A) experimental because he isn’t sure and is testing out
F = m*a
30 N = (ma + mb) * a
30 = 5*a
a = 6 m/s ^2
F de B em A
30 - F de B,A = ma * a
30 - F de B em A = 3 * 6
30 - 18 = F de B em A
12 = F de B em A
Resposta: 6 m/s^2 e 12N
Bate com o gabarito, man? Ou eu tô viajando aqui?
Abç!
Theres: the vacuole, nucleus, rough endoplamid reticulum, smooth endoplasmic reticulum, cell memebrane, cell wall, chloroplast, mitochondria, golgi apperatus, lysosomes, and ribosomes
To reach a vertical height of 13.8 ft against gravity, which has an acceleration of 32 ft/s^2, the required vertical speed can be calculated from the equation:
vi^2 - vf^2 = 2*g*h
Given that it has vf = 0 (it is not moving vertically at its maximum height), g = 32, and h = 13.8, we can solve for vi:
vi^2 = 29.72 ft/s
This is only its vertical speed, so this is equivalent to its original speed multiplied by the sine of the angle:
29.72 ft/s = (v_original)*(sin 42.2<span>°</span>)
v_original = 44.24 ft/s
Converting to m/s, this can be divided by 3.28 to get 13.49 m/s.
