Answer:
the diameter of the second pipe is 2.52 in
Explanation:
Given the data in the question;
We know that; the rate of flow is the same;
so
Av1 = Av2
v ∝ √h
=
= √(
)
( π/4.D1² / π/4.D2² ) = √( )
( D1² / D2² ) = √( ) since second is double of first
so
( D1² / D2² ) = √( )
3² / D2² = √2
D2²√2 = 9
D2² = 9/√2
D2² = 9 / 1.4142
D2² = 6.364
D2 = √ 6.364
D2 = 2.52 in
Therefore, the diameter of the second pipe is 2.52 in
Answer:
The two forces are;
1) Force 1 with magnitude of approximately 183.013 N, acting 30° to the left of the resultant force
2) Force 2 with magnitude of approximately 129.41 N acting at an inclination of 45° to the right of the resultant force
Explanation:
The given parameters are;
The (magnitude) of the resultant of two forces = 250 N
The angle of inclination of the two forces to the resultant = 30° and 45°
Let, F₁ and F₂ represent the two forces, we have;
F₁ is inclined 30° to the left of the resultant force and F₂ is inclined 45° to the right of the resultant force
The components of F₁ are = -F₁ × sin(30°)·i + F₁ × cos(30°)·j
The components of F₂ are = F₂ × sin(45°)·i + F₂ × cos(45°)·j
The sum of the forces = F₂ × sin(45°)·i + F₂ × cos(45°)·j + (-F₁ × sin(30°)·i + F₁ × cos(30°)·j) = 250·j
The resultant force, R = 250·j, which is in the y-direction, therefore, the component of the two forces in the x-direction cancel out
We have;
F₂ × sin(45°)·i = F₁ × sin(30°)·i
F₂ ·√2/2 = F₁/2
∴ F₁ = F₂ ·√2
∴ F₂ × cos(45°)·j + F₁ × cos(30°)·j = 250·j
Which gives;
F₂ × cos(45°)·j + F₂ ·√2 × cos(30°)·j = 250·j
F₂ × ((cos(45°) + √2 × cos(30°))·j = 250·j
F₂ × ((√2)/2 × (1 + √3))·j = 250·j
F₂ × ((√2)/2 × (1 + √3))·j = 250·j
F₂ = 250·j/(((√2)/2 × (1 + √3))·j) ≈ 129.41 N
F₂ ≈ 129.41 N
F₁ = √2 × F₂ = √2 × 129.41 N ≈ 183.013 N
F₁ ≈ 183.013 N
The two forces are;
A force with magnitude of approximately 183.013 N is inclined 30° to the left of the resultant force and a force with magnitude of approximately 129.41 N is inclined 45° to the right of the resultant force.
