Assuming no air resistance, how far will a 0.0010 kg raindrop have fallen when it hits the ground 30.0 s later. assume g = 9.8 m/s2.

a)294 m
b)4410 m
c)8820 m
d)44,000 km

Option b, 4410 meters.

Explanation: If there is no air resistance, the acceleration experienced by the raindrop is equal to the gravitational acceleration:

a = -9.8m/s^2

for the velocity, we integrate over time, and there is no constant because we don't have initial velocity.

v = - (9.8m/s^2)*t

For the position, we integrate again over time, and we add a constant of integration that will be equal to the initial height of the raindrop.

r = (-9.8m/s^2)(t^2)/2 + H

if at t= 30s the raindrop reaches the ground, we have that:

r(30) = 0 = (-4.9*900)m + H

H = 4.9*900m = 4410m

So the initial height of the raindrop s 4410 meters.

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.

photocells convert light energy to electricity