5.1 m
Explanation:
Let's set the ground as our reference point. Let's also call the dropped ball to be ball #1 and its height above the ground at any time t is given by
(1)
where 10 represents its initial height or displacement of 10 m above the ground. At the same time, the displacement of the second ball with respect to the ground 
is given by
(2)
At the instant the two balls collide, they will have the same displacement, therefore
or
Solving for t, we get
We can use either Eqn(1) or Eqn(2) to hind the height where they collide. Let's use Eqn(1):
Answer:
2.029×10^-18 J
Explanation:
E=hv
so
E=(3.06×10^15)*(6.63×10^-34)
E=2.029×10^-18 J
Answer:
t = 22.2 s
Explanation:
angular distance covered in the 36.0 s is
θ = ω(avg)t = ½(10.0 + 30.0)36 = 720 radians
720/2 = 360 radians
α = Δω/t = (30 - 10)/36 = 5/9 rad/s²
θ = ω₀t + ½αt²
360 = 10.0t + ½(5/9)t²
0 = (5/18)t² + 10.0t - 360
0 = t² + 36t - 1296
t = (-36 ±√(36² - 4(1)(-1296))) / 2
t = (-36 ±√(6480)) / 2
t = -18 ±√1620
we ignore the negative time result as it occurs before we care.
t = -18 + √1620 = 22.249223... s
A single reflection, like shouting at the side of a mountain and hearing
your voice come back to you, is an 'echo'.
Multiple reflection, like clapping your hands once inside a large room,
is 'reverberation'.
