Answer:
Explanation:
Given conditions
1)The stress on the blade is 100 MPa
2)The yield strength of the blade is 175 MPa
3)The Young’s modulus for the blade is 50 GPa
4)The strain contributed by the primary creep regime (not including the initial elastic strain) was 0.25 % or 0.0025 strain, and this strain was realized in the first 4 hours.
5)The temperature of the blade is 800°C.
6)The formula for the creep rate in the steady-state regime is dε /dt = 1 x 10-5 σ4 exp (-2 eV/kT)
where: dε /dt is in cm/cm-hr σ is in MPa T is in Kelvink = 8.62 x 10-5 eV/K
Young Modulus, E = Stress,
/Strain, ∈
initial Strain, 


creep rate in the steady state


but Tinitial = 0


solving the above equation,
we get
Tfinal = 2459.82 hr
Answer:
2)
3) 
Explanation:
1) Expressing the Division as the summation of the quotient and the remainder
for
118, knowing it is originally a decimal form:
118:2=59 +(0), 59/2 =29 + 1, 29/2=14+1, 14/2=7+0, 7/2=3+1, 3/2=1+1, 1/2=0+1

2) 
Similarly, we'll start the process with the absolute value of -49 since we want the positive value of it. Then let's start the successive divisions till zero.
|-49|=49
49:2=24+1, 24:2=12+0,12:2=6+0,6:2=3+0,3:2=1+1,1:2=0+1
100011

3) 
The first step on that is dividing by 16, and then dividing their quotient again by 16, so on and adding their remainders. Simply put:

Answer:
0.19s
Explanation:
Queueing delay is the time a job waits in a queue before it can be executed. it is the difference in time betwen when the packet data reaches it destination and the time when it was executed.
Queueing delay =(N-1) L /2R
where N = no of packet =93
L = size of packet = 4MB
R = bandwidth = 1.4Gbps = 1×10⁹ bps
4 MB = 4194304 Bytes
(93 - 1)4194304 / 2× 10⁹
queueing delay =192937984 ×10⁻⁹
=0.19s
Following are the two sources of data:
- Internal Source. When data are collected from reports and records of the organisation itself, it is known as the internal source. ...
- External Source. When data are collected from outside the organisation, it is known as the external source.
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