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Mass The measure of the amount of matter in an object.
Answer:
Explanation:
volume of 20.9 N
= 20.9 / 11.5 m³
= 1.8174 m³
In one hour 1.8174 m³ flows
in one second volume flowing = 1.8174 / 60 x 60
= 5 x 10⁻⁴ m³
Rate of volume flow = 5 x 10⁻⁴ m³ / s .
Answer:
Answer D
Explanation:
They need to be checked monthly
Answer:
(N-1) × (L/2R) = (N-1)/2
Explanation:
let L is length of packet
R is rate
N is number of packets
then
first packet arrived with 0 delay
Second packet arrived at = L/R
Third packet arrived at = 2L/R
Nth packet arrived at = (n-1)L/R
Total queuing delay = L/R + 2L/R + ... + (n - 1)L/R = L(n - 1)/2R
Now
L / R = (1000) / (10^6 ) s = 1 ms
L/2R = 0.5 ms
average queuing delay for N packets = (N-1) * (L/2R) = (N-1)/2
the average queuing delay of a packet = 0 ( put N=1)
Answer:
The volume flow rate necessary to keep the temperature of the ethanol in the pipe below its flashpoint should be greater than 1.574m^3/s
Explanation:
Q = MCp(T2 - T1)
Q (quantity of heat) = Power (P) × time (t)
Density (D) = Mass (M)/Volume (V)
M = DV
Therefore, Pt = DVCp(T2 - T1)
V/t (volume flow rate) = P/DCp(T2 - T1)
P = 20kW = 20×1000W = 20,000W, D(rho) = 789kg/m^3, Cp = 2.44J/kgK, T2 = 16.6°C = 16.6+273K = 289.6K, T1 = 10°C = 10+273K = 283K
Volume flow rate = 20,000/789×2.44(289.6-283) = 20,000/789×2.44×6.6 = 1.574m^3/s (this is the volume flow rate at the flashpoint temperature)
The volume flow rate necessary to keep the ethanol below its flashpoint temperature should be greater than 1.574m^3/s
