Explanation:
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Answer:
The volume percentage of graphite is 10.197 per cent.
Explanation:
The volume percent of graphite is the ratio of the volume occupied by the graphite phase to the volume occupied by the graphite and ferrite phases. The weight percent in the cast iron is 3.2 wt% (graphite) and 96.8 wt% (ferrite). The volume percentage of graphite is:
Where:
- Volume occupied by the graphite phase, measured in cubic centimeters.
- Volume occupied by the graphite phase, measured in cubic centimeters.
The expression is expanded by using the definition of density and subsequently simplified:
Where:
, 
- Masses of the ferrite and graphite phases, measured in grams.
- Densities of the ferrite and graphite phases, measured in grams per cubic centimeter.
If 
, 
, 
and 
, the volume percentage of graphite is:
The volume percentage of graphite is 10.197 per cent.
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:
The specific heat capacity of substance A is 1.16 J/g
Explanation:
The substances A and B come to a thermal equilibrium, therefore, the heat given by the hotter substance B is absorbed by the colder substance A.
The equation becomes:
Heat release by Substance B = Heat Gained by Substance A
The heat can be calculated by the formula:
Heat = mCΔT
where,
m = mass of substance
C = specific heat capacity of substance
ΔT = difference in temperature of substance
Therefore, the equation becomes:
(mCΔT) of A = (mCΔT) of B
<u>FOR SUBSTANCE A:</u>
m = 6.01 g
ΔT = Final Temperature - Initial Temperature
ΔT = 46.1°C - 20°C = 26.1°C
C = ?
<u>FOR SUBSTANCE B:</u>
m = 25.6 g
ΔT = Initial Temperature - Final Temperature
ΔT = 52.2°C - 46.1°C = 6.1°C
C = 1.17 J/g
Therefore, eqn becomes:
(6.01 g)(C)(26.1°C) = (25.6 g)(1.17 J/g)(6.1°C)
C = (182.7072 J °C)/(156.861 g °C)
<u>C = 1.16 J/g</u>
