Answer:
4.5kg/min
Explanation:
Given parameters

if we take
The mass flow rate of the second stream = 
The mass flow rate of mixed exit stream = 
Now from mass conservation


The temperature of the mixed exit stream given as

Therefore the mass flow rate of second stream will be 4.5 kg/min.
Answer:
Resistor B
Explanation:
Since resistance is the opposition to the flow of current in a circuit,
first let assume the two resistors are connected in parallel to the voltage, recall that when connection is in parallel, the different amount of current pass through the resistors depending on the value with the small resistor having a lower resistance effect hence higher current will pass through
The energy dissipated in each resistor can be calculated as
.
from the formula we can conclude that the energy value will be higher for the resistor with small resistance value. hence more heating effect which will cause it to be warm.
Also when connected individually the current flow from the voltage source will pass through the resistor which when we calculate the energy dissipated, the resistor with smaller value will be higher because it will draw more current which will in turn lead to a heating effect and cause the resistor to be warm. Hence we can conclude that the resistance B has greatest resistance value.
Answer:
See the attached file for the design.
Explanation:
Find attached for the explanation.
Answer:

Explanation:
Given data:
Ammonia Nitrogen 30 mg/L
pH = 8.5
-log[H +] = 8.5
[H +] = 10^{-8.5}

Rate constant is given as
...........1

Total ammonia as NItrogen is given as 30 mg/l
![\%NH_4^{+} = \frac{ [NH_4] \times 100}{[NH_4^{+}] + [NH_3]}](https://tex.z-dn.net/?f=%5C%25NH_4%5E%7B%2B%7D%20%3D%20%5Cfrac%7B%20%5BNH_4%5D%20%5Ctimes%20100%7D%7B%5BNH_4%5E%7B%2B%7D%5D%20%2B%20%5BNH_3%5D%7D)
= 
.....2
from equation 1 we have
{10^{8.5}}
plug this value in equation 2 we get

Total ammonia as N = 30 mg/lt

Answer:
10 ft or 3.048 m (for unqualified person)
1.09 ft or 0.33 m (for qualified person)
Explanation:
OSHA has prescribed "Minimum Approach Distance - MAD" to ensure safety of personnel working with electrical power lines.
For a unqualified person:
The minimum approach distance is 10 ft or 3.048 m
For a qualified person:
For a voltage in the range of 300V to 750V (Phase to Ground or Phase to Phase) the minimum approach distance is 1.09 ft or 0.33 m
Since 600V lies in the above range hence it is applicable on 600V.
OSHA has developed a full list of minimum approach distance according to the level of voltages (50V to 72000V)