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
Answer: 1.22 m
Explanation:
The equation of motion in this situation is:
(1)
Where:
is the final height of the ball
is the initial height of the ball
is the vertical component of the initial velocity (assuming the ball was thrown vertically and there is no horizontal velocity)
is the time at which the ball lands
is the acceleration due gravity
So, with these conditions the equation is rewritten as:
(2)
(3)
Finally:
Answer: A.) Regulations at a local level.
When a force is applied to the box , this will cause an acceleration to the box.
(force =mass×acceleration)
So the box has a constant acceleration and a changing velocity.
Water travels through long, thin tubes running up from the roots through the stems and leaves called xylem.Water moves up the xylem through a process called capillary action.
Capillary action allows water to be pulled through the thin tubes because the molecules of the water are attracted to the molecules that make up the tube. The water molecules at the top are pulled up the tube and the water molecules below them are pulled along because of their attraction to the water molecules above them.
