Answer:
0.89 cos (st +116.57°v)
Explanation:
Please kindly check attachment for the step by step solution of the given problem.
Answer:
The service level for each component must be 97.91%
Explanation:
If we want a 90% confidence of starting on time, that means we need
As the probability of each component being ready is independent from the others, that means that the probability of the 5 components being ready is equal to multiply each probability:
The probability of being ready on time is equal to the service level (in fraction), and all 5 are equal so we can say:
![0.9=(\mbox{service level(in fraction)})^5\\\\\sqrt[5]{0.9} =\mbox{service level(in fraction)}=0.9791\\\mbox{In percentage}: \mbox{service level (in fraction)}*100 = 97.91\%](https://tex.z-dn.net/?f=0.9%3D%28%5Cmbox%7Bservice%20level%28in%20fraction%29%7D%29%5E5%5C%5C%5C%5C%5Csqrt%5B5%5D%7B0.9%7D%20%3D%5Cmbox%7Bservice%20level%28in%20fraction%29%7D%3D0.9791%5C%5C%5Cmbox%7BIn%20percentage%7D%3A%20%5Cmbox%7Bservice%20level%20%28in%20fraction%29%7D%2A100%20%3D%2097.91%5C%25)
Answer: 112 + 19.27
Explanation:
Super elevation is an inward transverse slope provided through out the length of the horizontal curve which ends up serving as a counteract to the centrifugal force and checks tendency of overturning. It changes from infinite radius to radius of a transition curve.
Super curve elevation (e) = 4%
4/100= 0.04
e= V^2/gR
Make R the subject of the formula.
egR= V^2
R= V^2/eg
V= 45mph
=45 × 0.44704m/s
=20.1168m/s
g (force due to gravity) =9.81
Therefore,
R= (20.1168)^2/9.81 × 0.04
= 1031.31m
Tangent Length( T) = PI - PC
Tangent Length= 10875 - 10500
=375m
T= R Tan(I/2)
375= 1031.31 × Tan(I/2)
I= 39.96
Also,
L= πRI/180
= 719.27m
Station PT= Stat PC+ L
10500 + 719.27
=11219.27
=112 + 19.27
Answer:
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What soil horizon contains mostly clay silt and sand?
E Horizon - This eluviation (leaching) layer is light in color; this layer is beneath the A Horizon and above the B Horizon. It is made up mostly of sand and silt, having lost most of its minerals and clay as water drips through the soil (in the process of eluviation).
Answer: the final temperature of the steam 581.5 °C
Explanation:
Given that;
P₁ = 11 MPa
T₁ = 600°C
exit at; P₂= 5.5 MPa
Now from superheated steam table( p=11 MPa, T=600°C)
h₁ = 3615 kJ/kg
h₁ = h₂ ( by throttling process and adiabatic isentholpic )
from saturated steam table at; ( h= 3615 kJ/kg, P= 5.5 MPa )
Temperature = 581.5 °C
Therefore the final temperature of the steam 581.5 °C
