Respuesta :
Answer:
1. 4042.5
2. 5.687 m>2
Explanation:
Question 1: pressure difference = height × density × g , p = h × p × g
Question 2: Δ(K.E.)=12(ρA 1v1Δt)(v22−v21).
The only possible source for this increase in energy is the work done by pressure in pushing the fluid into the upward
Taking the pressure on area A1 to be P1, the total force on A1 is P1A 1. In the time Δt, this force acts through a distance v1Δt, and hence does work = force × distance =P1A1v1Δt. .
So this is work done on our chunk of fluid by the fluid pushing it from behind — but that’s not the end of the story, because our chunk of fluid itself does work pushing the fluid in front of it, so to find the total increase in our chunk’s energy, we must subtract off the external work it does. That is, the total work done by pressure on our fluid is
P1A 1v1Δt−P2A2v2Δt=(P1−P2)A 1v1Δt
remembering that A 1v1=A2v2.
This work done must equal the change in kinetic energy, so
(P1−P2)A 1v1Δt=12(ρA 1v1Δt)(v∧2−v2∧1)
from which
P1+12ρv21=P2+12ρv22.
p1-p2=1/2p2-p1(v2-v1)2
4042.5=1/2*100-750(v2-v1)
8085/250=(v2-v1)
32.34=(v2-v1)2
v2-v1=32.34[tex]\sqrt{32.34}[/tex]
The density of the oil is less than the density of water, therefore, the length
of the column of oil will be longer the length of an equal weight of water.
The correct responses are
- (a) The difference in the height of the two liquids surfaces is approximately 1.24 × 10⁻² m.
- (b) The speed of the air being blown across the left arm is approximately 13.71 m/s.
Reasons:
Given parameters;
Density of the oil = 750 kg/m³
The height of the column of oil, L = 5.00 cm = 0.05 m
(a) Required;
The height difference between the liquid surfaces
Solution:
The height of water equivalent to 5.00 m of the oil is found as follows;
Pressure, P = ρ·g·h
Where;
ρ = The density of the fluid
g = Acceleration due to gravity = 9.81 m/s²
h = The height of the liquid
Therefore;
At equilibrium, we have;
Pressure of water, [tex]P_{water}[/tex] = Pressure of oil, [tex]P_{oil}[/tex]
[tex]P_{water}[/tex] = [tex]\rho _{water}[/tex] × g × [tex]h_{water}[/tex]
[tex]\rho _{water}[/tex] = 997 kg/m³
[tex]P_{oil}[/tex] = [tex]\rho _{oil}[/tex] × g × [tex]h_{oil}[/tex]
[tex]h_{oil}[/tex] = 0.05 m = [tex]\frac{1}{20}[/tex] m
∴ [tex]\rho _{water}[/tex] × g × [tex]h_{water}[/tex] = [tex]\mathbf{\rho _{oil}}[/tex] × g × [tex]\mathbf{h_{oil}}[/tex]
997 × 9.81 × [tex]h_{water}[/tex] = 750 × 9.81 × 0.05 = 367.875
[tex]h_{water} = \dfrac{367.875}{997 \times 9.81} = \mathbf{\dfrac{75}{1994}}[/tex]
The height of the water column = [tex]\mathbf{\frac{75}{1994}}[/tex] meters
The difference in height, Δh = [tex]\mathbf{h_{oil}}[/tex] - [tex]\mathbf{h_{water}}[/tex]
Therefore;
[tex]\Delta h= \dfrac{1}{20} - \dfrac{75}{1994} = \dfrac{247} {19940} \approx 1.24 \times 10^{-2}[/tex]
- The difference in height, Δh ≈ 1.24 × 10⁻² m
b) When the right arm is shielded from air, and air is blown across the left
column, until the height of the two liquids are the same, we have;
Dynamic pressure due to the air = Difference in pressure due to water and
oil of the same length, L
Pressure due to the air = [tex]\rho _{water}[/tex] × g × L - [tex]\rho _{oil}[/tex] × g × L
[tex]\mathrm{Pressure \ due \ to \ the \ air} = \mathbf{ \dfrac{1}{2} \cdot \rho_{air} \cdot v^2}[/tex]
Therefore;
[tex]\mathbf{\dfrac{1}{2} \times 1.29 \times v^2} =997 \times 9.81 \times 0.05 - 750 \times 9.81 \times 0.05 = 121.1535[/tex]
[tex]v^2 =\dfrac{121.1535 \times 2}{1.29} \approx \mathbf{187.83}[/tex]
v ≈ √(187.83) ≈ 13.71
- The speed of the air being blown across the left arm, v ≈ 13.71 m/s
Learn more here:
https://brainly.com/question/18892830
