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:
White light entering a prism is bent, or refracted, and the light separates into its constituent wavelengths. Each wavelength of light has a different colour and bends at a different angle. The colours of white light always emerge through a prism in the same order—red, orange, yellow, green, blue, indigo, and violet.
Change of a liquid to a gas
Answer:
1.) 11 km/s
2.) 9.03 × 10^-5 metres
Explanation:
Given that an electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric field, which has magnitude E = 48 N/C.
Electron q = 1.6×10^-19 C
Electron mass = 9.11×10^-31 Kg
(a) What is the speed of the electron 1.3 ns after entering this region?
E = F/q
F = Eq
Ma = Eq
M × V/t = Eq
Substitute all the parameters into the formula
9.11×10^-31 × V/1.3×10^-9 = 48 × 1.6×10^-19
V = 7.68×10^-18 /7.0×10^-22
V = 10971.43 m/s
V = 11 Km/s approximately
(b) How far does the electron travel during the 1.3 ns interval?
The initial velocity U = 64 km/s
S = ut + 1/2at^2
S = 64000×1.3×10^-6 + 1/2 × 8.4×10^12 × ( 1.3×10^-9)^2
S =8.32×10^-5 + 7.13×10^-6
S = 9.03 × 10^-5 metres
Answer:
<em>Fn = 50 N</em>
Explanation:
<u>Net Force</u>
The net force is the sum of all the forces acting on an object.
When all the forces act in a single line, the direction of the forces is given by their signs. Positive signs are assumed to be up and left on both axes.
The box being raised by a force has a weight of W=-125 N. The negative sign indicates the weight points down. The accelerating force goes up and its value is F=175 N. The positive sign indicates this force pushes the box up.
The forces acting on the box are:
The weight W=-125 N
The accelerating force F=175 N
The net force is
Fn = W + F = -125N + 175 N
Fn = 50 N
