The fatal current is 51 mA = 0.051 Ampere.
The resistance is 2,050Ω .
Voltage = (current) x (resistance)
= (0.051 Ampere) x (2,050 Ω) = 104.6 volts .
==================
This is what the arithmetic says IF the information in the question
is correct.
I don't know how true this is, and I certainly don't plan to test it,
but I have read that a current as small as 15 mA through the
heart can be fatal, not 51 mA .
If 15 mA can do it, and the sweaty electrician's resistance is
really 2,050 Ω, then the fatal voltage could be as little as 31 volts !
The voltage at the wall-outlets in your house is 120 volts in the USA !
THAT's why you don't want to stick paper clips or a screwdriver into
outlets, and why you want to cover unused outlets with plastic plugs
if there are babies crawling around.
Answer:
220 A
Explanation:
The magnetic force on the floating rod due to the rod held close to the ground is F = BI₁L where B = magnetic field due to rod held close the ground = μ₀I₂/2πd where μ₀ = permeability of free space = 4π × 10⁻⁷ H/m, I₂ = current in rod close to ground and d = distance between both rods = 11 mm = 0.011 m. Also, I₁ = current in floating rod and L = length of rod = 1.1 m.
So, F = BI₁L
F = (μ₀I₂/2πd)I₁L
F = μ₀I₁I₂L/2πd
Given that the current in the rods are the same, I₁ = I₂ = I
So,
F = μ₀I²L/2πd
Now, the magnetic force on the floating rod equals its weight , W = mg where m = mass of rod = 0.10kg and g = acceleration due to gravity = 9.8 m/s²
So, F = W
μ₀I²L/2πd = mg
making I subject of the formula, we have
I² = 2πdmg/μ₀L
I = √(2πdmg/μ₀L)
substituting the values of the variables into the equation, we have
I = √(2π × 0.011 m × 0.1 kg × 9.8 m/s²/[4π × 10⁻⁷ H/m × 1.1 m])
I = √(0.01078 kgm²/s²/[2 × 10⁻⁷ H/m × 1.1 m])
I = √(0.01078 kgm²/s²/[2.2 × 10⁻⁷ H])
I = √(0.0049 × 10⁷kgm²/s²H)
I = √(0.049 × 10⁶kgm²/s²H)
I = 0.22 × 10³ A
I = 220 A
Acceleration due to gravity
I think your question should be:
An industrial laser is used to burn a hole through a piece of metal. The average intensity of the light is

What is the rms value of (a) the electric field and
(b) the magnetic field in the electromagnetic wave emitted by the laser
Answer:
a) 
b) 
Explanation:
To find the RMS value of the electric field, let's use the formula:

Where
;
;

Therefore
![E_r_m_s = sqrt*{(1.239*10^9W/m^2) / [(3.00*10^8m/s)*(8.85*10^-^1^2C^2/N.m^2)]}](https://tex.z-dn.net/?f=%20E_r_m_s%20%3D%20sqrt%2A%7B%281.239%2A10%5E9W%2Fm%5E2%29%20%2F%20%5B%283.00%2A10%5E8m%2Fs%29%2A%288.85%2A10%5E-%5E1%5E2C%5E2%2FN.m%5E2%29%5D%7D%20)

b) to find the magnetic field in the electromagnetic wave emitted by the laser we use:
;
;

Answer:
its soul for that context not sole also rip and i have no idea
Explanation: