Answer: The answer is A. The company is trying to transfer intellectual capital to a knowledge management system
Answer:
The maximum water pressure at the discharge of the pump (exit) = 496 kPa
Explanation:
The equation expressing the relationship of the power input of a pump can be computed as:

where;
m = mass flow rate = 120 kg/min
the pressure at the inlet
= 96 kPa
the pressure at the exit
= ???
the pressure
= 1000 kg/m³
∴




400000 = P₂ - 96000
400000 + 96000 = P₂
P₂ = 496000 Pa
P₂ = 496 kPa
Thus, the maximum water pressure at the discharge of the pump (exit) = 496 kPa
Answer:
it is reducely very iloretable chance for a software engineer to give an end to this question
Answer & Explanation:
function Temprature
NYC=[33 33 18 29 40 55 19 22 32 37 58 54 51 52 45 41 45 39 36 45 33 18 19 19 28 34 44 21 23 30 39];
DEN=[39 48 61 39 14 37 43 38 46 39 55 46 46 39 54 45 52 52 62 45 62 40 25 57 60 57 20 32 50 48 28];
%AVERAGE CALCULATION AND ROUND TO NEAREST INT
avgNYC=round(mean(NYC));
avgDEN=round(mean(DEN));
fprintf('\nThe average temperature for the month of January in New York city is %g (F)',avgNYC);
fprintf('\nThe average temperature for the month of January in Denvar is %g (F)',avgDEN);
%part B
count=1;
NNYC=0;
NDEN=0;
while count<=length(NYC)
if NYC(count)>avgNYC
NNYC=NNYC+1;
end
if DEN(count)>avgDEN
NDEN=NDEN+1;
end
count=count+1;
end
fprintf('\nDuring %g days, the temprature in New York city was above the average',NNYC);
fprintf('\nDuring %g days, the temprature in Denvar was above the average',NDEN);
%part C
count=1;
highDen=0;
while count<=length(NYC)
if NYC(count)>DEN(count)
highDen=highDen+1;
end
count=count+1;
end
fprintf('\nDuring %g days, the temprature in Denver was higher than the temprature in New York city.\n',highDen);
end
%output
check the attachment for additional Information
Answer:
19063.6051 g
Explanation:
Pressure = Atmospheric pressure + Gauge Pressure
Atmospheric pressure = 97 kPa
Gauge pressure = 500 kPa
Total pressure = 500 + 97 kPa = 597 kPa
Also, P (kPa) = 1/101.325 P(atm)
Pressure = 5.89193 atm
Volume = 2.5 m³ = 2500 L ( As m³ = 1000 L)
Temperature = 28 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (28.2 + 273.15) K = 301.15 K
Using ideal gas equation as:
PV=nRT
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
5.89193 atm × 2500 L = n × 0.0821 L.atm/K.mol × 301.15 K
⇒n = 595.76 moles
Molar mass of oxygen gas = 31.9988 g/mol
Mass = Moles * Molar mass = 595.76 * 31.9988 g = 19063.6051 g