Answer:
Explanation:
Usar motores eléctricos en aviones ofrece numerosas ventajas reales. A diferencia de los motores de combustión interna los motores eléctricos no necesitan aire para funcionar, lo que significa que pueden mantener toda su capacidad y potencia incluso a altitudes elevadas donde el aire es más tenue.
Answer:
4.8°C
Explanation:
The rate of heat transfer through the wall is given by:
![q=\frac{Ak}{L}dT](https://tex.z-dn.net/?f=q%3D%5Cfrac%7BAk%7D%7BL%7DdT)
![\frac{q}{A}=\frac{k}{L}dT](https://tex.z-dn.net/?f=%5Cfrac%7Bq%7D%7BA%7D%3D%5Cfrac%7Bk%7D%7BL%7DdT)
Assumptions:
1) the system is at equilibrium
2) the heat transfer from foam side to interface and interface to block side is equal. There is no heat retention at any point
3) the external surface of the wall (concrete block side) is large enough that all heat is dissipated and there is no increase in temperature of the air on that side
![{k_{fi}= 0.03 W/m.K](https://tex.z-dn.net/?f=%7Bk_%7Bfi%7D%3D%200.03%20W%2Fm.K)
![{L_{fi}= 5 cm = 0.05 m](https://tex.z-dn.net/?f=%7BL_%7Bfi%7D%3D%205%20cm%20%3D%200.05%20m)
![{T_{fi}= 25 \°C](https://tex.z-dn.net/?f=%7BT_%7Bfi%7D%3D%2025%20%5C%C2%B0C)
![{k_{cb} = 0.5 W/m.K](https://tex.z-dn.net/?f=%7Bk_%7Bcb%7D%20%3D%200.5%20W%2Fm.K)
![{L_{cb}= 20 cm = 0.20 m](https://tex.z-dn.net/?f=%7BL_%7Bcb%7D%3D%2020%20cm%20%3D%200.20%20m)
![{T_{cb}= 0 \°C](https://tex.z-dn.net/?f=%7BT_%7Bcb%7D%3D%200%20%5C%C2%B0C)
temperature at the interface
Solving for
will give the temperature at the interface:
![\frac{q}{A}=\frac{k_{fi} }{L_{fi} }(T_{fi} -T_{m})=\frac{k_{cb} }{L_{cb} }(T_{m} -T_{cb})](https://tex.z-dn.net/?f=%5Cfrac%7Bq%7D%7BA%7D%3D%5Cfrac%7Bk_%7Bfi%7D%20%7D%7BL_%7Bfi%7D%20%7D%28T_%7Bfi%7D%20-T_%7Bm%7D%29%3D%5Cfrac%7Bk_%7Bcb%7D%20%7D%7BL_%7Bcb%7D%20%7D%28T_%7Bm%7D%20-T_%7Bcb%7D%29)
![\frac{0.03}{0.05 }(25 -T_{m})=\frac{0.5}{0.2}(T_{m} -0})](https://tex.z-dn.net/?f=%5Cfrac%7B0.03%7D%7B0.05%20%7D%2825%20-T_%7Bm%7D%29%3D%5Cfrac%7B0.5%7D%7B0.2%7D%28T_%7Bm%7D%20-0%7D%29)
![15 -0.6T_{m}=2.5T_{m}](https://tex.z-dn.net/?f=15%20-0.6T_%7Bm%7D%3D2.5T_%7Bm%7D)
![3.1T_{m}=15](https://tex.z-dn.net/?f=3.1T_%7Bm%7D%3D15)
![T_{m}=4.8](https://tex.z-dn.net/?f=T_%7Bm%7D%3D4.8)
Answer:C 0.12 V
Explanation:
Given
Concentration of ![Fe^{2+} M_1=0.40 M](https://tex.z-dn.net/?f=Fe%5E%7B2%2B%7D%20M_1%3D0.40%20M)
Concentration of ![Ni^{2+} M_2=0.002 M](https://tex.z-dn.net/?f=Ni%5E%7B2%2B%7D%20M_2%3D0.002%20M)
Standard Potential for Ni and Fe are
and ![V_1=-0.44 V](https://tex.z-dn.net/?f=V_1%3D-0.44%20V)
![\Delta V=V_2-V_1-\frac{0.0592}{2}\log (\frac{M_1}{M_2})](https://tex.z-dn.net/?f=%5CDelta%20V%3DV_2-V_1-%5Cfrac%7B0.0592%7D%7B2%7D%5Clog%20%28%5Cfrac%7BM_1%7D%7BM_2%7D%29)
![\Delta V=-0.25-(-0.44)-\frac{0.0592}{2}\log (\frac{0.4}{0.002})](https://tex.z-dn.net/?f=%5CDelta%20V%3D-0.25-%28-0.44%29-%5Cfrac%7B0.0592%7D%7B2%7D%5Clog%20%28%5Cfrac%7B0.4%7D%7B0.002%7D%29)
![\Delta V=0.12\ V](https://tex.z-dn.net/?f=%5CDelta%20V%3D0.12%5C%20V)
Answer:
D
Explanation: She hopes to be able to make this, however she hasn't yet...therefore she is thinking of a concept and it's development
Answer:
γ
=0.01, P=248 kN
Explanation:
Given Data:
displacement = 2mm ;
height = 200mm ;
l = 400mm ;
w = 100 ;
G = 620 MPa = 620 N//mm²; 1MPa = 1N//mm²
a. Average Shear Strain:
The average shear strain can be determined by dividing the total displacement of plate by height
γ
= displacement / total height
= 2/200 = 0.01
b. Force P on upper plate:
Now, as we know that force per unit area equals to stress
τ = P/A
Also, τ = Gγ![_{xy}](https://tex.z-dn.net/?f=_%7Bxy%7D)
By comapring both equations, we get
P/A = Gγ
------------ eq(1)
First we need to calculate total area,
A = l*w = 400 * 100= 4*10^4mm²
By putting the values in equation 1, we get
P/40000 = 620 * 0.01
P = 248000 N or 2.48 *10^5 N or 248 kN