Apoint charge with a charge q1 = 3.60 μc is held stationary at the origin. a second point charge with a charge q2 = -4.60 μc moves from the point x= 0.100 m , y= 0 to the point x= 0.250 m , y= 0.250 m .how much work w is done by the electric force on the moving point charge?

W = -1.069 J

Explanation:

To find the work (W) of the electric force, we can use the formula:

W = q.k₀.Q(1/r₁ - 1/r₂)

q = moving point charge

k₀ = 9.10⁹ Nm²/C²

Q = stationary charge

r₁ = distance from origin to starting point

r₂ = distance from origin to ending point

r₁ = 0.100 - 0 = 0.1

r₂ = √((0.25-0)² + (0.25-0)²) = √(0.25² + 0.25²) = √0.125 = 0.3536

W = -4.6*10⁻⁶*9*10⁹*3.6*10⁻⁶(1/0.1 - 1/0.3536)

W = -149.04*10⁻³(10 - 2.8284)

W = -149.04*10⁻³*7.1716 = -1068.85*10⁻³ = -1.069 J

28. evolution

29. environment, survive, offspring  

explanation:

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.