Answer:
32000 bits/seconds
Explanation:
Given that :
there are 16 signal combinations (states) = 2⁴
bits n = 4
and a baud rate (number of signals/second) = 8000/second
Therefore; the number of bits per seconds can be calculated as follows:
Number of bits per seconds = bits n × number of signal per seconds
Number of bits per seconds = 4 × 8000/second
Number of bits per seconds = 32000 bits/seconds
Answer:
The pressure drop across the pipe also reduces by half of its initial value if the viscosity of the fluid reduces by half of its original value.
Explanation:
For a fully developed laminar flow in a circular pipe, the flowrate (volumetric) is given by the Hagen-Poiseulle's equation.
Q = π(ΔPR⁴/8μL)
where Q = volumetric flowrate
ΔP = Pressure drop across the pipe
μ = fluid viscosity
L = pipe length
If all the other parameters are kept constant, the pressure drop across the circular pipe is directly proportional to the viscosity of the fluid flowing in the pipe
ΔP = μ(8QL/πR⁴)
ΔP = Kμ
K = (8QL/πR⁴) = constant (for this question)
ΔP = Kμ
K = (ΔP/μ)
So, if the viscosity is halved, the new viscosity (μ₁) will be half of the original viscosity (μ).
μ₁ = (μ/2)
The new pressure drop (ΔP₁) is then
ΔP₁ = Kμ₁ = K(μ/2)
Recall,
K = (ΔP/μ)
ΔP₁ = K(μ/2) = (ΔP/μ) × (μ/2) = (ΔP/2)
Hence, the pressure drop across the pipe also reduces by half of its initial value if the viscosity of the fluid reduces by half of its value.
Hope this Helps!!!
Answer:
A fluid flowing along a flat plate will stick to it at the point of contact
Explanation:
and this is known as the no-slip condition. ... This is the precise reason why shear stress in a fluid can also be interpreted as the flux of momentum.
Answer:
b) The null hypothesis should be rejected.
Explanation:
The null hypothesis is that the mean shear strength of spot welds is at least
3.1 MPa
H0: u ≥3.1 MPa against the claim Ha: u< 3.1 MPa
The alternate hypothesis is that the mean shear strength of spot welds is less than 3.1 MPa.
This is one tailed test
The critical region Z(0.05) < ± 1.645
The Sample mean= x`= 3.07
The number of welds= n= 15
Standard Deviation= s= 0.069
Applying z test
z= x`-u/s/√n
z= 3.07-3.1/0.069/√15
z= -0.03/0.0178
z= -1.68
As the calculated z= -1.68 falls in the critical region Z(0.05) < ± 1.645 the null hypothesis is rejected and the alternate hypothesis is accepted that the mean shear strength of spot welds is less than 3.1 MPa
Answer:
Explanation:
The Turnaround time is the amount of time that elapses between the job arriving and completing. We assume that all jobs arrive at time 0, the turnaround time will simply be the time that they complete.
Round Robin:
we assume that the time quantum of the scheduler is 1 second.The table below gives a break down of which jobs will be processed during each time quantum. A asterisk(*) indicates that the job completes during that quantum.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
A B C D E A B C* D E A B D E A B D* E A B E A B* E A E A E* A A
C* = 8
D*=17
B*=23
E*=28
AVERAGE TURNAROUND = (8+17+23+28+30)/5 =106/5 = 21.2 MINUTES
B) PRIORITY SCHEDULING:
1-6 7-14 15-24 25-26 27-30
B E A C D
AVERAGETURNAROUND =(6+14+24+26+30)/5 = 100/5 = 20 MINUTES.
C)FCFS
1-10 11-16 17-18 19-22 23-30
A B C D E
AVERAGE TURNAROUND =(10+16+18+22+30)/5 = 96/5=19.2 MINTUES
D)SJF
1-2 3-6 7-12 13-20 21-30
C D B E A
AVERAGE TURNAROUND - (2+6+12+20+30)/5 =70/5 =14 MINUTES.
