Frost will disturb the smooth flow of air over the wing, unpleasantly
distressing its lifting competence. In other words, this spoils the even flow
of air over the wings, by this means decreasing lifting capability. Also, frost
may avoid the airplane from becoming flying at normal departure speed.
Answer:
a) F = 680 N, b) W = 215 .4 J
, c) F = 1278.4 N
Explanation:
a) Hooke's law is
F = k x
To find the displacement (x) let's use the elastic energy equation
= ½ k x²
k = 2
/ x²
k = 2 85.0 / 0.250²
k = 2720 N / m
We replace and look for elastic force
F = 2720 0.250
F = 680 N
b) The definition of work is
W = ΔEm
W =
- 
W = ½ k (
² - x₀²)
The final distance
= 0.250 +0.220
= 0.4750 m
We calculate the work
W = ½ 2720 (0.47² - 0.25²)
W = 215 .4 J
We calculate the strength
F = k 
F = 2720 0.470
F = 1278.4 N
Answer:
Interviewing and observation are two methods of collecting qualitative data as part of research. ... Interviews vary from structured, in which a set list of questions is asked of every interviewee, to unstructured, which is open-ended.
Answer:
The distance that the racecar traveled is indeed 500m. But at the end of the lap, it is right back where it started. So overall, it has been displaced 0m.
Explanation:
The answer is.......................................STEEL!!!!!!!!!!