. For this particular exercise, we are dealing with cross-sectional areas of circular pipes. The problem involves splitting one area into two equal areas, with the total area remaining constant. Here's how to set it up:
\[ \text{Area of the aorta} = \frac{\pi d_a^2}{4} \]
where \(d_a\) is the diameter of the aorta. Each of the two branches also has an area given by \(\frac{\pi d_b^2}{4}\), where \(d_b\) is the diameter of one branch. Since their combined area equals the aorta’s area, we write:
\[ 2 \times \frac{\pi d_b^2}{4} = \frac{\pi d_a^2}{4} \]
This sets the foundation for calculating the diameter of each branch.
Understanding the problem
Understanding the problem involves interpreting the given physical scenario correctly. Here, the aorta is described as branching into two equal-sized arteries. The key aspect is that the combined cross-sectional area of these branches remains equal to the original aorta's area.
This implies a concept of conservation of area when dividing the aorta. The challenge is to determine the diameter of each branch given that the total area doesn't change. Breaking down what we need to find helps: It's the diameter each smaller artery needs so their combined area equals the original area of the aorta.
Geometric relationships
Understanding the geometric relationships within circles helps simplify solving the problem at hand. The cross-sectional area of any circular object, like an artery, is given by the formula \(\pi\left(\frac{d}{2}\right)^2\).
This formula is pivotal because it shows that the area of a circle is proportional to the square of its radius or diameter. Applying this to our problem, if we have two smaller circles (arteries) that must together equal the area of one larger circle (the aorta), understanding this proportional relationship is key.
Thus, the task involves solving how to equate the squared diameter terms from different circles using this relationship.
Diameter conversion
Converting and comparing diameters underlie the solution to the exercise. Once we established that\( \frac{\pi d_a^2}{4} = 2 \times \frac{\pi d_b^2}{4} \), simplifying leads to comparing the diameters.
When solving \(\frac{d_a^2}{4} = \frac{d_b^2}{2}\), the math simplifies to \(d_a^2 = 2d_b^2\). We aim to express \(d_b\) in terms of \(d_a\). Upon further simplification by isolating \(d_b\), we find \(d_b^2 = \frac{d_a^2}{2}\).
Taking the square root, concerning conversion, leads us to the formula for \(d_b = \frac{d_a}{\sqrt{2}}\). This calculation accurately converts the known aorta diameter to the diameter of each resulting artery. Understanding how to perform these conversions is essential in translating areas to diameters in geometric problems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) If a flea can jump straight up to a height of \(22.0 \mathrm{cm},\) what is its initial speed (in \(\mathrm{m} / \mathrm{s} )\) as it leaves the ground, neglecting air resistance? (b) How long is it in the air? (c) What are the magnitude and direction of its acceleration while it is (i) moving upward? (ii) moving downward? (iii) at the highest point?

If the contraction of the left ventricle of the heart lasts 250 \(\mathrm{ms}\) and the rate of flow of the blood in the aorta (the large artery leaving the heart) is \(1.0 \mathrm{m} / \mathrm{s},\) what is the average acceleration of a red blood cell as it leaves the heart? A. 250 \(\mathrm{m} / \mathrm{s}^{2}\) B. 25 \(\mathrm{m} / \mathrm{s}^{2}\) C. 40 \(\mathrm{m} / \mathrm{s}^{2}\) D. 4.0 \(\mathrm{m} / \mathrm{s}^{2}\)

A test driver at Incredible Motors, Inc., is testing a new model car having a speedometer calibrated to read m/s rather than mi/h. The following series of speedometer readings was obtained during a test run: $$\begin{array}{llllllllll}{\text { Time (s) }} & {0} & {2} & {4} & {6} & {8} & {10} & {12} & {14} & {16} \\ {\text { Velocity }(\mathrm{m} / \mathrm{s})} & {0} & {0} & {2} & {5} & {10} & {15} & {20} & {22} & {22}\end{array}$$ (a) Compute the average acceleration during each 2 s interval. Is the acceleration constant? Is it constant during any part of the test run? (b) Make a velocity-time graph of the data shown, using scales of \(1 \mathrm{cm}=1\) s horizontally and \(1 \mathrm{cm}=\) 2 \(\mathrm{m} / \mathrm{s}\) vertically. Draw a smooth curve through the plotted points. By measuring the slope of your curve, find the magnitude of the instantaneous acceleration at times \(t=9 \mathrm{s}, 13 \mathrm{s}\) and 15 \(\mathrm{s}\) .

After its journey to the head, arms, and legs, a round trip distance of about \(3 \mathrm{m},\) the red blood cell returns to the left ventricle after 1 minute. What is the average velocity for the trip? A. \(0.05 \mathrm{m} / \mathrm{s},\) downward B. \(0.5 \mathrm{m} / \mathrm{s},\) downward C. \(0 \mathrm{m} / \mathrm{s},\) round trip D. 0.05 \(\mathrm{m} / \mathrm{s}\) , upward

The "reaction time" of the average automobile driver is about 0.7 s. (The reaction time is the interval between the perception of a signal to stop and the application of the brakes.) If an automobile can slow down with an acceleration of 12.0 \(\mathrm{ft} / \mathrm{s}^{2}\) , compute the total distance covered in coming to a stop after a signal is observed (a) from an initial velocity of 15.0 \(\mathrm{mi} / \mathrm{h}(\mathrm{in}\) a school zone) and (b) from an initial velocity of 55.0 \(\mathrm{mi} / \mathrm{h}\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.