The force of attraction / repultion of two charges is given by F = KQ1Q2 / d^2; where K is the Coulumbs constant, Q1 is the charge on charge 1, Q2 is the charge on charge 2, d is the distance of the two chardes apart.
Thus,
KQ1Q2 / 4 = 2 x 10^-5
KQ1Q2 = 4 x 2 x 10^-5 = 8 x 10^-5
When the distance between the charges was reduced to 1 meter.
F = KQ1Q2 / d^2 = 8 x 10^-5 / 1^2 = 8 x 10^-5
Therefore, option C is the correct answer.
The standard cell potential helps to determine the oxidative and reductive strength of species. All the species that lie below the Standard Hydrogen Electrode in the standard cell potential table are stong oxidizing agents and are highly electronegative while those that are above the Standard Hydrogen Electrode are either weak oxidants or reducing agents
Answer:
69.69 g
Explanation:
Evaporation of water will take out latent heat of vaporization. Let the mass of water be m and latent heat of vaporization of water be 2260000 J per kg
Heat taken up by evaporating water
= 2260000 x m J
Heat lost by body
= mass x specific heat of body x drop in temperature
60 x 3500 x .750 ( specific heat of human body is 3.5 kJ/kg.k)
= 157500 J
Heat loss = heat gain
2260000 m= 157500
m = .06969 kg
= 69.69 g
so the 1st on is the one on the left, middle is right and the 3rd one is the right one
