Which of these credit building options do you personally think is the easiest method that you can see yourself doing? Explain yo
1 answer:
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Answer:
33.429 N-m
Explanation:
Given :
Inclination angle of two shaft, α = 20°
Speed of shaft A, = 1000 rpm
Mass of flywheel, m = 30 kg
Radius of Gyration, k =100 mm
= 0.1 m
Now we know that for maximum velocity,
= 1064.1 rpm
Now we know
Mass of flywheel, m = 30 kg
Radius of Gyration, k =100 mm
= 0.1 m
Therefore moment of inertia of flywheel, I = m.
=30 X
= 0.3 kg-
Now torque on the output shaft
T₂ = I x ω
= 0.3 X 1064.2 rpm
=
= 33.429 N-m
Torque on the Shaft B is 33.429 N-m
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:
13.4 mm
Explanation:
Given data :
Load amplitude ( F ) = 22,000 N
factor of safety ( N )= 2.0
Take ( Fatigue limit stress amplitude for this alloy ) б = 310 MPa
<u>calculate the minimum allowable bar diameter to ensure that fatigue failure will not occur</u>
minimum allowable bar diameter = 13.4 * 10^-3 m ≈ 13.4 mm
<em>attached below is a detailed solution</em>
