Recent content by Alcyon

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    Resistor Change from Aorta to Capillaries

    Of course...:D I can't believe I'm going to be a doctor a few years from now. Thank you very much, you have been of great help. Greetings from Holland! :)
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    Resistor Change from Aorta to Capillaries

    Okay, so R_total = R * 9*10^9 ?
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    Resistor Change from Aorta to Capillaries

    Okay...but there is R1 - R9billion...I can't possibly do that! Ah...so: R1 * 9billion?
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    Resistor Change from Aorta to Capillaries

    Edit: My bad again. 1:R_total = 1/R1 + 1/R2 ... right?
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    Resistor Change from Aorta to Capillaries

    Yes...so the resistance per unit length of a single capillary is to be raised to the power of 9 billion?
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    Resistor Change from Aorta to Capillaries

    I don't fully understand though...what about the 8.89 *10^-8? Edit: My bad...but what's the final result?
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    Resistor Change from Aorta to Capillaries

    The radius is: sqrt (A/pi), isn't it? (thanks again for your help...really appreciate it. :))
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    Resistor Change from Aorta to Capillaries

    Ah, no...I can't really. We have the "resistance" of one circular tube now, right? So we have to raise it to the power of 9 billion?
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    Resistor Change from Aorta to Capillaries

    Ya, you're right. But I have to write down something, so let's just pretend that we've got lamina flow here. Thank you by the way...:)
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    Resistor Change from Aorta to Capillaries

    The Hagen-Poiseuille law is the key...
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    Resistor Change from Aorta to Capillaries

    Come on guys...you use to talk about things like pulling the Moon towards Earth...now, here's a real question and you don't know what to say anymore? You can't be serious on that one...
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    Resistor Change from Aorta to Capillaries

    The aorta branches into up to 30-40 billion capillary vessels, 8-10 billion of which are used effectively. As a consequence the cross-sectional area increases by a factor of 800. If we start from the assumption that there are 9 billion capillary vessels (all of which are equally thick) – how...
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