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2 years ago

9

Vehicles arrive at a single toll booth beginning at 8:00 A.M. They arrive and depart according to a uniform deterministic distri

bution. However, the toll booth does not open until 8:10 A.M. The average arrival rate is 8 veh/min, and the average departure rate is 10 veh/min. Assuming D/D/1 queuing, when does the initial queue clear and what are the total delay, the average delay per vehicle, longest queue length (in vehicles), and the wait time of the 100th vehicle to arrive (assuming first-in-first-out)?

1 answer:

jok3333 [9.3K]2 years ago

6 0

Answer:

Explanation:

answers attached bellow

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The fatigue data for a brass alloy are given as follows: Stress Amplitude (MPa) Cycles to Failure 170 3.7 × 104 148 1.0 × 105 13

prohojiy [21]

Answer:

i) S–N plot is attached

ii) fatigue strength = 100 MPa

iii) fatigue life = 5.62 x 10^(5) cycles

Explanation:

i) I have attached the S–N plot (stress amplitude versus logarithm of cycles to failure)

ii) The question says we should find the fatigue strength at 4 × 10^(6) cycles.

So let's find the log of this and trace it on the graph attached.

Log(4 × 10^(6)) = 6.6

From the graph attached, at log of cycle value of 6.6, the fatigue strength is approximately 100 MPa

iii) The question says we should find the fatigue life for 120 MPa.

Thus, from the graph, at stress amplitude of 120 MPa, the log of cycles is approximately 5.75.

Thus,the fatigue life will be the inverse log of 5.75.

Thus, fatigue life = 10^(5.75)

Fatigue life = 5.62 x 10^(5)

8 0

2 years ago

An annealed copper strip of 228 mm wide and 25 mm thick being rolled to a thickness of 20 mm, in one pass. The roll radius is 30

sdas [7]

Answer:

The roll force is 1.59 MN

The power required in this operation is 644.96 kW

Explanation:

Given;

width of the annealed copper, w = 228 m

thickness of the copper, h₀ = 25 mm

final thickness, hf = 20 mm

roll radius, R = 300 mm

The roll force is given by;

$F = LwY_{avg}$

where;

w is the width of the annealed copper

$Y_{avg}$ is average true stress of the strip in the roll gap

L is length of arc in contact, and for frictionless situation it is given as;

$L = \sqrt{R(h_o-h_f)} \\\\L = \sqrt{300(25-20)}\\\\L = 38.73 \ mm$

Now, determine the average true stress, $Y_{avg}$ , for the annealed copper;

The absolute value of the true strain, ε = ln(25/20)

ε = 0.223

from true stress vs true strain graph; at true strain of 0.223, the true stress is 280 MPa.

Then, the average true stress = ¹/₂(280 MPa.) = 180 MPa

Finally determine the roll force;

$F = LwY_{avg}$

$F = (\frac{38.73 }{1000})(\frac{228}{1000})*180 \ MPa\\\\F = 1.59 \ MN$

The power required in this operation is given by;

$P = \frac{2\pi FLN}{60}\\\\P = \frac{2\pi (1.59*10^6)(0.03873)(100)}{60}\\\\P = 644955.2 \ W\\\\P = 644.96 \ kW$

5 0

2 years ago

Use the predicate specifications(x, y): x beats yF (x): x is an (American) football teamQ(x, y): x is quarterback of yL(x,y): x

attashe74 [19]

Answer:

a) ∀y∃x(Q(x, y))

b) (B(Jayhawks, W ildcats)→¬∀y(L(Jayhawks, y)))

c) ∃x(B(Wildcats, x) ∧ B(x, Jayhawks))

Explanation:

a) The statement can be rewritten as "For all football teams, there exists a quarterback" which is written in logical symbols.

b) The statement is an implication and thus have a premise and a conclusion. The premise states "Jayhawks beat the Wildcats" which is translated using B(x, y). The conclusion can be rewritten as "It is not the case that Jayhawks lose to all football teams".

c) The statement is a simple conjunction which can be written as "There exists a team x such that the Wildcats beats x and x beats Jayhawks"

7 0

2 years ago

A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 3.5 mm (0.14 in

RideAnS [48]

To resolve this problem we have,

$R=3.5mm\\F_f1=950N\\L_1=50mm\\b=12mm\\L_2=40mm$

$F_{f2}$ is unknown.

With these dates we can calculate the Flexural strenght of the specimen,

$\sigma{fs}=\frac{F_{f1}L}{\pi R^3}\\\sigma{fs}=\frac{(950)(50*10^{-3})}{\pi 3.5*10^{-3}}\\\sigma{fs}=352.65Mpa$

After that, we can calculate the flexural strenght for the square cross section using the previously value.

$\sigma{fs}=\frac{F_{f2}L}{\pi R^3}\\(352.65*10^6)=\frac{3Ff(40*10^{-3})}{2(12*10^{-10})}\\F_{f2}=\frac{352.65*10^6}{34722.22}\\F_{f2}=10156.32N\\F_{f2}=10.2kN$

6 0

2 years ago

The extruder head in a fused- deposition modeling setup has a diameter of 1.25 mm (0.05 in) and produces layers that are 0.25mm

Angelina_Jolie [31]

Answer:

The time taken will be "1 hour 51 min". The further explanation is given below.

Explanation:

The given values are:

Number of required layers:

= $\frac{38}{0.25}$

= $152 \ layers$

Diameter (d):

= 1.25 mm

Velocity (v):

= 40 mm/s

Now,

The area of one layer will be:

= $38\times 38 \ mm^2$

= $1444 \ mm^2$

The area covered every \second will be:

= $d\times v$

= $1.25\times 40$

= $50 \ mm^2$

The time required to deposit one layer will be:

= $\frac{1444}{50}$

= $28.88 \ sec$

The time required for one layer will be:

= $15 \ sec$

∴ Total times required for one layer will be:

= $15+28.88$

= $43.88 \ sec$

So,

Number of layers = 152

Therefore,

Total time will be:

= $152\times 43.88$

= $6669.76 \ sec$

= $1 \ hour \ 51 \ min$

6 0

2 years ago