The tiny ripples on the soup are not only similar to wind-generated
waves ... they ARE wind-generated waves. This is a big part of the
reason why they bear such an uncanny resemblance.
Answer:
532 millimeters of mercury
Explanation:
In order to convert the pressure from atm to millimeters of mercury (mm Hg), we should remind the conversion factor between the two units:
1 atm = 760 mm Hg
Therefore, we can solve the problem by setting up the following proportion:

Solving for x, we find

Answer:
Block A
Explanation:
Block A will float higher in the water compared to the second Block.
The density of water is 1g/cm³.
According to the principle of floatation "an object that floats in a liquid will displace equal amount of fluid to the weight of the object".
A body will become more submerged in water if it has more density because density is the mass per volume of body.
An object with a higher density than another will sink in the liquid of the one with lesser density.
- Object A has lesser density and will float higher up and displace very little water.
- Object B has higher density and will be more submerged.
Answer:
K_a = 8,111 J
Explanation:
This is a collision exercise, let's define the system as formed by the two particles A and B, in this way the forces during the collision are internal and the moment is conserved
initial instant. Just before dropping the particles
p₀ = 0
final moment
p_f = m_a v_a + m_b v_b
p₀ = p_f
0 = m_a v_a + m_b v_b
tells us that
m_a = 8 m_b
0 = 8 m_b v_a + m_b v_b
v_b = - 8 v_a (1)
indicate that the transfer is complete, therefore the kinematic energy is conserved
starting point
Em₀ = K₀ = 73 J
final point. After separating the body
Em_f = K_f = ½ m_a v_a² + ½ m_b v_b²
K₀ = K_f
73 = ½ m_a (v_a² + v_b² / 8)
we substitute equation 1
73 = ½ m_a (v_a² + 8² v_a² / 8)
73 = ½ m_a (9 v_a²)
73/9 = ½ m_a (v_a²) = K_a
K_a = 8,111 J