Answer:
1. documenting, tinkering, testing
Explanation:
Technological design is defined as the process of study, design and development of new technologies.
There are some action in the methodical tests and refinements specific to technological design include documenting, tinkering, testing.
<u>Documenting </u><u>includes collecting all the information about the design and develop the product, </u><u>tinkering</u><u> involves repairing or adjust the issues found in the development, and </u><u>testing </u><u>helps to evaluate if the product is ready to work as it is supposed to.</u>
Hence, the correct answer is "1."
To protect the patents of those they work . Patents are legal rights of ownership to something that you have made or created.
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Answer:
The most important function of the legislative branch is its lawmaking authority. In order for a law to be created, a bill must be introduced by either a member of the House or Senate. Once introduced, the bill is brought to a committee for review. ... Each committee is organized around a specific policy function.
Explanation:
1) At tne same temperature and with the same volume, initially the chamber 1 has the dobule of moles of gas than the chamber 2, so the pressure in the chamber 1 ( call it p1) is the double of the pressure of chamber 2 (p2)
=> p1 = 2 p2
Which is easy to demonstrate using ideal gas equation:
p1 = nRT/V = 2.0 mol * RT / 1 liter
p2 = nRT/V = 1.0 mol * RT / 1 liter
=> p1 / p2 = 2.0 / 1.0 = 2 => p1 = 2 * p2
2) Assuming that when the valve is opened there is not change in temperature, there will be 1.00 + 2.00 moles of gas in a volumen of 2 liters.
So, the pressure in both chambers (which form one same vessel) is:
p = nRT/V = 3.0 mol * RT / 2liter
which compared to the initial pressure in chamber 1, p1, is:
p / p1 = (3/2) / 2 = 3/4 => p = (3/4)p1
So, the answer is that the pressure in the chamber 1 decreases to 3/4 its original pressure.
You can also see how the pressure in chamber 2 changes:
p / p2 = (3/2) / 1 = 3/2, which means that the pressure in the chamber 2 decreases to 3/2 of its original pressure.