Two metal balls have charges of 7.1 × 10-6  coulombs and  6.9 × 10-6  coulombs. they are 5.7 × 10-1  meters apart. what is the force of interaction between the two balls?   (k  = 9.0 × 109  newton·meters2/coulomb2) 5.7 newtons 6.9 newtons 7.1 newtons 1.4 newtons

Hey there, Your answer wold be 1.36 newtons rounded to 1.4 newton. This is how!

force = k (q1 * q2) / r^2 

= (9.0 × 10^9 Nm^2/C^2) [(7.1 × 10^-6 C) (6.9 × 10^-6 C)] / (5.7 × 10^-1 m)^2 

= [ 9.0 × 7.1 × 6.9 / (5.7)^2 ] × 10^-1 N 

= 13.6 × 10^-1 N 

= 1.36 N

ANSWER-1.4 

HOPE I HELPED

Answer: buoyancy forceexplanation: in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.