I think the correct answer would be D. The tap water in the experiment is one the three test conditions of the independent variable, the type of water. The independent variable in a experiment is the one being manipulated or the one being changed. In this case, it is the type of water.
<h2>Answer:</h2><h3>D. ability to react with oxygen</h3><h2>Explanation:</h2>
<em>Im</em><em> </em><em>not</em><em> </em><em>sure</em><em> </em><em>this</em><em> </em><em>in</em><em> </em><em>your</em><em> </em><em> </em><em>choices</em><em> </em><em>but</em><em> </em><em>if</em><em> </em><em>it</em><em> </em><em>is</em><em>,</em><em> </em><em>this</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>answer</em><em>. </em>
<em>I</em><em> </em><em>hope</em><em> </em><em>I've</em><em> </em><em>helped</em><em>. </em>
Answer:
Amplitude = 8 Volts
Frequency = 0.067 kHz
Explanation:
Note: The missing picture in question is attached for your review.
Given:
Volts/Div = 2 V/div
Time/Div = 5 msec/div
Finding Amplitude:
Now, as you can see in the attached picture, there are 4 division between two peaks of the waveform, so,

(Multiplying by 2 V/div because oscilloscope dial is set at 2 V/div)
Finding Frequency:
As can be seen in attached picture, 3 division are there for one complete cycle of waveform,so,

Since,

Not totally sure but i would say a normal? its not refraction or incidence if its perpendicular and i dont think its a mirror if its an imaginary line so yeah normal (normals are always perpendicular to their surface too i think so)
The question here would be what is the volume of the room. The density of air that is given has no use. We simply multiply the dimensions given of the room to determine the volume.
<span>43.0m × 18.0m × 15.0m = 11610m^3 ( 3.28 ft / 1 m)^3 = 4.09 x 10^5 ft^3</span>