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sergeinik [125]
3 years ago
12

When you hold your hands at your sides, you may have noticed that the veins sometimes bulge--the height difference between your

heart and your hands produces increased pressure in the veins. The same thing happens in the arteries. If the average arterial pressure at your heart is a typical 100 mmHg, what is the average arterial pressure in your hands when they are held at your side? Assume your hands are 60 cm below your heart.
Physics
1 answer:
hodyreva [135]3 years ago
7 0

To solve this problem it is necessary to use the concepts related to pressure and pressure, absolute and atmospheric.

Average arterial pressure in the hands,

P = 100mmHg+ h*\rho_{blood}g

Where,

P = Pressure

h = height (at this case the length of the arm)

Replacing with our values

P = 100mm(Hg)+(600mm)(\frac{\rho_{blood}}{\rho_{mercury}})(Hg)

P = 100mmHg+600*\frac{1060}{13600}mmHg

P = 100mmHg+46.765mmHg

P = 146.765mmHg

Where,

\rho_{blood} = 1060Kg/m^3

\rho_{mercury}=13600Kg/m^3

Therefore the pressure is 146.765mmHg

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Answer:

a

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b

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From the question we are told that

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     Resistivity is  \rho  =  1.69* 10^{-8} \Omega \cdot m

     The  length of each wire  is  l =  6.25 \  m

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    R  =  \frac{\rho *  l  }{A}

=> R  =  \frac{  1.69* 10^{-8} * 6.25 }{ 7.0695 *10^{-6} }

=> R  = 0.0149 \ \Omega

Generally the resistance of the first wire is mathematically represented as

    R_1  =  \frac{\rho *  l  }{A_1}

=> R_1  =  \frac{  1.69* 10^{-8} * 6.25 }{5.5899 *10^{-6} }

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