I think it would be Reflection because the light in the ray reflects off of any flat or shiny object.... Hope this helps ^-^....
Answer: Transverse
Explanation: Transverse waves possess a vertical wave motion and a horizontal particle motion.
It must exist in p<span>lasma state.</span>
Answer:
1. 
2. 
Explanation:
1. According to Newton's law of motion, the puck motion is affected by the acceleration, which is generated by the push force F.
In Newton's 2nd law: F = ma
where m is the mass of the object and a is the resulted acceleration. So in the 2nd experiment, if we double the mass, a would be reduced by half.

Since the puck start from rest, in the 1st experiment, to achieve speed of v it would take t time

Now that acceleration is halved:


You would need to push for twice amount of time 
2. The distance traveled by the puck is as the following equation:

So if the acceleration is halved while maintaining the same d:

As
, then
. Also 



So t increased by 1.14
I am pretty sure that <span>the following whihc cannot be determined by looking at the phase diagram is definitely </span>D. system pressure. I consider this one to be correct because only this point is not included into<span> phase diagram and can't be determined itself. Hope it will help! Regards!</span>