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
The density of the object is approximately 1.91 kg per m³.
42 kg is a measure of mass, and 22 m³ is a measure of volume. Knowing this, you can use the relationship

to solve for the object's density.
42 kg

22 m³

1.91 kg per m³.
Answer:
Explanation:
We shall apply law of conservation of momentum to know velocity after collision . Let it be v .
total momentum before collision = total momentum after collision
15 x 1.5 - 12 x .75 = ( 15 + 12 ) v
v = .5 m /s
kinetic energy before collision
1/2 x 15 x 1.5² + 1/2 x 12 x .75²
= 16.875 + 3.375
= 20.25 J
kinetic energy after collision
= 1/2 x ( 15 + 12 ) x .5²
= 3.375 J
Loss of energy = 16.875 J
This energy appear as heat and sound energy that is produced during collision .
Answer:
v’= 9.74 m / s
Explanation:
The Doppler effect is due to the relative movement of the sound source and the receiver of the sound, in this case we must perform the exercise in two steps, the first to find the frequency that the bat hears and then the frequency that the audience hears that also It is sitting.
Frequency shift heard by the murciela, in case the source is still and the observer (bat) moves closer
f₁ ’= f₀ (v + v₀)/v
Frequency shift emitted by the speaker in the bat, in this case the source is moving away from the observer (public sitting) that is at rest
f₂’= f₁’ v/(v - vs)
Note that in this case the bat is observant in one case and emitter in the other, called its velocity v’
v’= vo = vs
Let's replace
f₂’= f₀ (v + v’)/v v/(v -v ’)
f₂’= f₀ (v + v’) / (v -v ’)
(v –v’ ) f₂’ / f₀ = v + v ’
v’ (1+ f₂’ /f₀) = v (f₂’/fo - 1)
v’ (1 + 1.059) = 340 (1.059 - 1)
v’= 20.06 / 2.059
v’= 9.74 m / s