Question:
1) test the model and analyze the results of the test
2) building the model and observing it
3) observing the model and reporting results
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4) designing the model and drawing conclusions
Answer:
The correct option is;
1) Test the model and analyze the results of the test
Explanation:
Based on the flowchart, a model improvement process involves the implementation of a process or model improvement cycle such as the Plan Do Study Act, PDSA cycle, however feedback to the process will be be gotten from testing the model and analyzing the results of the tests. When grey areas or aspects of the model are found that cause the system to malfunction are determined, steps should then be taken to improve the performance of the model.
Answer:
Given:
high temperature reservoir 
low temperature reservoir 
thermal efficiency 
The engines are said to operate on Carnot cycle which is totally reversible.
To find the intermediate temperature between the two engines, The thermal efficiency of the first heat engine can be defined as

The thermal efficiency of second heat engine can be written as

The temperature of intermediate reservoir can be defined as

Answer:
<em>a) 4.51 lbf-s^2/ft</em>
<em>b) 65.8 kg</em>
<em>c) 645 N</em>
<em>d) 23.8 lb</em>
<em>e) 65.8 kg</em>
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Explanation:
Weight of the man on Earth = 145 lb
a) Mass in slug is...
32.174 pound = 1 slug
145 pound =
slug
= 145/32.174 = <em>4.51 lbf-s^2/ft</em>
b) Mass in kg is...
2.205 pounds = 1 kg
145 pounds =
kg
= 145/2.205 = <em>65.8 kg</em>
c) Weight in Newton = mg
where
m is mass in kg
g is acceleration due to gravity on Earth = 9.81 m/s^2
Weight in Newton = 65.8 x 9.81 = <em>645 N</em>
d) If on the moon with acceleration due to gravity of 5.30 ft/s^2,
1 m/s^2 = 3.2808 ft/s^2
m/s^2 = 5.30 ft/s^2
= 5.30/3.2808 = 1.6155 m/s^2
weight in Newton = mg = 65.8 x 1.6155 = 106
weight in pounds = 106/4.448 = <em>23.8 lb</em>
e) The mass of the man does not change on the moon. It will therefore have the same value as his mass here on Earth
mass on the moon = <em>65.8 kg</em>
Answer:
6.5 × 10¹⁵/ cm³
Explanation:
Thinking process:
The relation 
With the expression Ef - Ei = 0.36 × 1.6 × 10⁻¹⁹
and ni = 1.5 × 10¹⁰
Temperature, T = 300 K
K = 1.38 × 10⁻²³
This generates N₀ = 1.654 × 10¹⁶ per cube
Now, there are 10¹⁶ per cubic centimeter
Hence, 
Answer:
Yes, it is possible to maintain a pressure of 10 kPa in a condenser that is being cooled by river water that is entering at 20 °C because this temperature (20 °C) of the external cooling water is less than the saturation temperature of steam which is which is 45.81 °C, and heated by a boiler; as a result of this condition, coupled with the assumption that the turbine, pump, and interconnecting tube are adiabatic, and the condenser exchanges its heat with the external cooling river water, it possible to maintain a pressure of 10 kPa.