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Step2247 [10]
3 years ago
13

Variations in the resistivity of blood can give valuable clues about changes in various properties of the blood. Suppose a medic

al device attaches two electrodes into a 1.5-mm-diameter vein at positions 5.0 cm apart. What is the blood resistivity if a 9.0 V potential difference causes a 230 μA current through the blood in the vein?
Physics
1 answer:
Elena-2011 [213]3 years ago
7 0

Answer:

Answer:

1.1 x 10^9 ohm metre

Explanation:

diameter = 1.5 mm

length, l = 5 cm

Potential difference, V = 9 V

current, i = 230 micro Ampere = 230 x 10^-6 A

radius, r = diameter / 2 = 1.5 / 2 = 0.75 x 10^-3 m

Let the resistivity is ρ.

Area of crossection

A = πr² = 3.14 x 0.75 x 0.75 x 10^-6 = 1.766 x 10^-6 m^2

Use Ohm's law to find the value of resistance

V =  i x R

9 = 230 x 10^-6 x R

R = 39130.4 ohm

Use the formula for the resistance

R=\rho \frac{l}{A}

\rho =\frac{RA}{l}

\rho =\frac{39130.4\times 0.05}{1.766\times 10^{-6}}

ρ = 1.1 x 10^9 ohm metre

Explanation:

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Read 2 more answers
The escape velocity of any object from Earth is 11.2 km/s. (a) Express this speed in m/s and km/h. (b) At what temperature would
natima [27]

Answer:

a ) 11.1 *10^3 m/s = 39.96 Km/h

b) T_{o2} =1.58*10^5 K

Explanation:

a)v_{es} =v_{rms}= 11.1 km/s =11.1 *10^3 m/s = 39.96 Km/h

b)

M_O2 = 32.00 g/mol =32.0*10^{-3} kg/mol

gas constant R = 8.31 j/mol.K

v_{rms} = \sqrt{ \frac{3RT}{M}}

So, v_{rms,o2} =\sqrt{ \frac{3RT_{o2}}{M_{o2}}}

multiply each side by M_{o2}, so we have

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solving for temperature T_{o2}

T_{o2} = \frac{v_{rms,o2}^2 *M_{o2}}{3R}

In the question given,v_{rms} =v_{es}

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T_{o2} =1.58*10^5 K

7 0
3 years ago
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