Answer:
I = 1205.69 Lx
Explanation:
The irradiation or intensity of the solar radiation on the earth is maximum for the vertical fire, with a value I₀
I = I₀ sin θ
in this case with the initial data we can calculate the initial irradiance
I₀ =
I₀ = 1600 /sin 53
I₀ = 2003.42 lx
for when the angle is θ = 37º
I = 2003.42 sin 37
I = 1205.69 Lx
Answer: Coefficient= 0.35 per day
Explanation:
To find the bio degradation reaction rate coefficient, we have
k= ![\frac{(Cin)(Qin)-(Cout)(Qout)}{(Clagoon)V}](https://tex.z-dn.net/?f=%5Cfrac%7B%28Cin%29%28Qin%29-%28Cout%29%28Qout%29%7D%7B%28Clagoon%29V%7D)
Here, the C lagoon= 20 mg/L
Q in= Q out= 8640 m³/d
C in= 100 mg/L
C out= 20 mg/L
V= 10 ha* 1* 10
V= 10⁵ m³
So, k= ![\frac{8640*100-8640*20}{20*10^5}](https://tex.z-dn.net/?f=%5Cfrac%7B8640%2A100-8640%2A20%7D%7B20%2A10%5E5%7D)
k= 0.35 per day
Answer:
<em><u>1</u></em>
<em><u>1What is the output of 2 Input XNOR gate if both the inputs are same? Explanation: The output of 2 Input XNOR gate is 1 if both the inputs are same. The output of the XNOR gate is 1 if both the inputs are logic 0 or logic 1. This is why they are called as equality detector.</u></em>
Answer:
V = 0.5 m/s
Explanation:
given data:
width of channel = 4 m
depth of channel = 2 m
mass flow rate = 4000 kg/s = 4 m3/s
we know that mass flow rate is given as
![\dot{m}=\rho AV](https://tex.z-dn.net/?f=%5Cdot%7Bm%7D%3D%5Crho%20AV)
Putting all the value to get the velocity of the flow
![\frac{\dot{m}}{\rho A} = V](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cdot%7Bm%7D%7D%7B%5Crho%20A%7D%20%3D%20V)
![V = \frac{4000}{1000*4*2}](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B4000%7D%7B1000%2A4%2A2%7D)
V = 0.5 m/s
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)