Answer:
a) In a parallel plate capacitor the capacitance is
So, in order to halve the capacitance the plate separation should be twice the original distance.
b) Same question as (a).
c) According to the above equations, the charge stored in a capacitor is directly proportional to the plate separation.
In order to halve the charge stored the plate separation should be half the original distance.
d) Same question as (c).
e) The energy stored in a capacitor is
In order to halve the energy stored in the capacitor, the plate separation should be twice the original distance.
f) Same question as (e).
g) The energy density is given by the following equation
As it turns out, the energy density is independent from the geometric factors. This makes sense, because the electric field in a parallel field capacitor is independent from the distance.
Answer:
T = 4.42 10⁴ N
Explanation:
this is a problem of standing waves, let's start with the open tube, to calculate the wavelength
λ = 4L / n n = 1, 3, 5, ... (2n-1)
How the third resonance is excited
m = 3
L = 192 cm = 1.92 m
λ = 4 1.92 / 3
λ = 2.56 m
As in the resonant processes, the frequency is maintained until you look for the frequency in this tube, with the speed ratio
v = λ f
f = v / λ
f = 343 / 2.56
f = 133.98 Hz
Now he works with the rope, which oscillates in its second mode m = 2 and has a length of L = 37 cm = 0.37 m
The expression for standing waves on a string is
λ = 2L / n
λ = 2 0.37 / 2
λ = 0.37 m
The speed of the wave is
v = λ f
As we have some resonance processes between the string and the tube the frequency is the same
v = 0.37 133.98
v = 49.57 m / s
Let's use the relationship of the speed of the wave with the properties of the string
v = √ T /μ
T = v² μ
T = 49.57² 18
T = 4.42 10⁴ N
C) Hypnosis Certification this is only showing your certification and not a method of hypnosis
Answer:
I don't now sorry HHHAHAH GOOD LUCK
