Acalifornia condor is approaching its nest with a large chunk of carrion in its beak. as it approaches, it makes an upward swoop, achieving a momentary upward velocity of 12.8 m/s when the carrion falls from its mouth, hitting a cliff outcropping 32.1 m below. determine the speed of the carrion upon hitting the outcropping. (28.2 m/s)

Since the bird is initially going up, therefore upon falling form its mouth, the carrion was also going up at similar velocity of 12.8 m/s before it falls down. So calculate the distance taken during this time of momentarily going up:

(v = final velocity at the peak before it falls = 0, v0 = initial velocity = 12.8 m/s, g = gravity = 9.81 m/s^2, d is height taken)

v^2 = v0^2 - 2 g d

0 = 12.8^2 - 2 * 9.81 * d

d = 8.35 m

SO total distance it fell is:

d* = 8.35 m + 32.1 m = 40.45 m

So calculating for final velocity v (v0 = 0 since it stops at the peak before falling down):

v^2 = v0^2 + 2 g d

v^2 = 0 + 2 * 9.81 * (32.1 + 8.35)

v^2 = 793.642

v = 28.17 m/s = 28.2 m/s

beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge