California I think not sure
Before a person walks through burning coal, the person will make sure their feet are very wet. When they start walking on the coal, this moisture will evaporate and form a protective gas layer underneath the person's feet. You can see examples of this if you happen to drip some water on a hot stove or any very hot surface. The water will very easily glide around on top of a newly formed layer of air underneath it -- like air hockey pucks on an air hockey table. Note that when someone walks through burning coal, typically this is also done very quickly to prevent a great deal of exposure to possible harm. By walking quickly, thinking positively, and letting the water cushion you from immediate danger over a short distance, such a task is possible. You may have also heard of physics teachers demonstrating how this principle works by sticking their hand first in a bucket of water and then quickly in a bucket of boiling molten lead. In the lead, their hand is protected briefly by a layer of gas from the evaporated water (the water vapor). I'm fairly sure that there is a name for this particular layer of gas, but I'm afraid the name is beyond me at the moment. In other words, water vapor has a low heat capacity and poor thermal conduction. Very often, the coals or wood embers that are used in fire walking also have a low heat capacity. Sweat produced on the bottom of people's feet also helps form a protective water vapor. All of this together makes it possible, if moving quickly enough, to walk across hot coals without getting burned. WARNING: Do not attempt to perform any of the actions described above. You can seriously injure yourself. Answered by: Ted Pavlic, Electrical Engineering Undergrad Student, Ohio St. (citing my source)
It accelerates in the y component (bc of gravity) AND the x-component (b/c of the friction force).
Density is the mass per unit volume of any object. It is calculated by dividing the mass of an object by its volume. This is:
ρ = m/V
ρ = 4.05 g / 12 mL
ρ = 0.3375 g/mL
<h3>
ρ ≅ 0.338 g/mL</h3>
OPTION A
