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Aneli [31]
4 years ago
8

The resistance of the body varies from approximately 500 kΩ (when it is very dry) to about 1.20 kΩ (when it is wet). The maximum

safe current is about 4.90 mA. At 10.0 mA or above, muscle contractions can occur that may be fatal.
(a) What is the largest potential difference that a person can safely touch if his body is wet?
(b) Is this result within the range of common household voltages?

Physics
2 answers:
Vesnalui [34]4 years ago
6 0

Explanation:

Below is an attachment containing the solution.

scoray [572]4 years ago
4 0

Answer:

a) The maximum potential difference the wet human body can take = 5.88 V

b) This value is very much lower than the normal voltage of household outlets (120 V) and isn't in the same range at all.

This means one should be very careful operating electrical appliances (even at home) while the body is wet.

Explanation:

From Ohm's law, V = IR

The resistance of the body when the body is wet = 1.2 kΩ = 1200 Ω

Maximum safe current that the body can withstand = 4.90 mA = 0.0049 A

Maximum voltage the body can withstand while wet = (Maximum safe current that the body can withstand) × (Resistance of the body when the body is wet) = 0.0049 × 1200 = 5.88 V

b) This value is still very much lower than the normal voltage of household outlets (120 V) and isn't in the same range at all.

This means one should be very careful operating electrical appliances (even at home) while the body is wet.

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F=\dfrac{kq^2}{r^2}\\\\F=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{(0.005)^2}\\\\F=9.21\times 10^{-24}\ N

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3 years ago
When two long parallel wires carry unequal currents, the magnitude of the magnetic force that one wire exerts on the other is F.
LuckyWell [14K]

Answer:

The magnitude of the new magnetic force on each wire will be 4 times of the initial force.    

Explanation:

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If the current in both wires is now doubled, new force per unit length is given by :

(\dfrac{F}{l})'=\dfrac{\mu_o (2I_1)(2I_2)}{2\pi r}\\\\(\dfrac{F}{l})'=4\times \dfrac{\mu_o I_1I_2}{2\pi r}\\\\(\dfrac{F}{l})'=4\times \dfrac{F}{l}

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A 10.0-kg box starts at rest and slides 3.5 m down a ramp inclined at an angle of 10° with the horizontal. If there is no fricti
Alenkinab [10]

Answer:

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3 years ago
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