1.) equal volume of different substances have "different" masses.
2.)The more closely packed arrangement the particles of a substance have, "increases" its density.
3.)the SI unit of power is "Watts".
4.)an iron nail sinks in water but floats on " mercury ".
5.)balloons used for advertisements are filled with " helium" gas.
6.)"Conduction" is the primary mode of heat transfer in liquid and gases.
I hope this helps you...
Answer: C. Metals are found on the left side of the periodic table.
Explanation:
The periodic table contains vertical columns called as groups and horizontal rows called as periods.
Period 2 contains 8 elements which are lithium, beryllium , boron , carbon, nitrogen, oxygen , fluorine and neon. Only Lithium and beryllium are metals.
Group 18 contains all the noble gases which are all non metals.
Metals are the elements which loose electrons easily and form positive ions. Non-metals are the elements which can gain electrons easily and form negative ions.
Metals are present on left side of the periodic table and as we move to right side of the periodic table , the metallic character decreases and thus non metals are found on the right side of the periodic table.
Answer:B. Increased the amount of charge.
Explanation:
The gas planets usually have extremely high gravitational pulls, the surface isn't solid (since its a gas planet), and gas planets are larger than the inner planets.
<span>Similarities- These planets all have moons and they both revolve around the sun (obviously).
Hope this helps.</span>
Answer:

Explanation:
Given data:



Let the distance traveled by the object in the second case be 
In the given problem, work done by the forces are same in both the cases.
Thus,




