Answer:
value of the rate constant is 80 mg/L-hr
Explanation:
given data
concentration of compound Cao = 90 mg/l
Ca = 10 mg/L
time = 1 hour
solution
we use here zero order reaction rate flow
-rA = K Ca
= K
-d Ca = k dt
now we will integrate it on the both side by Cao to ca we get
solve it we get
cao - Ca = K t
put here value and we get K
90 - 10 = K 1
K = 80 mg/L-hr
so value of the rate constant is 80 mg/L-hr
Serviceable wheel bearings can be repacked by removing the dust cap and the cotter pin must be replaced with a new one every time it is removed. Therefore, both Tech A and B are correct.
Sealed bearing assemblies are typically prefilled with lubricant. They also make a more reliable and consistent installation process due to the fact that every into comes preset at the proper clearance.
Furthermore, the serviceable wheel bearings can be repacked by removing the dust cap, filling it with grease, and reinstalling it. Also, cotter pin must be replaced with a new one every time it is removed.
Therefore, both tech A and B are correct.
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Answer:
Hello your question is incomplete attached below is the complete question
answer:
Considering Laminar flow
Q ( heat ) will be independent of diameter
Considering Turbulent flow
The heat transfer will increase with decreasing "dia" for the turbulent
heat transfer = f(d^-0.8 )
Explanation:
attached below is the detailed solution
Considering Laminar flow
Q ( heat ) will be independent of diameter
Considering Turbulent flow
The heat transfer will increase with decreasing "dia" for the turbulent
heat transfer = f(d^-0.8 )
The carriage handwheel is used to manually position and/or hand feed the carriage in the longitudinal or Z axis.
IamSugarBee
Answer:
magnitude of thrust uis 11061.65 lb/ft
location is 5 ft from bottom
Explanation:
Given data:
Height of vertical wall is 15 ft
OCR is 1.5
![\phi = 33^o](https://tex.z-dn.net/?f=%5Cphi%20%3D%2033%5Eo)
saturated uit weight![\gamma_{sat} = 115.0 lb/ft^3](https://tex.z-dn.net/?f=%20%5Cgamma_%7Bsat%7D%20%3D%20115.0%20lb%2Fft%5E3)
coeeficent of earth pressure ![K_o](https://tex.z-dn.net/?f=K_o)
![K_o = 1 -sin \phi](https://tex.z-dn.net/?f=K_o%20%3D%201%20-sin%20%5Cphi)
= 1 - sin 33 = 0.455
for over consolidate
![K_{con} = K_o \times OCR](https://tex.z-dn.net/?f=K_%7Bcon%7D%20%3D%20K_o%20%5Ctimes%20OCR)
![= 0.455 \times 1.5 = 0.683](https://tex.z-dn.net/?f=%20%3D%200.455%20%5Ctimes%201.5%20%3D%200.683)
Pressure at bottom of wall is
![P =K_{con} \times (\gamma_{sat} - \gamma_{w}) + \gamma_w \times H](https://tex.z-dn.net/?f=P%20%3DK_%7Bcon%7D%20%5Ctimes%20%28%5Cgamma_%7Bsat%7D%20-%20%5Cgamma_%7Bw%7D%29%20%2B%20%5Cgamma_w%20%5Ctimes%20H)
![= 0.683 \times (115 - 62.4) \times 15 + 62.4 \times 15](https://tex.z-dn.net/?f=%3D%200.683%20%5Ctimes%20%28115%20-%2062.4%29%20%5Ctimes%2015%20%2B%2062.4%20%5Ctimes%2015)
P = 1474.88 lb/ft^3
Magnitude pf thrust is
![F= \frac{1}{2} PH](https://tex.z-dn.net/?f=F%3D%20%5Cfrac%7B1%7D%7B2%7D%20PH)
![=\frac{1}{2} 1474.88\times 15 = 11061.65 lb/ft](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B2%7D%201474.88%5Ctimes%2015%20%3D%2011061.65%20lb%2Fft)
the location must H/3 from bottom so
![x = \frac{15}{3} = 5 ft](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B15%7D%7B3%7D%20%3D%205%20ft)