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Masja [62]
3 years ago
8

The elastic settlement of an isolated single pile under a working load similar to that of piles in the group it represents, is p

redicted to be 0.25 inches. What is the expected settlement for the pile group given the following information?
1.Group: 16 piles in a 4x4 group
2.Pile Diameter: 12 inches
3.Pile Center to Center Spacing: 3 feet

Engineering
1 answer:
Ganezh [65]3 years ago
3 0

Answer:

The expected settlement for the pile group using the given information is 19.92mm or 0.79 inch

Explanation:

In this question, we are asked to calculate the expected settlement for the pole group given some information.

Please check attachment for complete solution and step by step explanation

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Assume a program requires the execution of 50 x 10^6 FP instructions, 110 x 10^6 INT instructions, 80 x 10^6 Load/Store (L/S) in
Svetradugi [14.3K]

Answer:

We can not improve CPI of FP instructions when we run the program two times faster because it would be negative.

Explanation:

Processor clock rate = 2 GHz

Execution Time =   ∑  (\frac{Clock cyles}{Clock rate})

Clock cycles can be determined using following formula

Clock cycles = (CPI_{FP} x  No. FP instructions )+ ( CPI_{INT} x No. INT instructions) + ( CPI_{L/S}  x No. L/S instructions ) + ( CPI_{branch} x No. branch instructions)

Clock cycles = ( 50 x 10^{6} x 1) + (  110 x 10^{6} x 1) + ( 80 x 10^{6} x 4) + ( 16 x 10^{6} x 2)

Clock cycles = 512 x 10⁶

So,Initial Execution time for FP instructions is,

    = \frac{512(10^{6}) }{2(10^{9}) }

 Initial execution Time =  256 x 10⁻³

For 16 processors ,

clock cycle = 512 x 10⁶

Execution Time = 256 x 10⁻³

To run the program two times faster, half the number of clock cycles

(\frac{Clockcycles}{2} )=   (CPI_{FP} x  No. FP instructions )+ ( CPI_{INT} x No. INT instructions) + ( CPI_{L/S}  x No. L/S instructions ) + ( CPI_{branch} x No. branch instructions)

CPI_{FP improved} x No. FP instructions  =  (\frac{Clockcycles}{2} ) -[ ( CPI_{INT} x No. INT instructions) + ( CPI_{L/S}  x No. L/S instructions ) + ( CPI_{branch} x No. branch instructions)]

CPI_{FP improved} x 50 x 10^{6}  = ( \frac{512(10)^{6} }{2} ) - [ (  110 x 10^{6} x 1) + ( 80 x 10^{6} x 4) + ( 16 x 10^{6} x 2)]

CPI_{FP improved} x 50 x 10^{6}  =  - 206 x 10^{6}

CPI_{FP improved}  = - 206 x 10^{6} / 50 x 10^{6}

CPI_{FP improved} = - 4.12 < 0

3 0
4 years ago
Print reading for industry unit 9 review questions
RideAnS [48]
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4 0
3 years ago
Read 2 more answers
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
3 years ago
A 3.5-m3 rigid tank initially contains air whose density is 2 kg/m3 . The tank is connected to a high-pressure supply line throu
Mumz [18]

Answer:

Explanation:

First, we find the mass of the air originally in the tank.

Density is given as mass divided by volume. It is given as:

Density = \frac{mass}{volume}

Therefore, mass is:

mass = denisty *volume

Density of air = 2 kg/m^3; Volume of the tank =  3.5 m^3

=> Mass = 3.5 * 2 = 7 kg

The mass of the air initially in the tank is 7 kg.

After air is allowed to enter, the mass changes.

New density = 6.5 kg/m^3

The new mass will be:

Mass = 6.5 * 3.5 = 22.75 kg

We can now find the mass of air that has entered the tank:

Mass of air that entered tank = New mass of air - Original mass of air

M = 22.75 - 7.0 = 15.75 kg

The mass of air that entered the tank is 15.75 kg.

6 0
3 years ago
Vending machine controller (adapted from Katz, "Contemporary Logic Design") Design and implement a finite state machine that con
babunello [35]

Answer:

Check the explanation

Explanation:

A vending machine controller is that type of machine that comes with a single serial port on the same chip as the programmable processor. The controller comprises of a port arbitrator that selectively attaches or links one of a number of serially communicating devices to this single serial port.

Kindly check the attached image to get the step by step explanation to the above question.

6 0
3 years ago
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