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Chapter 9: Problem 42

If the average volume flow of blood through the aorta is $8.5 \times 10^{-5} \mathrm{m}^{3} / \mathrm{s}\( and the cross-sectional area of the aorta is \)3.0 \times 10^{-4} \mathrm{m}^{2},$ what is the average speed of blood in the aorta?

Short Answer

Expert verified
Answer: The average speed of blood in the aorta is approximately \(0.283\bar{3} \mathrm{m} / \mathrm{s}\).

Step by step solution

01

Write down the given values

We have the following given values: - Average volume flow of blood through the aorta (Flow Rate) = \(8.5 \times 10^{-5} \mathrm{m}^{3} / \mathrm{s}\) - Cross-sectional area of the aorta (Area) = \(3.0 \times 10^{-4} \mathrm{m}^{2}\)
02

Use the flow rate formula to find speed

The flow rate formula is: Flow Rate = Area × Speed Plug in the given values: \(8.5 \times 10^{-5} \mathrm{m}^{3} / \mathrm{s} = 3.0 \times 10^{-4} \mathrm{m}^{2} \times \text{Speed}\) Now, we need to solve for 'Speed'.
03

Rearrange the equation and solve for speed

Divide both sides of the equation by the cross-sectional area (\(3.0 \times 10^{-4} \mathrm{m}^{2}\)): Speed \(=\dfrac{8.5 \times 10^{-5} \mathrm{m}^{3} / \mathrm{s}}{3.0 \times 10^{-4} \mathrm{m}^{2}}\) Now, calculate the speed: Speed \(=\dfrac{8.5 \times 10^{-5}}{3.0 \times 10^{-4}} \mathrm{m} / \mathrm{s} = 0.283\bar{3} \mathrm{m} / \mathrm{s}\)
04

Write the final answer

The average speed of blood in the aorta is approximately \(0.283\bar{3} \mathrm{m} / \mathrm{s}\).

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