How does ionic bonding work?
If you look at a periodic chart, you will see some columns. The elements all the way over to the right live a charmed life. They have EXACTLY the number of electrons that they want in their outer shell (presumes previous knowledge of valence shell electrons). They are totally happy with their lot in life. Now, look at the column just to the left. Those elements are SO CLOSE to perfect that they just can't stand it! They wish so badly that they could get one more electron, and then their lives would be perfect like the noble gases over there on the end... We'll come back to them in a minute. Now, look all the way over on the left hand side. Those elements are SO CLOSE to perfect that they just can't stand it! They wish so badly that they could GET RID OF just one electron, and their lives would be perfect too! Do you see an opportunity here? Well, the one atom can give its electron to the other one, and everybody can be happy, right? Cool! Well, what happens to your charge when you give away an electron? (You get a positive one charge.) And what happens to your charge when you receive an electron? (You get a negative one charge.) So these two atoms now have opposite charges - and what do we know about how opposite charges act? (They are attracted to one another.) So these two atoms just "hang out" together because of their opposite charges. This is an ionic bond.
Can you help me understand how to use the Rule of Nines?
The "rule of nines" is used to determine the coverage of burns to determine how critical the patient's condition is. A patient arrives at the emergency room with second degree burns on the front of his torso along with the front of one leg. What percentage of his body has second degree burns? And is his burn considered critical? In the rule of nines, the torso equals 18% of the body's surface, or two nines. The front of the leg is 9%. So the total coverage of this burn is 27%. If 25% or more of a patient's body is covered with second degree burns, it is considered a critical burn.
How do I go about finding the velocity of a projectile as it leaves the firing apparatus?
Projectile motion is everywhere around us, from launching a satellite (or manned space vehicle) into orbit, to military artillery fire, to the neighborhood Nerf gun war. In order to be successful in any of these, one must first know the velocity of a projectile as it leaves the launching apparatus before being able to take aim and fire. Let's develop a strategy for determining the velocity of a projectile as it leaves its launching apparatus. First, we have to remember that when making calculations for projectile motion that the vertical and horizontal movement are completely independent of one another. So we have to work those problems separately. Let's set our apparatus on a surface above the surface of the earth. Let's fire it horizontally just to simplify things! First, let's think about time in the air. The time in the air is dependent only upon the vertical drop of the projectile. From the moment the projectile is fired, it falls at exactly the same rate as if it were just dropped from a stationary position. So to calculate the time in the air, we simply measure the height of the projectile and do a free-fall calculation (assumes student knows how to calculate the time for an object in free fall). Then, once we have the amount of time in the air, we can turn our attention to the horizontal movement! We can fire the projectile and measure the distance the projectile travels. At that point, we know a displacement and a time. It is a simple matter to use the formula v = d/t to find the velocity of the projectile as it leaves the launching apparatus.