/*! 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 6 The aorta is a major artery, ris... [FREE SOLUTION] | 91Ó°ÊÓ

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The aorta is a major artery, rising upward from the left ventricle of the heart and curving down to carry blood to the abdomen and lower half of the body. The curved artery can be approximated as a semicircular arch whose diameter is \(5.0 \mathrm{cm} .\) If blood flows through the aortic arch at a speed of \(0.32 \mathrm{m} / \mathrm{s}\) what is the magnitude (in \(\mathrm{m} / \mathrm{s}^{2}\) ) of the blood's centripetal acceleration?

Short Answer

Expert verified
The centripetal acceleration is \(4.096\, \mathrm{m/s^2}\).

Step by step solution

01

Understand the Problem

We are tasked with finding the centripetal acceleration of blood flowing through a semicircular aortic arch. We know the diameter of the semicircle is \(5.0\, \mathrm{cm}\), and the flow speed of the blood is \(0.32\, \mathrm{m/s}\). The centripetal acceleration is given by the formula \(a_c = \frac{v^2}{r}\), where \(v\) is the speed and \(r\) is the radius of the circular path.
02

Convert Diameter to Radius

Given the diameter is \(5.0\, \mathrm{cm}\), we need to convert this to meters to match the units of the flow speed. The radius \(r\) is half the diameter: \[ r = \frac{5.0\, \mathrm{cm}}{2} = 2.5\, \mathrm{cm} = 0.025\, \mathrm{m}. \]
03

Apply the Formula for Centripetal Acceleration

Substitute the given values into the centripetal acceleration formula: \[ a_c = \frac{v^2}{r} = \frac{(0.32\, \mathrm{m/s})^2}{0.025\, \mathrm{m}}. \]
04

Calculate Centripetal Acceleration

Perform the calculations: \[ a_c = \frac{0.1024\, \mathrm{m^2/s^2}}{0.025\, \mathrm{m}} = 4.096\, \mathrm{m/s^2}. \]
05

State the Final Result

The magnitude of the blood's centripetal acceleration in the aortic arch is \(4.096\, \mathrm{m/s^2}\).

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Key Concepts

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

Aortic Arch
The aortic arch is an essential part of the human circulatory system. It is the curved section of the aorta, the main artery carrying oxygenated blood away from the heart to the rest of the body. Forming a semicircular shape, the aortic arch connects the ascending aorta, which rises from the heart, to the descending aorta, which travels down toward the abdomen. This unique structure provides two critical functions for blood flow. First, it efficiently redirects blood without interrupting the strong forward momentum generated by the heart. Second, its curvature allows for the distribution of blood to the upper regions of the body, including the head and arms, through branching arteries. Understanding the aortic arch's role is crucial in studying how blood moves in the circulatory system.
Blood Flow
Blood flow through the aortic arch involves the movement of blood driven by the pumping action of the heart. As blood exits the left ventricle, it enters the aorta and passes through the semicircular arch at a given speed. This flow speed is typically measured in meters per second (m/s). The consistent speed of blood flow ensures that nutrients and oxygen are delivered efficiently to cells throughout the body. During this process, the blood is subject to centripetal force, which is the inward force necessary to keep it moving along the curved path of the aortic arch. This force contributes to centripetal acceleration, which can be determined by analyzing the blood's velocity and the curvature of the path it takes. By understanding these dynamics, one can appreciate the complexity and elegance of the cardiovascular system.
Radius Conversion
When dealing with circular or semicircular paths, it is often necessary to convert diameters to radii to use in calculations. The radius of a circle is half its diameter. In this exercise, the aortic arch has a diameter of 5.0 cm. To find the radius in meters, we first divide the diameter by two. Thus, the radius would be \( r = \frac{5.0\, \mathrm{cm}}{2} = 2.5\, \mathrm{cm} \). To convert centimeters to meters, a simple conversion is required: multiply by 0.01. Therefore, \( 2.5\, \mathrm{cm} \) becomes \( 0.025\, \mathrm{m} \). Using the correct units is vital in physics problems to ensure accurate calculations and avoid errors.
Physics Problem Solving
Physics problem-solving involves a step-by-step approach to understanding and calculating various physical quantities. In this instance, we're tasked with determining the centripetal acceleration of blood flowing through a curved aortic arch. This process involves several key steps:
  • Identify known values: diameter (5.0 cm) and blood flow speed (0.32 m/s).
  • Convert any necessary units, such as diameter to radius in meters.
  • Apply relevant formulas: use \( a_c = \frac{v^2}{r} \) for centripetal acceleration.
  • Complete the calculations, ensuring all units are consistent.
In this problem, these steps led to the calculation of a centripetal acceleration of 4.096 m/s². Each step builds upon the previous, making it vital to approach such exercises methodically and with careful attention to detail. This methodical strategy is key to successfully solving complex physics problems, enabling students to confidently tackle similar challenges they might encounter in their studies.

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Most popular questions from this chapter

Two banked curves have the same radius. Curve \(A\) is banked at an angle of \(13^{\circ},\) and curve \(\mathrm{B}\) is banked at an angle of \(19^{\circ} .\) A car can travel around curve A without relying on friction at a speed of \(18 \mathrm{m} / \mathrm{s}\). At what speed can this car travel around curve B without relying on friction?

On a banked race track, the smallest circular path on which cars can move has a radius of \(112 \mathrm{m},\) while the largest has a radius of \(165 \mathrm{m},\) as the drawing illustrates. The height of the outer wall is \(18 \mathrm{m}\). Find \((\mathrm{a})\) the smallest and (b) the largest speed at which cars can move on this track without relying on friction.

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In a skating stunt known as crack-the-whip, a number of skaters hold hands and form a straight line. They try to skate so that the line rotates about the skater at one end, who acts as the pivot. The skater farthest out has a mass of \(80.0 \mathrm{kg}\) and is \(6.10 \mathrm{m}\) from the pivot. He is skating at a speed of \(6.80 \mathrm{m} / \mathrm{s}\). Determine the magnitude of the centripetal force that acts on him.

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