Answer:
c) site preparation
Explanation:
A construction process can be defined as a series of important physical events (processes) that must be accomplished during the execution of a construction project.
Generally, in the construction of any physical asset such as offices, hospitals, schools, stadiums etc, the first step of the construction process is site preparation. Site preparation refers to processes such as clearing, blasting, levelling, landfilling, surveying, cutting, excavating and demolition of all unwanted objects on a piece of land, so as to make it ready for use.
This ultimately implies that, site preparation should be the first task to be accomplished in the construction process.
Hence, the construction process typically begins with site preparation before other activities such as the laying of foundation can be done.
Additionally, construction costs can be defined as the overall costs associated with the development of a built asset, project or property. The construction costs is classified into two (2) main categories and these are; capital and operational costs.
Answer: Pull.
Because it's all about height width and Breadth!
Answer:
The tube surface temperature immediately after installation is 120.4°C and after prolonged service is 110.8°C
Explanation:
The properties of water at 100°C and 1 atm are:
pL = 957.9 kg/m³
pV = 0.596 kg/m³
ΔHL = 2257 kJ/kg
CpL = 4.217 kJ/kg K
uL = 279x10⁻⁶Ns/m²
KL = 0.68 W/m K
σ = 58.9x10³N/m
When the water boils on the surface its heat flux is:
For copper-water, the properties are:
Cfg = 0.0128
The heat flux is:
qn = 0.9 * 18703.42 = 16833.078 W/m²
The tube surface temperature immediately after installation is:
Tinst = 100 + 20.4 = 120.4°C
For rough surfaces, Cfg = 0.0068. Using the same equation:
ΔT = 10.8°C
The tube surface temperature after prolonged service is:
Tprolo = 100 + 10.8 = 110.8°C
Answer: 24 pA
Explanation:
As pure silicon is a semiconductor, the resistivity value is strongly dependent of temperature, as the main responsible for conductivity, the number of charge carriers (both electrons and holes) does.
Based on these considerations, we found that at room temperature, pure silicon resistivity can be approximated as 2.1. 10⁵ Ω cm.
The resistance R of a given resistor, is expressed by the following formula:
R = ρ L / A
Replacing by the values for resistivity, L and A, we have
R = 2.1. 10⁵ Ω cm. (10⁴ μm/cm). 50 μm/ 0.5 μm2
R = 2.1. 10¹¹ Ω
Assuming that we can apply Ohm´s Law, the current that would pass through this resistor for an applied voltage of 5 V, is as follows:
I = V/R = 5 V / 2.1.10¹¹ Ω = 2.38. 10⁻¹¹ A= 24 pA
