Answer: the most potential energy == 5 kg book, 2 m from the ground= 98 Joules
Explanation:
potential energy = m g h
m = mass
g = acceleration due gravity = 9.8 m/s²
h = distance above ground
1. Pe₁ = 1 kg x 2 m x g = 2 g
2. Pe₂ = 5 kg x 2 m x g = 10 g = 10 kg m x 9,8 m/s² = 98 Joules
3. Pe₃ = 1 kg x 0,5 m x g = 0,5 g
4. Pe₄ = 5 kg x 0.5 m x g = 2,5 g
10 > 2,5 > 2 >0,5
20/9.8 = 2.0 seconds. The ball stops after 2 seconds.
The problem is solved and the questions are answered below.
Explanation:
a. To calculate the speed of the 0.66 kg ball just before the collision
V₀ + K₀ = V₁ + K₁
= mgh₀ = 1/2 mv₁²
where, h= r - r cosθ
V = 
V = 2.42 m/s
b. Calculate the speed of the 0.22 kg ball immediately after the collision
y = y₀ + Vy₀t - 1/2 gt²
0 = 1.2 - 1/2 gt²
t = 0.495 s
x = x₀ + Vx₀t
1.4 = 0 + vx₀ (0.495)
Vx₀ = 2.83 m/s
C. To Calculate the speed of the 0.66 kg ball immediately after the collision
m₁ v₁ = m₁ v₃ + m₂ v₄
(0.66)(2.42) = (0.66) v₃ + (0.22)(2.83)
V₃ = 1.48 m/s
D. To Indicate the direction of motion of the 0.66 kg ball immediately after the collision is to the right.
E. To Calculate the height to which the 0.66 kg ball rises after the collision
V₀ + k₀ = V₁ + k₁
1/2 mv₀² = mgh₁
h₁ = v₀²/2 g
= 0.112 m
F. Based on your data, No the collision is not elastic.
Δk = 1/2 m₁v₃² =1/2 m₂v₄² - 1/2 m₁v₁²
= 1/2 (0.66)(1.48)² + 1/2 (0.22)(2.83)² - 1/2 (0.66)(2.42)²
= - 0.329 J
Hence, kinetic energy is not conserved.
Answer:
- The velocity component in the flow direction is much larger than that in the normal direction ( A )
- The temperature and velocity gradients normal to the flow are much greater than those along the flow direction ( b )
Explanation:
For a steady two-dimensional flow the boundary layer approximations are The velocity component in the flow direction is much larger than that in the normal direction and The temperature and velocity gradients normal to the flow are much greater than those along the flow direction
assuming Vx ⇒ V∞ ⇒ U and Vy ⇒ u from continuity equation we know that
Vy << Vx