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Aneli [31]
3 years ago
8

The average bulk resistivity of the human body (apart from surface resistance of the skin) is about 5.0 Ω⋅m. The conducting path

between the hands can be represented approximately as a cylinder 1.6 m long and 0.10 m in diameter. The skin resistance can be made negligible by soaking the hands in salt water (a) What is the resistance between the hands if the skin resistance is negligible? (b) if skin resistance is negligible, what potential difference between the hands is needed for lethal shock current 100 mA?
Physics
1 answer:
Varvara68 [4.7K]3 years ago
7 0

Answer:

A) R = 1019.11 Ω

B) V = 101.91 V

Explanation:

a) The resistance is given by the formula;

R = ρL/A

Where;

ρ is bulk density = 5.0 Ωm

L is length of cylinder = 1.6m

A is area = (π/4)D² = (π/4) x 0.1² = 0.007854 m²

Plugging in the relevant values, we obtain ;

R = (5 Ωm x 1.6m)/(0.007854 m²)

R = 1019.11 Ω

b) potential difference is given by;

V = RI

Thus, plugging in relevant values, we obtain;

V = 1019.11 Ω ∙ 0.1A

V = 101.91 V

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