A seesaw remains stationary when two students of equal weight sit on the ends
c
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
higher temp = higher energy = higher frequency = shorter wavelength
The answer is true hope that helped!!
Answer:
The mass of the other worker is 45 kg
Explanation:
The given parameters are;
The gravitational potential energy of one construction worker = The gravitational potential energy of the other construction worker
The mass of the lighter construction worker, m₁ = 90 kg
The height level of the lighter construction worker's location = h₁
The height level of the other construction worker's location = h₂ = 2·h₁
The gravitational potential energy, P.E., is given as follows;
P.E. = m·g·h
Where;
m = The mass of the object at height
g = The acceleration due to gravity
h = The height at which is located
Let P.E.₁ represent the gravitational potential energy of one construction worker and let P.E.₂ represent the gravitational potential energy of the other construction worker
We have;
P.E.₁ = P.E.₂
Therefore;
m₁·g·h₁ = m₂·g·h₂
h₂ = 2·h₁
We have;
m₁·g·h₁ = m₂·g·2·h₁
m₁ = 2·m₂
90 kg = 2 × m₂
m₂ = (90 kg)/2 = 45 kg
The mass of the other construction worker is 45 kg.
