Answer:
The major effects of ice accretion on the aircraft is that it disturbs the flow of air and effects the aircraft's performance.
Explanation:
The ice accretion effects the longitudinal stability of an aircraft as:
1. The accumulation of ice on the tail of an aircraft results in the reduction the longitudinal stability and the elevator's efficacy.
2. When the flap is deflected at 
with no power there is an increase in the longitudinal velocity.
3. When the angle of attack is higher close to the stall where separation occurs in the early stages of flow, the effect of ice accretion are of importance.
4. When the situation involves no flap at reduced power setting results in the decrease in aircraft's longitudinal stability an increase in change in coefficient of pitching moment with attack angle.
Given Information:
Inductance = L = 5 mH = 0.005 H
Time = t = 2 seconds
Required Information:
Current at t = 2 seconds = i(t) = ?
Energy at t = 2 seconds = W = ?
Answer:
Current at t = 2 seconds = i(t) = 735.75 A
Energy at t = 2 seconds = W = 1353.32 J
Explanation:
The voltage across an inductor is given as
The current flowing through the inductor is given by
Where L is the inductance and i(0) is the initial current in the inductor which we will assume to be zero since it is not given.
![i(t) = \frac{1}{0.005} \int_0^t \mathrm{5(1-e^{-0.5t}}) \,\mathrm{d}t \,+ 0\\\\i(t) = 200 \int_0^t \mathrm{5(1-e^{-0.5t}}) \,\mathrm{d}t \\\\i(t) = 200 \: [ {5\: (t + \frac{e^{-0.5t}}{0.5})]_0^t \\i(t) = 200\times5\: \: [ { (t + 2e^{-0.5t} + 2 )] \\](https://tex.z-dn.net/?f=i%28t%29%20%3D%20%5Cfrac%7B1%7D%7B0.005%7D%20%5Cint_0%5Et%20%5Cmathrm%7B5%281-e%5E%7B-0.5t%7D%7D%29%20%5C%2C%5Cmathrm%7Bd%7Dt%20%5C%2C%2B%200%5C%5C%5C%5Ci%28t%29%20%3D%20200%20%5Cint_0%5Et%20%5Cmathrm%7B5%281-e%5E%7B-0.5t%7D%7D%29%20%5C%2C%5Cmathrm%7Bd%7Dt%20%5C%5C%5C%5Ci%28t%29%20%3D%20200%20%5C%3A%20%5B%20%7B5%5C%3A%20%28t%20%2B%20%5Cfrac%7Be%5E%7B-0.5t%7D%7D%7B0.5%7D%29%5D_0%5Et%20%5C%5Ci%28t%29%20%3D%20200%5Ctimes5%5C%3A%20%5C%3A%20%5B%20%7B%20%28t%20%2B%202e%5E%7B-0.5t%7D%20%2B%202%20%29%5D%20%5C%5C)
So the current at t = 2 seconds is
The energy stored in the inductor at t = 2 seconds is
Answer:
A detailed structure diagram
Explanation:
Schematics include blueprints and diagrams and they help people design buildings.
Answer:
a) 4.7 kΩ, +/- 5%
b) 2.0 MΩ, +/- 20%
Explanation:
a) If the resistor has the following combination of color bands:
1) Yellow = 1st digit = 4
2) Violet = 2nd digit = 7
3) Red = multiplier = 10e2
4) Gold = tolerance = +/- 5%
this means that the resistor has 4700 Ω (or 4.7 kΩ), with 5% tolerance.
b) Repeating the process for the following combination of color bands:
1) Red = 1st digit = 2
2) Black = 2nd digit = 0
3) Green = multiplier = 10e5
4) Nothing = tolerance = +/- 20%
This combination represents to a resistor of 2*10⁶ Ω (or 2.0 MΩ), with +/- 20% tolerance.
Answer:
a) P ≥ 22.164 Kips
b) Q = 5.4 Kips
Explanation:
GIven
W = 18 Kips
μ₁ = 0.30
μ₂ = 0.60
a) P = ?
We get F₁ and F₂ as follows:
F₁ = μ₁*W = 0.30*18 Kips = 5.4 Kips
F₂ = μ₂*Nef = 0.6*Nef
Then, we apply
∑Fy = 0 (+↑)
Nef*Cos 12º - F₂*Sin 12º = W
⇒ Nef*Cos 12º - (0.6*Nef)*Sin 12º = 18
⇒ Nef = 21.09 Kips
Wedge moves if
P ≥ F₁ + F₂*Cos 12º + Nef*Sin 12º
⇒ P ≥ 5.4 Kips + 0.6*21.09 Kips*Cos 12º + 21.09 Kips*Sin 12º
⇒ P ≥ 22.164 Kips
b) For the static equilibrium of base plate
Q = F₁ = 5.4 Kips
We can see the pic shown in order to understand the question.
