Answer:
<em>the % recovery of aluminum product is 80.5%</em>
<em>the % purity of the aluminum product is 54.7%</em>
<em></em>
Explanation:
feed rate to separator = 2500 kg/hr
in one hour, there will be 2500 kg/hr x 1 hr = 2500 kg of material is fed into the machine
of this 2500 kg, the feed is known to contain 174 kg of aluminium and 2326 kg of rejects.
After the separation, 256 kg is collected in the product stream.
of this 256 kg, 140 kg is aluminium.
% recovery of aluminium will be = mass of aluminium in material collected in the product stream ÷ mass of aluminium contained in the feed material
% recovery of aluminium = 140kg/174kg x 100% = <em>80.5%</em>
% purity of the aluminium product = mass of aluminium in final product ÷ total mass of product collected in product stream
% purity of the aluminium product = 140kg/256kg
x 100% = <em>54.7%</em>
Answer:
have you heard of gnoogle?
Explanation:have you heard of goongle?
Answer:
The answer is below
Explanation:
Given that:
Diameter (D) = 0.03 mm = 0.00003 m, length (L) = 2.4 mm = 0.0024 m, longitudinal tensile strength
, Fracture strength
![(\sigma_f)=5100\ MPa=5100*10^6\ Pa,fiber-matrix\ stres(\sigma_m)=17.5\ MPa=17.5*10^6\ Pa,matrix\ strength=\tau_c=17\ MPa=17 *10^6\ Pa](https://tex.z-dn.net/?f=%28%5Csigma_f%29%3D5100%5C%20MPa%3D5100%2A10%5E6%5C%20Pa%2Cfiber-matrix%5C%20stres%28%5Csigma_m%29%3D17.5%5C%20MPa%3D17.5%2A10%5E6%5C%20Pa%2Cmatrix%5C%20strength%3D%5Ctau_c%3D17%5C%20MPa%3D17%20%2A10%5E6%5C%20Pa)
a) The critical length (
) is given by:
![L_c=\sigma_f*(\frac{D}{2*\tau_c} )=5100*10^6*\frac{0.00003}{2*17*10^6}=0.0045\ m=4.5\ mm](https://tex.z-dn.net/?f=L_c%3D%5Csigma_f%2A%28%5Cfrac%7BD%7D%7B2%2A%5Ctau_c%7D%20%29%3D5100%2A10%5E6%2A%5Cfrac%7B0.00003%7D%7B2%2A17%2A10%5E6%7D%3D0.0045%5C%20m%3D4.5%5C%20mm)
The critical length (4.5 mm) is greater than the given length, hence th composite can be produced.
b) The volume fraction (Vf) is gotten from the formula:
![\sigma_{cd}=\frac{L*\tau_c}{D}*V_f+\sigma_m(1-V_f)\\\\V_f=\frac{\sigma_{cd}-\sigma_{m}}{\frac{L*\tau_c}{D}-\sigma_{m}} \\\\Substituting:\\\\V_f=\frac{630*10^6-17.5*10^6}{\frac{0.0024*17*10^6}{0.00003} -17.5*10^6} \\\\V_f=0.456](https://tex.z-dn.net/?f=%5Csigma_%7Bcd%7D%3D%5Cfrac%7BL%2A%5Ctau_c%7D%7BD%7D%2AV_f%2B%5Csigma_m%281-V_f%29%5C%5C%5C%5CV_f%3D%5Cfrac%7B%5Csigma_%7Bcd%7D-%5Csigma_%7Bm%7D%7D%7B%5Cfrac%7BL%2A%5Ctau_c%7D%7BD%7D-%5Csigma_%7Bm%7D%7D%20%20%5C%5C%5C%5CSubstituting%3A%5C%5C%5C%5CV_f%3D%5Cfrac%7B630%2A10%5E6-17.5%2A10%5E6%7D%7B%5Cfrac%7B0.0024%2A17%2A10%5E6%7D%7B0.00003%7D%20-17.5%2A10%5E6%7D%20%5C%5C%5C%5CV_f%3D0.456)
Answer:
Observational Skills
Explanation:
Observing the area also known as scanning the scene
Answer:
battery life in year = 9 years and 48 days
Explanation:
given data
Battery Ampere-hours = 1.5
Pulse voltage = 2 V
Pulse width = 1.5 m sec
Pulse time period = 1 sec
Electrode heart resistance = 150 Ω
Current drain on the battery = 1.25 µA
to find out
battery life in years
solution
we get first here duty cycle that is express as
duty cycle =
...............1
duty cycle = 1.5 × ![10^{-3}](https://tex.z-dn.net/?f=10%5E%7B-3%7D)
and applied voltage will be
applied voltage = duty energy × voltage ...........2
applied voltage = 1.5 ×
× 2
applied voltage = 3 mV
so current will be
current =
................3
current = ![\frac{3}{150}](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B150%7D)
current = 20 µA
so net current will be
net current = 20 - 1.25
net current = 18.75 µA
so battery life will be
battery life = ![\frac{1.5}{18.75*10^{-6}}](https://tex.z-dn.net/?f=%5Cfrac%7B1.5%7D%7B18.75%2A10%5E%7B-6%7D%7D)
battery life = 80000 hours
battery life in year = ![\frac{80000}{8760}](https://tex.z-dn.net/?f=%5Cfrac%7B80000%7D%7B8760%7D)
battery life in year = 9.13 years
battery life in year = 9 years and 48 days