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.
Answer: E-cycle used computer equipment. Recycle Printer Cartridges and Drum Units. Properly Dispose of Old Technology.
Hope this helps!
we have
x<5
x>c
we know that
The solution is the intersection of both solution sets of the given inequalities.
The solutions of the compound inequality must be solutions of both inequalities.
The value of c could be 5 or any number greater than 5, such that there are no solutions to the compound inequality
Because
A number cannot be both less than 5 and greater than 5 at the same time
therefore
the answer is
for c_> there are no solutions to the compound inequality
Is miraculous ladybug one
Answer:
a. 0.4544 N
b. ![5.112 \times 10^{-5 M}](https://tex.z-dn.net/?f=5.112%20%5Ctimes%2010%5E%7B-5%20M%7D)
Explanation:
For computing the normality and molarity of the acid solution first we need to do the following calculations
The balanced reaction
![H_2SO_4 + 2NaOH = Na_2SO_4 + 2H_2O](https://tex.z-dn.net/?f=H_2SO_4%20%2B%202NaOH%20%3D%20Na_2SO_4%20%2B%202H_2O)
![NaOH\ Mass = Normality \times equivalent\ weight \times\ volume](https://tex.z-dn.net/?f=NaOH%5C%20Mass%20%3D%20Normality%20%5Ctimes%20equivalent%5C%20weight%20%5Ctimes%5C%20volume)
![= 0.3200 \times 40 g \times 21.30 mL \times 1L/1000mL](https://tex.z-dn.net/?f=%3D%200.3200%20%5Ctimes%2040%20g%20%5Ctimes%2021.30%20mL%20%5Ctimes%20%201L%2F1000mL)
= 0.27264 g
![NaOH\ mass = \frac{mass}{molecular\ weight}](https://tex.z-dn.net/?f=NaOH%5C%20mass%20%3D%20%5Cfrac%7Bmass%7D%7Bmolecular%5C%20weight%7D)
![= \frac{0.27264\ g}{40g/mol}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B0.27264%5C%20g%7D%7B40g%2Fmol%7D)
= 0.006816 mol
Now
Moles of
needed is
![= \frac{0.006816}{2}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B0.006816%7D%7B2%7D)
= 0.003408 mol
![Mass\ of\ H_2SO_4 = moles \times molecular\ weight](https://tex.z-dn.net/?f=Mass%5C%20of%5C%20H_2SO_4%20%3D%20moles%20%5Ctimes%20molecular%5C%20weight)
![= 0.003408\ mol \times 98g/mol](https://tex.z-dn.net/?f=%3D%200.003408%5C%20mol%20%5Ctimes%2098g%2Fmol)
= 0.333984 g
Now based on the above calculation
a. Normality of acid is
![= \frac{acid\ mass}{equivalent\ weight \times volume}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7Bacid%5C%20mass%7D%7Bequivalent%5C%20weight%20%5Ctimes%20volume%7D)
![= \frac{0.333984 g}{49 \times 0.015}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B0.333984%20g%7D%7B49%20%5Ctimes%200.015%7D)
= 0.4544 N
b. And, the acid solution molarity is
![= \frac{moles}{Volume}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7Bmoles%7D%7BVolume%7D)
![= \frac{0.003408 mol}{15\ mL \times 1L/1000\ mL}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B0.003408%20mol%7D%7B15%5C%20mL%20%5Ctimes%20%201L%2F1000%5C%20mL%7D)
= 0.00005112
=![5.112 \times 10^{-5 M}](https://tex.z-dn.net/?f=5.112%20%5Ctimes%2010%5E%7B-5%20M%7D)
We simply applied the above formulas