In a vacuum, two particles have charges of q1 and q2, where q1 = +3.3c. they are separated by a distance of 0.24 m, and particle 1 experiences an attractive force of 4.1 n. what is the value of q2, with its sign?

-7.95 pC

Explanation:

If the particle experiences an attractive force towards a positive charge it must be negatively charged.

Coulomb's law:

With ε0 = 8.85*10^-12 C^2/(N*m^2)

Then:

c.the charges if the proton and electron and gravity

answer: 802.3 kilo joules of energy represents 191.75 kilo calorie of energy.

explanation:

energy released during combustion of one mole methane = 802.3 kilo joules

if 1 kilo joule = 0.239006   kilo calorie

then ,802.3 kilo joule will be = 802.30.239006 kilo calorie

802.3 kilo joule = 191.75 kilo calorie

hence, the energy in kilo calorie is 191.75 kilo calorie.