Answer:
Pyrolysis
It is a biomass in anaerobic conditions results in the production of solid charcoal, liquid bio-oil and fuel gases. . Pyroolysis can be categorized into three groups depending on environment conditions, namely conventional pyrolysis, fast pyrolysis and flash pyrolysis.
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Answer:
see attached
Explanation:
Assuming flow is uniform across the cross section of the artery, the mass flow rate is the product of the volumetric flow rate and the density.
(5 cm³/s)(1.06 g/cm³) = 5.3 g/s
If we assume the blood splits evenly at the bifurcation, then the downstream mass flow rate in each artery is half that:
(5.3 g/s)/2 = 2.65 g/s
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The average velocity will be the ratio of volumetric flow rate to area. Upstream, that is ...
(5 cm³/s)/(π(0.25 cm)²) ≈ 25.5 cm/s
Downstream, we have half the volumetric flow and a smaller area.
(2.5 cm³/s)/(π(0.15 cm)²) ≈ 35.4 cm/s
Answer:
A place where code is written.
The amount of settlement that would occur at the end of 1.5 year and 5 year are 7.3 cm and 13.14 cm respectively.
<h3>How to determine the amount of settlement?</h3>
For a layer of 3.8 m thickness, we were given the following parameters:
U = 50% = 0.5.
Sc = 7.3 cm.
For Sf, we have:
Sf = Sc/U
Sf = 7.3/0.5
Sf = 14.6
Therefore, Sf for a layer of 38 m thickness is given by:
Sf = 14.6 × 38/3.8
Sf = 146 cm.
At 50%, the time for a layer of 3.8 m thickness is:
= 1.5 year.
At 50%, the time for a layer of 38 m thickness is:
= 1.5 × (38/3.8)²
= 150 years.
For the thickness of 38 m, U₂ is given by:
![\frac{U_1^2}{U_2^2} =\frac{(T_v)_1}{(T_v)_2} = \frac{t_1}{t_2} \\\\U_2^2 = U_1^2 \times [\frac{t_2}{t_1} ]\\\\U_2^2 = 0.5^2 \times [\frac{1.5}{150} ]\\\\U_2^2 = 0.25 \times 0.01\\\\U_2=\sqrt{0.0025} \\\\U_2=0.05](https://tex.z-dn.net/?f=%5Cfrac%7BU_1%5E2%7D%7BU_2%5E2%7D%20%3D%5Cfrac%7B%28T_v%29_1%7D%7B%28T_v%29_2%7D%20%3D%20%5Cfrac%7Bt_1%7D%7Bt_2%7D%20%5C%5C%5C%5CU_2%5E2%20%3D%20U_1%5E2%20%5Ctimes%20%5B%5Cfrac%7Bt_2%7D%7Bt_1%7D%20%5D%5C%5C%5C%5CU_2%5E2%20%3D%200.5%5E2%20%5Ctimes%20%5B%5Cfrac%7B1.5%7D%7B150%7D%20%5D%5C%5C%5C%5CU_2%5E2%20%3D%200.25%20%20%5Ctimes%200.01%5C%5C%5C%5CU_2%3D%5Csqrt%7B0.0025%7D%20%5C%5C%5C%5CU_2%3D0.05)
The new settlement after 1.5 year is:
Sc = U₂Sf
Sc = 0.05 × 146
Sc = 7.3 cm.
For time, t₂ = 5 year:
![U_2^2 = U_1^2 \times [\frac{t_2}{t_1} ]\\\\U_2^2 = 0.5^2 \times [\frac{5}{150} ]\\\\U_2^2 = 0.25 \times 0.03\\\\U_2=\sqrt{0.0075} \\\\U_2=0.09](https://tex.z-dn.net/?f=U_2%5E2%20%3D%20U_1%5E2%20%5Ctimes%20%5B%5Cfrac%7Bt_2%7D%7Bt_1%7D%20%5D%5C%5C%5C%5CU_2%5E2%20%3D%200.5%5E2%20%5Ctimes%20%5B%5Cfrac%7B5%7D%7B150%7D%20%5D%5C%5C%5C%5CU_2%5E2%20%3D%200.25%20%20%5Ctimes%200.03%5C%5C%5C%5CU_2%3D%5Csqrt%7B0.0075%7D%20%5C%5C%5C%5CU_2%3D0.09)
The new settlement after 5 year is:
Sc = U₂Sf
Sc = 0.09 × 146
Sc = 13.14 cm.
Read more on clay layer here: brainly.com/question/22238205