Answer:
D. Test a piece of treated iron and a piece of untreated iron under the same conditions.
Explanation:
A controlled experiment is defined as an experiment where the experimenter or the researcher tries to test or experiment his result by changing one variable at a time and keeping the other factors under the same conditions and sees how it effects the result.
The factors which can be changed in an experiment is called its variables.
In the context, testing a piece of treated iron as well as a untreated piece of iron under the same working conditions forms a controlled experiment.
<span>Charles Darwin and Alfred Russel Wallace were the ones who had the idea of evolution in 1858!</span>
Position is measured in meters (m), so it is a base quantity.
<h3>What is base quantity?</h3>
A base or fundamental quantity is a physical quantity, in which other quantities are derived from.
Example of fundamental quantities;
- Mass
- Length (position)
- Time
- Temperature
- Amount of substance
<h3>What is a derived quantity?</h3>
Derived quantities are those quantities obtained or expressed from fundamental quantities.
Example of derived quantities;
- Speed
- Acceleration
- Volume
- Area
- Density, etc
Thus, we can conclude that position measured in meters (m) is a base quantity.
Learn more about base quantities here: brainly.com/question/14480063
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Answer:
C. All planets orbit the Sun in the same direction.
E. The orbits of the planets are evenly distributed in distance from the Sun.
F. All planets orbit the Sun in a roughly flat plane.
Explanation:
As we can see that due to big bang theory of planet formation we can say all planets revolve around the sun in circular orbit with almost in same plane
Size of the planets are unevenly distributed and they all revolve around the sun in same directions
so here correct options are
C. All planets orbit the Sun in the same direction.
E. The orbits of the planets are evenly distributed in distance from the Sun.
F. All planets orbit the Sun in a roughly flat plane.
Answer:
a) 
Explanation:
a) the cross-sectional area of the hose would be the square of radius times pi. And since the sectional radius is half of its diameter d. We can express the cross-sectional area A1 in term of diameter d1
