Answer:

Explanation:
<u>Dimensional Analysis</u>
It's given the relation between quantities A, B, and C as follows:

and the dimensions of each variable is:



Substituting the dimensions into the relation (the coefficient is not important in dimension analysis):

Operating:


Equating the exponents:


Adding both equations:

Solving:


Answer:

I’d say b is the better option because d is starving yourself which creates unhealthy habits, c would cause yo yo dieting or binging because you treat food as a reward which is a toxic mindset, and a same explanation. I wish there was an answer to just track what you eat, have daily exercise and enjoy in moderation, though.
When you're out of breath, you feel dizzy, lung pain and maybe even nausea or side pain.
Answer:
Since the hockey puck is moving at constant velocity
So here we will have

where we have
a = acceleration of the object
m = mass of object
so here since we know that acceleration is defined as rate of change in velocity
so here we will say that


so we have

so we will say that hockey puck is in equilibrium as there is no net force of it
Neon light recharge them self with more light and electricity, they rapidly give out
the energy they absorbed to get themselves back to normal again. They
do this by giving out tiny packets of light energy
called photons. I;m not sure if this helps but you gave no info really on what you were asking like if there were choices that would have been helpful to see :p