Answer:
See attachment and explanation.
Explanation:
- The following question can be solved better with the help of a MATLAB program as follows. The code is given in the attachment.
- The plot of the graph is given in attachment.
- The code covers the entire spectrum of the poly-tropic range ( 1.2 - 1.6 ) and 20 steps ( cases ) have been plotted and compared in the attached plot.
Answer:
The level of service of of compound grade freeway is LOSB.
Explanation:
Find the provided attachments for explanation
According to O*NET, the common work contexts for Licensing Examiners and Inspectors include:
- Telephone
- Face-to-face discussions
- Contact with others
- Importance of being exact or accurate.
O*NET is an acronym for occupational information network and it refers to a free resource center or online database that is updated from time to time with several occupational definitions, so as to help the following categories of people understand the current work situation in the United States of America:
- Workforce development professionals
- Human resource (HR) managers
On O*NET, work contexts are typically used to describe the physical and social elements that are common to a particular profession or occupational work. Also, the less common work contexts are listed toward the bottom while common work contexts are listed toward the top.
According to O*NET, the common work contexts for Licensing Examiners and Inspectors include:
1. Telephone
2. Face-to-face discussions
3. Contact with others
4. Importance of being exact or accurate.
Read more on work contexts here: brainly.com/question/22826220
Answer:
0.0406 m/s
Explanation:
Given:
Diameter of the tube, D = 25 mm = 0.025 m
cross-sectional area of the tube = (π/4)D² = (π/4)(0.025)² = 4.9 × 10⁻⁴ m²
Mass flow rate = 0.01 kg/s
Now,
the mass flow rate is given as:
mass flow rate = ρAV
where,
ρ is the density of the water = 1000 kg/m³
A is the area of cross-section of the pipe
V is the average velocity through the pipe
thus,
0.01 = 1000 × 4.9 × 10⁻⁴ × V
or
V = 0.0203 m/s
also,
Reynold's number, Re = 
where,
ν is the kinematic viscosity of the water = 0.833 × 10⁻⁶ m²/s
thus,
Re = 
or
Re = 611.39 < 2000
thus,
the flow is laminar
hence,
the maximum velocity = 2 × average velocity = 2 × 0.0203 m/s
or
maximum velocity = 0.0406 m/s