A human being can be electrocuted if a current as small as 55 mA passes near the heart. An electrician working with sweaty hands makes good contact with the two conductors he is holding. If his resistance is 2300 Ω, what might the fatal voltage in volts be?

The lethal voltage for the electrician under those conditions is 126.5 V.

Explanation:

To discover what is the lethal voltage to the electrician we need to find out what is the voltage that produces 55 mA = 0.055 A when across a resistance of 2300 Ohms (Electrician's body resistancy). For that we'll use Ohm's Law wich is expressed by the following equation:

V = i*R

Where V is the voltage we want to find out, i is the current wich is lethal to the electrician and R is his body resistance. By applying the given values we have:

V = 0.055*2300 = 126.5 V.

The lethal voltage for the electrician under those conditions is 126.5 V.

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.