Answer:
= -32.53 m / s
this velocity is directed downwards
Explanation:
This is a free fall exercise, let's use the expression
= v_{oy}^{2} + 2 g (y -yo)
where we are assuming that there is friction with the air, as the body falls its initial velocity is zero
v_{oy} = √ 2g (y - y₀)
let's calculate
v_{y} = √ (2 9.8 (0-54.0))
= -32.53 m / s
this velocity is directed downwards
To solve this exercise we will need the concepts concerning impedance and capacitive reactance.
The potential in terms of impedance is given by,
Where,
Impedance is equal to
For definition we know that Impedance is equal also to
Where f is the frequency and c the capacitive reactance.
Re-arrange for c, we have,
Explanation and answer:
This type of question can be clarified and sometimes solved by drawing a proper diagram or sketch. (see below)
Solution:
Since we do not know the reaction of the support, we can take moments about the support (thereby eliminating its involvement).
CCW moment = 0.900(5.00/2 - x) kg-m
CW moment = 0.300*(5.00/2-2.00)) = 0.150 kg-m
At equilibrium, CCW moment = CW moment, so
0.900(2.50-x) = 0.150
Expand and solve
2.25 - 0.900x = 0.150
0.900x = 2.25-0.15 = 2.10
x = 2.10 / 0.900 = 2.33 m (to nearest cm)
