Answer: The total vehicle delay is
39sec/veh
Explanation: we shall define only the values that are important to this question, so that the solution will be very clear for your understanding.
Effective red time (r) = 25sec
Arrival rate (A) = 900veh/h = 0.25veh/sec
Departure rate (D) = 1800veh/h = 0.5veh/sec
STEP1: FIND THE TRAFFIC INTENSITY (p)
p = A ÷ D
p = 0.25 ÷ 0.5 = 0.5
STEP 2: FIND THE TOTAL VEHICLE DELAY AFTER ONE CYCLE
The total vehicle delay is how long it will take a vehicle to wait on the queue, before passing.
Dt = (A × r^2) ÷ 2(1 - p)
Dt = (0.25 × 25^2) ÷ 2(1 - 0.5)
Dt = 156.25 ÷ 4 = 39.0625
Therefore the total vehicle delay after one cycle is;
Dt = 39
Answer:
a. 2x/3
b. 8
Explanation:
fundamental period can be defined to mean that at after every period of 2π radians or 360° the value of graph is repeated. For such functions the fundamental period is the period after which they repeat themselves.
It van also be looked as The fundamental period of cos(θ) is 2π. That is (for example) cos(0) to cos(2π) represents one full period.
Please see attachment for the step by step solution.
Answer:
attached below
Explanation:
a) G(s) = 1 / s( s+2)(s + 4 )
Bode asymptotic magnitude and asymptotic phase plots
attached below
b) G(s) = (s+5)/(s+2)(s+4)
phase angles = tan^-1 w/s , -tan^-1 w/s , tan^-1 w/4
attached below
c) G(s)= (s+3)(s+5)/s(s+2)(s+4)
solution attached below
Answer:

Explanation:
From the information given:
Life requirement = 40 kh = 40 
Speed (N) = 520 rev/min
Reliability goal
= 0.9
Radial load
= 2600 lbf
To find C10 value by using the formula:

where;


The Weibull parameters include:



∴
Using the above formula:


![C_{10} = 3640 \times \bigg[\dfrac{1248}{0.9933481582}\bigg]^{\dfrac{3}{10}}](https://tex.z-dn.net/?f=C_%7B10%7D%20%3D%203640%20%5Ctimes%20%5Cbigg%5B%5Cdfrac%7B1248%7D%7B0.9933481582%7D%5Cbigg%5D%5E%7B%5Cdfrac%7B3%7D%7B10%7D%7D)

Recall that:
1 kN = 225 lbf
∴

