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jeka94
3 years ago
6

Two routes connect an origin and a destination. Routes 1 and 2 have performance functions t1 = 2 + X1 and t2 = 1 + X2, where the

t's are in minutes and the x's are in thousands of vehicles per hour. The travel times on the routes are known to be in user equilibrium. If an observation for route 1 finds that the gaps between 30% of the vehicles are less than 6 seconds. Estimate the volume and average travel times for the two routes
Engineering
1 answer:
Musya8 [376]3 years ago
7 0

Solution :

Given

$t_1=2+x_1$

$t_2=1+x_2$

Now,

$P(h

$0.4=1-P(h \geq5)$

$0.6=P(h \geq5)$

$0.6= e^{\frac{-x_1 5}{3600}}$

Therefore,   $x_1=368 \ veh/h$

                        $=\frac{368}{1000} = 0.368$

Given,   $t_1=2+x_1$

                 = 2 + 0.368

                 = 2.368 min

At user equilibrium, $t_2=t_1$

∴  $t_2$ = 2.368 min

$t_2=1+x_2$

$2.368=1+x_2$

$x_2 = 1.368$

$x_2 = 1.368 \times 1000$

    = 1368 veh/h

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Create a project named CarDealer that contains a Form for an automobile dealer. Include options for at least three car models. A
PIT_PIT [208]

Answer:

/****************** The code for a form application that uses radioButtons * *and displays new form based on the user selection *****************/

Using system;

Using system windows forms;

namespace CarDealer

{

Public partial class Form1 : Form

{

public Form1()

{

//default constructor to initialize components

InitializeComponent();

}

//when user clicks on the details button

private void detail_Click

(Object sender, EventArgs e)

{

if (model1,Checked)

{

Forms2 f = new Form2()

f.ShowDialog()

model1.Checked = false

}

if (model2.Checked)

{

Form3 f = new Form3();

f.ShowDialog();

model2.Checked = false;

}

if (model3.Checked)

{

Form4 f = new Form4();

f.ShowDialog();

model3.Checked = false;

}

}

}

}

Explanation:

Program plan

- Design form: Place label controls with text as select car model and one empty label control to display total price. Change front type and size from each other labels properties window.

- Add three radio buttons with text Renault Kwit, Tata Tiago, Mahindra KUV 100.

- Add button with text View Details.

- Add 3 new windows form from project menu for each car model.

- Change form name and respective car models.

- Place picturebox control to each form by using image property add .jpeg image and place label control containing price of each model to each form

- When user select car model and click on view details, new form containing car model details is displayed.

Form Design is attached below

7 0
2 years ago
A pitfall cited in Section 1.10 is expecting to improve the overall performance of a computer by improving only one aspect of th
Oxana [17]

Answer:

a) For this case the new time to run the FP operation would be reduced 20% so that means 100-20% =80% from the original time

(1-0.2)*70 s =56s

The reduction on this case is 70-56 s=14s

And since the new total time would be given by 250-14=236 s

b) For this case the total time is reduced 20%  so that means that the new total time would be (1-0.2)=0.8 times the original total time (1-0.2) *250s =200 s

The original time for INT operations is calculated as:

250 = 70+85+40 +t_{INT}

t_{INT}=55s

For this part the only time that was changed is assumed the INT operations so then:

200 = 70+85+40 \Delta t_{INT}

And then: \Delta t_{INT}= 200-70-85-40=5 s

c) A reduction of the total time implies that the total time would be 205 s from the results above. And the time for FP is 70, for L/S is 85 and for INT operations is 55 s, so then if we add 70+85+55=210s, we see that 210>205 so then we cannot reduce the total time 20% just reducing the branch intructions.

Explanation:

From the info given we know that a computer running a program that requires 250 s, with 70 s spent executing FP instructions, 85 s executed L/S instructions and 40 s spent executing branch instructions.

Part 1

For this case the new time to run the FP operation would be reduced 20% so that means 100-20% =80% from the original time

(1-0.2)*70 s =56s

The reduction on this case is 70-56 s=14s

And since the new total time would be given by 250-14=236 s

Part 2

For this case the total time is reduced 20%  so that means that the new total time would be (1-0.2)=0.8 times the original total time (1-0.2) *250s =200 s

The original time for INT operations is calculated as:

250 = 70+85+40 +t_{INT}

t_{INT}=55s

For this part the only time that was changed is assumed the INT operations so then:

200 = 70+85+40 \Delta t_{INT}

And then: \Delta t_{INT}= 200-70-85-40=5 s

And we can quantify the decrease using the relative change:

\% Change = \frac{5s}{55 s} *100 = 9.09\% of reduction

Part 3

A reduction of the total time implies that the total time would be 205 s from the results above. And the time for FP is 70, for L/S is 85 and for INT operations is 55 s, so then if we add 70+85+55=210s, we see that 210>205 so then we cannot reduce the total time 20% just reducing the branch intructions.

8 0
3 years ago
A smooth ceramic sphere (SG 5 2.6) is immersed in a fl ow of water at 208C and 25 cm/s. What is the sphere diameter if it is enc
Aleks [24]

Answer:

a. 4\mu m

b. 1 m

Explanation:

According to the question, the data is as follows

The Density of water at 20 degrees celcius is 1000 kg/m^3

Viscosity is 0.001kg/m/.s

Velocity V = 25 cm/s

V = 0.25 m/s

Now

a. The creeping motion is

As we know that

Reynold Number = (Density of water × V × d) ÷ (Viscosity)

1 = (1,000 × 0.25 × d) ÷ 0.0001

d = (1 × 0.001) ÷ (1,000 × 0.25)

= 4E - 06^m

= 4\mu m

b. Now the sphere diameter is

Reynold Number = (Density of water × V × d) ÷ (Viscosity)

250,000 = (1,000 × 0.25 × d) ÷ 0.0001

d = (250,000 × 0.001) ÷ (1,000 × 0.25)

= 1 m

6 0
3 years ago
What is the value of the work interaction in this process?
Cloud [144]

Answer:

The answer is "-121\  \frac{KJ}{Kg}".

Explanation:

Please find the correct question in the attachment file.

using formula:

\to W=-P_1V_1+P_2V_2 \\\\When \\\\\to W= \frac{P_1V_1-P_2V_2}{n-1}\ \   or \ \  \frac{RT_1 -RT_2}{n-1}\\\\

W =\frac{R(T_1 -T_2)}{n-1}\\\\

    =\frac{0.287(25 -237)}{1.5-1}\\\\=\frac{0.287(-212)}{0.5}\\\\=\frac{-60.844}{0.5}\\\\=-121.688 \frac{KJ}{Kg}\\\\=-121 \frac{KJ}{Kg}\\\\

7 0
3 years ago
Engineers create a new metal that is stronger than steel but much lighter. This material is also significantly cheaper than what
zysi [14]

The best step for the engineers to make next is option D. Begin to design an airplane using this metal.

<h3>What is the metallic is plane parts?</h3>

Aluminum and its alloys are nevertheless very famous uncooked substances for the production of business planes, because of their excessive electricity at exceedingly low density. Currently, excessive-electricity alloy 7075, which includes copper, magnesium and zinc, is the only used predominantly withinside the plane industry.

The solution is D, due to the fact even as it's far crucial to marketplace the fabric and ensure humans are inquisitive about buying, they first want to attempt to layout aircraft the usage of this fabric. There isn't anyt any use promoting an aircraft constituted of this material_ if a aircraft can not be built.

Read more about the aircraft:

brainly.com/question/5055463

#SPJ1

7 0
2 years ago
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