348.34 m/s. When Superman reaches the train, his final velocity will be 348.34 m/s.
To solve this problem, we are going to use the kinematics equations for constant aceleration. The key for this problem are the equations 
and 
where 
is distance, 
is the initial velocity, 
is the final velocity, 
is time, and 
is aceleration.
Superman's initial velocity is 
, and he will have to cover a distance d = 850m in a time t = 4.22s. Since we know , and , we have to find the aceleration in order to find .
From the equation we have to clear , getting the equation as follows: 
.
Substituting the values:
To find we use the equation .
Substituting the values:
Answer:
Equation for SHM can be written
V = w A cos w t where w is the angular frequency and the velocity is a maximum at t = 0
V1 = w1 A cos w1 t
V2 = w2 A cos w2 t
V2 / V1 = w2 / w1 since cos X t = 1 if t = zero
V2 / V1 = 2 pi f2 / (2 pi f1) = f2 / f1 = T1 / T2
If the velocity is twice as large the period will be 1/2 long
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
