Explanation:
The given data is as follows.
Fluid is water so, density
Weight flow rate = 500 lbf/s = 2224.11 N/sec
Cross-sectional area (A) =
= 0.05184
Hence, weight flow rate will be given as follows.
w =
2224.11 N/sec =
V = m/s
= 4.373 m/s
Thus, we can conclude that average velocity in the given case is 4.373 m/s.
Answer:
r = <2.640 10⁶, 1.01 10⁸, 0> m
Explanation:
For this exercise we are going to solve it for each direction separately,
we locate a fixed reference frame in space at the height of the rocket, such that the position of the rocket is
r₀ = <4, 4, 0> 10³ m
X axis
the initial velocity of the ship on this axis is v₀ₓ = 0, when it passes through the point x₀ = 4 km it ignites the rockets, experiencing a force of Fₓ = 7.0 10⁵ N for 21 s and the rockets turn off
They ask where it is after one hour t = 1 h = 3600 s
Let's apply Newton's second law
Fₓ = m aₓ
aₓ = Fₓ / m
aₓ = 7.0 10⁵/2 10⁴
aₓ = 3.5 10¹ m / s
Let's use kinematics to find the distance
for the first t₁ = 21 s the movement is accelerated
x₁ = x₀ + v₀ t₁ + ½ aₓ t₁²
x₁ = x₀ + ½ aₓ t₁²
x₁ = 4000 + ½ 35 21² = 4000 + 7717,.5
x₁ = 11717.5 m
this instant has a speed of
vₓ = v₀ₓ + aₓ t
vₓ = aₓ t ₁
vₓ = 35 21
vₓ = 735 m / s
the rest of the time there is no acceleration so it is uniform motion at this speed
t₂ = 3600 - 21
t₂ = 3576 s
vₓ = x₂ / t₂
x₂ = vₓ t₂
x₂ = 735 3576
x₂ = 2,628 10⁶ m
the total distance traveled in this direction is
x_total = x₁ + x₂
x_total = 11717.5 + 2.628 10⁶
x_total = 2,640 10⁶ m
Y axis
on this axis it is in the initial position of y₀ = 4000 m, with an initial velocity of = 28 10³ m / s and there is no force on this axis F_{y} = 0
The movement in this axis is uniform,
v_{y} = y / t
y = v_{y} t
y = 28 10³/3600
y = 1.01 10⁸ m
the total distance is
y_total = y₀ + y
y_total = 4000 + 1.01 10⁸
y_total = 1.01 10⁸ m
Z axis
the initial position is z₀ = 0, with an initial velocity of v₀ = 0 and in this axis there is no force F_{z} = 0
the movement is uniform
z = 0
the final position of the rocket after 1 h is
r = <2.640 10⁶, 1.01 10⁸, 0> m