Answer:
h’ = 1/9 h
Explanation:
This exercise must be solved in parts:
* Let's start by finding the speed of sphere B at the lowest point, let's use the concepts of conservation of energy
starting point. Higher
Em₀ = U = m g h
final point. Lower, just before the crash
Em_f = K = ½ m
energy is conserved
Em₀ = Em_f
m g h = ½ m v²
v_b =
* Now let's analyze the collision of the two spheres. We form a system formed by the two spheres, therefore the forces during the collision are internal and the moment is conserved
initial instant. Just before the crash
p₀ = 2m 0 + m v_b
final instant. Right after the crash
p_f = (2m + m) v
the moment is preserved
p₀ = p_f
m v_b = 3m v
v = v_b / 3
v = ⅓ 
* finally we analyze the movement after the crash. Let's use the conservation of energy to the system formed by the two spheres stuck together
Starting point. Lower
Em₀ = K = ½ 3m v²
Final point. Higher
Em_f = U = (3m) g h'
Em₀ = Em_f
½ 3m v² = 3m g h’
we substitute
h’=
h’ =
h’ = 1/9 h
Answer:
Mechanical advantage = 4
Explanation:
Given the following data;
Distance of effort, de = 8m
Distance of ramp, dr = 2m
To find the mechanical advantage;
Mechanical advantage = de/dr
Substituting into the equation, we have;
Mechanical advantage = 8/2
Mechanical advantage = 4
According to periodic trends in the periodic table, the atomic radius decreases from left to right.
In period three, the element with the smallest atomic radius would be the element in the rightmost area. Protons increase as it goes to the right, which would mean they pull in electrons closer which decreases the size.
So in period 3, the element with the smallest atomic radius is Argon (Ar).