Part a by what distance do two objects carrying 1.0 c of charge each have to be separated before the electric force exerted on each object is 4.0 n ? express your answer with the appropriate units.

part b how much charge should be placed on each of two small objects separated by 1.0 m so that the electric force exerted on each is 4.0 n ? express your answer with the appropriate units.

Part A: 4.74×10⁴ m

Part B: 0.000211 C

Explanation:

Part A:

From coulombs law,

F  = kq²/r² Equation 1

Note: The two object are carrying equal charges.

Where F = Electric force, q = charge, r = distance of separation, k = proportionality constant

Given: F = 4.0 N, q = 1.0 C,

Constant: k = 1/4πe₀ = 9×10⁹ Nm²/C²

Substituting these values into equation 1,

4 = 9×10⁹(1²)/r²

r² = (9×10⁹)/4

r² = 2.25×10⁹

r = √(2.25×10⁹)

r = 4.74×10⁴ m.

Part 2:

F = kq²/r²

Making q the subject of the equation

q = √(Fr²/k) Equation 2

Where: F = 4.0 N, r = 1.0 m, k = 9×10⁹ Nm²/c²

Substituting these values into equation 2

q = √(4×1²/(9×10⁹ )

q = 2.11/10⁴

q = 0.000211 C.

friction

explanation:

they us to move forward and backwards

no. bullets are macro objects. wave nature is for light waves, extremely tiny particles. if a light particle was the size of your finger, the bullet would be the size of the universe. so basically, that huge difference in size makes the wave nature of matter insignificant.