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Equation of Continuity - Flow rate through an artery

  1. Dec 5, 2015 #1
    I'm really confused by this question, as it is different than the examples we did in class. I've compared my answer with a few classmates and I'm getting a different one, so I'm not sure if I've done it wrong or if they have. I'd really just like to know if I am on the right track! Thank you :)

    1. The problem statement, all variables and given/known data

    Blood flows at 0.465 cm/sec through an artery with an inside radius of 2.85 mm. What is the flow speed in another section where the artery radius increases to 4.35 mm? Round final answer to 3 significant figures.

    Givens:
    v1= 0.465 cm/sec
    r1 = 2.85 mm
    r2 = 4.35 mm

    2. Relevant equations
    A1v1=A2v2
    A=3.14(r)2

    3. The attempt at a solution
    Convert v1 = 0.465 cm/sec to 0.00465 m/sec
    Convert r1 = 2.85 mm to 0.00285 m
    Convert r2 = 4.35 mm to 0.00435 m
    Find A1→ A1=3.14(0.00285)2 = 2.550465x10-5
    Find A2→ A2=3.14(0.00435)2 = 5.941665x10-5

    A1v1 = A2v2
    v2 = (A1v1) ÷ A2
    = [2.550465x10-5(0.00465)] ÷ 5.941665x10-5
    = 1.996016647x10-3
    = 2.00x10-3 m/sec
     
  2. jcsd
  3. Dec 5, 2015 #2

    gneill

    User Avatar

    Staff: Mentor

    Hi Kate L, Welcome to Physics Forums.

    You've done all right, the result looks good.

    One thing you might keep in mind when using numerical values for constants like ##\pi## is to use more digits than the required number of significant figures. Otherwise rounding errors can creep into your results. The value of ##\pi## that you've used: 3.14, is already rounded (or truncated) to three significant figures. A better choice would be to use something like 3.1416. For this problem it doesn't make a difference because its influence cancels out when you divide one area by the other.
     
  4. Dec 5, 2015 #3
    Thank you, I will keep that in mind!
     
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