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:
Energy of wave will increase as the energy of wave is related to the amplitude of wave
Answer:
The motion in which all particles of a body move through the same distance in the same time.
Examples: A car moving along the road
A ball rolling on the ground
Answer:
817.5 Pa
Explanation:
From Bernoulli's equation, considering thst there is no height difference then
P1+½d(v1)²=P2+½d(v2)²
P1-P2=½d(v2²-v1²)
∆P=½d(v2²-v1²)
Where P represent pressure, d is density and v is velocity. Subscripts 1 and 2 represent inside and outside. ∆P is tge change in pressure
Given the speed at roof top as 128 km/h, we convert it to m/s as follows
128*1000/3600=35.555555555555=35.56 m/s
Velocity at the bottom of roof is 0 m/s
Density is given as 1.293 kg/m³
∆P=½*1.293*(35.56²-0)=817.5 Pa
