Two charged spheres on a frictionless horizontal surface are attached to opposite ends of a string & are in static equilibrium. The 18 kg red sphere has more charge than the 11 kg green sphere. The total charge magnitude on the spheres is 10 μC & they have the same polarity. As a result the tension is 2.5 N & the centers of the spheres are 0.06 m apart. Determine the charge magnitude on each sphere.

the red sphere has a charge of 9.899 10⁻⁶ C and

the green sphere 0.101 10⁻⁶ C

Explanation:

For this exercise we will use coulomb's law

          F = k q₁ q₂ / r²

where they indicate the value of the force F = 2.5 N, the distance between them r = 0.06 m

they also indicate the value of the total load

         q₁ + q₂ = 10 10⁻⁶ C

we substitute the values

         2.5 = 9 10⁹ q₁ q₂ / 0.06²

         2.5 0.06 2/9 10⁹ = q₁ q₂

         1 10⁻¹² = q₁ q₂

         10 10⁻⁶= q₁ + q₂

we have two equations with two i unknowns, so the system can be solved.

         q₁ = 1 10⁻¹² / q₂

we substitute in the other equation

         10 10⁻⁶ = 10⁻¹² / q₂ + q₂

let's solve the equation

         10 10⁻⁶ q₂ = 10⁻¹² + q₂²

         q₂² - 10⁻⁵ q₂ + 10⁻¹² = 0

          q₂ = [10⁻⁵ ± √(10⁻¹⁰ - 4 10⁻¹²)] / 2

          q₂ = [10⁻⁵ + - 0.9798 10⁻⁵] / 2 = 10⁻⁵ (1 + - 0.9798) / 2

          q₂ = 0.9899 10⁻⁵ C

          q₂ = 0.0101 10⁻⁵ C

let's find the charge of the other sphere

         q₂ = 0.9899 10⁻⁵ C

         q₁ = 10⁻⁵ - q₂

          q₁ = 10⁻⁵ - 0.9899 10⁻⁵

          q₁ = 0.0101   10⁻⁵    C

therefore the red sphere has a charge of 9.899 10⁻⁶ C and

the green sphere 0.101 10⁻⁶ C

answer: the correct answer id option(d).

explanation:

snell's law refraction:

the angle of refraction =

the angle of incidence =

refractive index of air = = 1

refractive index of glycerol = 1.47

from the given options, the most closest answer to 63.72° is 65° of option (d).

i sometimes feel that way too. i feel that i should hurt someone because they hurt me, but i wouldnt do it. that would make things so much worse i have so much experience with this stuff

explanation: