Explanation:
The two types of furnaces used in steel production are:
<u>Basic oxygen furnace </u>
In basic oxygen furnace, iron is combined with the varying amounts of the steel scrap and also small amounts of the flux in the Blast Furnace. Lance is introduced in vessel and blows about 99% of the pure oxygen causing rise in temperature to about 1700°C. This temperature melts scrap and the impurities are oxidized and results in the liquid steel.
<u>Electric arc furnace</u>
Electric arc furnace reuses existing steel. Furnace is charged with the steel scrap. It operates on basis of electrical charge between the two electrodes providing heat for process. Power is supplied through electrodes placed in furnace, which produce arc of the electricity through scrap steel which raises temperature to about 1600˚C. This temperature melts scrap and the impurities can be removed through use of the fluxes and results in the liquid steel.
Answer:
k = 0.1118 per min
Explanation:
Assume;
Initial number of bacteria = N0
Number of bacteria IN 'T' time = Nt
So,
![Nt=N0e^{-kt}\\\\in\ 6.2 min\\\\\\frac{N0}{2}= N0e^{-k(6.2)}\\\\ln\frac{1}{2} = -k[6.2]](https://tex.z-dn.net/?f=Nt%3DN0e%5E%7B-kt%7D%5C%5C%5C%5Cin%5C%206.2%20min%5C%5C%5C%5C%5C%5Cfrac%7BN0%7D%7B2%7D%3D%20N0e%5E%7B-k%286.2%29%7D%5C%5C%5C%5Cln%5Cfrac%7B1%7D%7B2%7D%20%3D%20-k%5B6.2%5D)
k = 0.1118 per min
Technician B I did this in my lesson and the right answer is B
Solution :
Given :
k = 0.5 per day
Volume, V 
Now, input rate = output rate + KCV ------------- (1)
Input rate 
The output rate 
= ( 40 + 0.5 ) x C x 1000
Decay rate = KCV
∴
= 1.16 C mg/s
Substituting all values in (1)
C = 4.93 mg/L
