Answer:
Evaluate the properties of water at 290 K = 272.73Kw
Explanation:
kindly check the attached file below
Answer:
938.7 milliseconds
Explanation:
Since the transmission rate is in bits, we will need to convert the packet size to Bits.
1 bytes = 8 bits
1 MiB = 2^20 bytes = 8 × 2^20 bits
5 MiB = 5 × 8 × 2^20 bits.
The formula for queueing delay of <em>n-th</em> packet is : (n - 1) × L/R
where L : packet size = 5 × 8 × 2^20 bits, n: packet number = 48 and R : transmission rate = 2.1 Gbps = 2.1 × 10^9 bits per second.
Therefore queueing delay for 48th packet = ( (48-1) ×5 × 8 × 2^20)/2.1 × 10^9
queueing delay for 48th packet = (47 ×40× 2^20)/2.1 × 10^9
queueing delay for 48th packet = 0.938725181 seconds
queueing delay for 48th packet = 938.725181 milliseconds = 938.7 milliseconds
Answer:
To* = 1074.9k ( stagnation temperature )
Po* = 0.467 MPa ( stagnation pressure )
Explanation:
Given data:
Po1 = 1.6 MPa
To = 430°c + 273 = 703 k
Mach number at inlet ( M1 )= 3
Mach number at exit ( M2 ) = 1
at M1 = 3
To1 / To* = 0.65398 ( value gotten from Rayleigh's flow gas tables )
therefore To* = 1074.9k
also
Po1 / Po* = 3.42445 ( value gotten from Rayleigh's flow gas tables )
therefore Po* = 0.467 MPa
at M2 = 1
To2 = To* = 1074.9k
Po2 = Po* = 0.467 MPa
using both Rayleigh's flow table and Isentropic Tables the values of stagnation temperature and stagnation pressure are the same at the exit
Answer:
C: Viscosity, the resistance to flow that fluids exhibit
Explanation:
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