Answer:
is the current through the body of the man.
energy dissipated.
Explanation:
Given:
- time for which the current lasted,

- potential difference between the feet,

- resistance between the feet,

<u>Now, from the Ohm's law we have:</u>


is the current through the body of the man.
<u>Energy dissipated in the body:</u>



Calcium ions have a +2 charge and sulfur ions have a -2 charge, so the formula is CaS
Answer:
I₃/Io % = 0.8.59
Explanation:
A polarizer is a complaint sheet for light in the polarization direction and blocks the perpendicular one. When we use two polarizers the transmission between them is described by Malus's law
I = I₀ cos² θ
Let's apply the previous exposures in our case, the light is indicatively not polarized, so the first polarized lets half of the light pass
I₁ = ½ I₀
The light transmitted by the second polarizer
I₂ = I₁ cos² θ
I₂ = (½ I₀) cos2 28
The transmission by the polarizing third is
I₃ = I₂ cos² θ₃
The angle of the third polarizer with respect to the second is
θ₃ = 90-28
θ₃ = 62º
I₃ = (½ I₀ cos² 28 cos² 62)
Let's calculate
I₃ = Io ½ 0.7796 0.2204
I₃ = Io 0.0859
I₃/Io= 0.0859 100
I₃/Io % = 0.8.59
Answer:
164.2°
Explanation:
speed of wind (w) = 45 km/h due south
speed of aircraft (a) = 165 km/h
in what direction (in degrees) should the aircraft head in so as to fly due west?
To get the direction the pilot should fly, we can form a triangle with the data available where
- the direction of the wind (due south) serves as the opposite side
- the direction the pilot would have to fly so he can end up at he west serves as the hypothenuse
- θ is the angle between the direction the pilot would have to fly and the direction the pilot wishes to fly.
- the direction the pilot wishes to fly ( west) will serve as the adjacent side
- all this can be seen from the attached diagram.
now sin θ = 
sin θ = 
θ =
0.2728
θ = 15.8°
since we are to use the counter-clockwise from east convention our measurement would have to be taken anticlockwise from the east direction, therefore the direction of the aircraft (Ф) = 180-15.8 = 164.2°