HELPP!
1 answer:
You might be interested in
Answer:
4.75 m/s
Explanation:
The computation of the velocity of the existing water is shown below:
Data provided in the question
Tall = 2 m
Inside diameter tank = 2m
Hole opened = 10 cm
Bottom of the tank = 0.75 m
Based on the above information, first we have to determine the height which is
= 2 - 0.75 - 0.10
= 2 - 0.85
= 1.15 m
We assume the following things
1. Compressible flow
2. Stream line followed
Now applied the Bernoulli equation to section 1 and 2
So we get
where,
P_1 = P_2 = hydrostatic
z_1 = 0
z_2 = h
Now
= 4.7476 m/sec
= 4.75 m/s
1) Analyze the equation
Vₓ = α + β t²
Vₓ = 3.00 + 0.100 t²
That is a quadratic equation, so the graph is a parabola.
2) Therefore, take into account that the main points to draw a parabola are:
i) vertex
ii) concavity
iii) y-intercepts
iv) x - intercepts
Also, for all graphs you need the domain and the range.
3) Find the y-intercept (t = 0)
t = 0 ⇒ Vₓ = 3.00 + 0.100 (0)² = 3.00
4) Find the x-intercepts (Vₓ = 0)
Vₓ = 0 ⇒ 0 = 3.00 + 0.100 t²
⇒ t² = - 3.00 / 0.100 = -30.0. Since t² cannot be negative, there is not x-intercepts.
5) Concavity
Since, the coefficient of t² is positive, the parabola open upwards.
6) Vertex
It is the local minimum of the equation. You can find it by the first derivative
Vₓ' = 2×0.100 t = 0 ⇒ t = 0
Vₓ = 3.00 + 0.100 (0)² = 3.00 m/s
⇒ vertex = (0,3.00)
7) The domain is given t ∈ [0,5.00]
8) You can also build a table with several points in the domain
t =0; Vₓ = 3.00 + 0.100 (0)² = 0
t = 1; Vₓ = 3.00 + 0.100 (1)² = 3.10
t = 2; Vₓ = 3.00 + 0.100 (2)² = 3.40
t = 3; Vₓ = 3.00 + 0.100 (3)² = 3.90
t = 4; Vₓ = 3.00 + 0.100 (4)² = 4.60
t = 5; Vₓ = 3.00 + 0.100 (5)² = 5.50
9) Range: Vₓ ∈ [ 3.00, 5.50]
10) All that information permits you to put several points in a coordinate sysment and sketch the same graph as the shown in the figure attached.
