2. A
1 answer:
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Answer:
minimum sight distance = 699 ft
Explanation:
given data
road lane = 4 divided road
median width = 12 ft
grade road = 5%
solution
we take here time gap factor for minor road vehicle when enter to major road from table
time gap = 8.1 sec
and for median width of 12 ft
time gap = 8.2 + 0.7 ( 1 + )
time gap = 9.5 second
so minimum sight distance will be
minimum sight distance = 1.47 × design speed × time gap
minimum sight distance = 1.47 × 50 × 9.5
minimum sight distance = 699 ft
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
Answer: The energy system related to your question is missing attached below is the energy system.
answer:
a) Work done = Net heat transfer
Q1 - Q2 + Q + W = 0
b) rate of work input ( W ) = 6.88 kW
Explanation:
Assuming CPair = 1.005 KJ/Kg/K
<u>Write the First law balance around the system and rate of work input to the system</u>
First law balance ( thermodynamics ) :
Work done = Net heat transfer
Q1 - Q2 + Q + W = 0 ---- ( 1 )
rate of work input into the system
W = Q2 - Q1 - Q -------- ( 2 )
where : Q2 = mCp T = 1.65 * 1.005 * 293 = 485.86 Kw
Q2 = mCp T = 1.65 * 1.005 * 308 = 510.74 Kw
Q = 18 Kw
Insert values into equation 2 above
W = 6.88 Kw
