difference between theory and practice?
Answer:
There is a huge difference between theory vs. practice. Theory assumes an outcome, while practice allows you to test the theory and see if it is accurate.
Theory and Practice Explained
Practice is the observation of disparate concepts (or a phenomenon) that needs explanation. A theory is a proposed explanation of the relationship between two or more concepts, or an explanation for how/why a phenomenon occurs.
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A 4-m-high and 6-m-long wall is constructed of twolarge 2-cm-thick steel plates (k 5 15 W/m·K) separated by1-cm-thick and 20-cm wide steel bars placed 99 cm apart. Theremaining space between the steel plates is filled with fiber-glass insulation (k 5 0.035 W/m·K). If the temperature dif-ference between the inner and the outer surfaces of the wallsis 22°C, determine the
Answer:
fart
Explanation:
A landowner and a contractor entered into a written contract under which the contractor agreed to build a building and pave an adjacent sidewalk for the landowner for $200,000. Later, while construction was proceeding, the landowner and the contractor entered into an oral modification under which the contractor was not obligated to pave the sidewalk but still would be entitled to $200,000 upon completion of the building. The contractor completed the building. The landowner, after discussions with his landscaper, demanded that the contractor pave the adjacent sidewalk. The contractor refused.
Has the contractor breached the contract?
(a) Calculate the heat flux through a sheet of steel that is 10 mm thick when the temperatures oneither side of the sheet are held constant at 300oC and 100oC, respectively.(b) Determine the heat loss per hour if the cross-sectional area of the sheet is 0.25 m2.(c) What will be the heat loss per hour if a sheet of soda-lime glass is used instead
Answer:
do the wam wam
Explanation:
The heat flux is =1038kW/m² , the heat lost per hour is =259.5 kW, the heat lost per hour using a sheet of soda- lime glass.
Calculation of heat fluxThe thickness of steel( t) = 10mm = 10× 10^-³m
The temperature difference on both sides = 300-100
∆T = 200°C
But the formula for heat flux = q = k∆T/t
Where K = thermal conductivity for steel = 51.9W/mK.
Substitute the variables into the formula for heat flux;
q = 51.9 × 200/10 × 10-³
q = 10380 × 10³/10
q = 10380000/10
q = 1038000 W/m² = 1038kW/m²
To calculate the heat lost per hour if the cross sectional area is = 0.25 m2 use the formula q × A
= 1038kW/m² × 0.25 m2
= 259.5 kW.
Learn more about heat flux here:
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