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)
Explanation:
(a) Hooke's law:
F = kx
7.50 N = k (0.0300 m)
k = 250 N/m
(b) Angular frequency:
ω = √(k/m)
ω = √((250 N/m) / (0.500 kg))
ω = 22.4 rad/s
Frequency:
f = ω / (2π)
f = 3.56 cycles/s
Period:
T = 1/f
T = 0.281 s
(c) EE = ½ kx²
EE = ½ (250 N/m) (0.0500 m)²
EE = 0.313 J
(d) A = 0.0500 m
(e) vmax = Aω
vmax = (0.0500 m) (22.4 rad/s)
vmax = 1.12 m/s
amax = Aω²
amax = (0.0500 m) (22.4 rad/s)²
amax = 25.0 m/s²
(f) x = A cos(ωt)
x = (0.0500 m) cos(22.4 rad/s × 0.500 s)
x = 0.00919 m
(g) v = dx/dt = -Aω sin(ωt)
v = -(0.0500 m) (22.4 rad/s) sin(22.4 rad/s × 0.500 s)
v = -1.10 m/s
a = dv/dt = -Aω² cos(ωt)
a = -(0.0500 m) (22.4 rad/s)² cos(22.4 rad/s × 0.500 s)
a = -4.59 m/s²
Fulcrum need to be positioned balanced with weight on both the sides following law of lever.
What is the physical law of the lever?
- It is the foundation for issues with weight and balance. According to this rule, a lever is balanced when the weight multiplied by the arm on one side of the fulcrum, which serves as the pivot point for the device, equals the weight multiplied by the arm on the opposing side.
- The lever is balanced, in other words, when the sum of the moments about the fulcrum is zero.
- The situation in which the positive moments (those attempting to turn the lever clockwise) equal the negative moments is known as this (those that try to rotate it counterclockwise).
- Moving the weights closer to or away from the fulcrum, as well as raising or lowering the weights, can alter the balance point, or CG, of the lever.
Learn more about the Fulcrum with the help of the given link:
brainly.com/question/16422662
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