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Luda [366]
4 years ago
8

A 500 tonne train collides with a station emergency buffer stop while travelling at 2 ms-1. The buffer stop has a mass of 1.5 to

nnes and is designed to act as a spring with stiffness K = 1 MN/m. (i) Estimate the extent that the buffer stop spring is compressed (Hint: consider the velocity of the train and buffer system after the collision – is there an easy approximation?) [~1.4 m]; (ii) Estimate the mean deceleration experienced by the train as it is halted. Is this deceleration likely to be a danger to the passengers? [~1.4 ms-2]
Engineering
1 answer:
Advocard [28]4 years ago
3 0

Answer:

a) Δx=1.41 m; b) a=1.41 m/(s^2)

Explanation:

a) To solve this problem, we should consider two processes:

1) inelastic collision between the train and the buffer;

2) motion of the pair to reach complete stop.

As the collision between the train and the buffer is inelastic, we have some energy wasted. However, the velocity after the collision can be determined using Momentum Conservation Law:

mtVt=(mt+mb)*V, where mt- mass of the train, mb- mass of the buffer, Vt- initial velocity of the train, V- velocity after the collision.

V=mtVt/(mt+mb)=1.994 m/s.

After train and buffer have collided, the motion can be described using Energy Conservation Law, where kinetic energy of the buffer-train couple is fully transmitted to the potential energy of elastic deformation of the spring. From this idea, the following equation can be derived:

\frac{(mt+mb)V^{2} }{2}=\frac{Kx^{2} }{2} , where x- displacement of the spring, K- spring stiffness.

From this equation, we can find: x=\sqrt{\frac{(mt+mb)V^{2} }{k} } =1.41 m

Regarding an approximation, we can assume, that the buffer is mass-less in comparison with the train and this assumption will yield very small error. If we consider mass-less buffer, there is no need to use mass of the buffer in calculations and there is no need to consider collision and finding velocity after the collision. To support this assumption, we can compare initial velocity of the train and final during the collision- train slowed down from 2m/s to 1.994 m/s- the change is insignificant, so we can directly apply Energy conservation law to the train-spring system and avoid buffer's mass.

b) To find mean acceleration, we can use the following known values: initial velocity is 1.994 m/s and the traveled length is 1.41 m. As we are looking for the mean value, we can assume, that the acceleration was constant and the following equation can be used: x=\frac{V^{2} }{2a}, where a- mean acceleration, V- velocity of the train and x- traveled distance.

From this equation, we can find acceleration as: a=\frac{V^{2} }{2x} =1.41 \frac{m}{s^{2} }

This acceleration is significantly smaller, than g (gravitational acceleration), so it is not dangerous to the passengers.

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Fantom [35]

Summary

Students learn about the variety of materials used by engineers in the design and construction of modern bridges. They also find out about the material properties important to bridge construction and consider the advantages and disadvantages of steel and concrete as common bridge-building materials to handle compressive and tensile forces.

This engineering curriculum aligns to Next Generation Science Standards (NGSS).

Engineering Connection

When designing structures such as bridges, engineers carefully choose the materials by anticipating the forces the materials (the structural components) are expected to experience during their lifetimes. Usually, ductile materials such as steel, aluminum and other metals are used for components that experience tensile loads. Brittle materials such as concrete, ceramics and glass are used for components that experience compressive loads.

Learning Objectives

After this lesson, students should be able to:

List several common materials used the design and construction of structures.

Describe several factors that engineers consider when selecting materials for the design of a bridge.

Explain the advantages and disadvantages of common materials used in engineering structures (steel and concrete).

Educational Standards

NGSS: Next Generation Science Standards - Science

Common Core State Standards - Math

International Technology and Engineering Educators Association - Technology

State Standards

Suggest an alignment not listed above

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Worksheets and Attachments

Strength of Materials Worksheet (doc)

Strength of Materials Worksheet (pdf)

Strength of Materials Worksheet Answers (doc)

Strength of Materials Worksheet Answers (pdf)

Strength of Materials Math Worksheet (doc)

Strength of Materials Math Worksheet (pdf)

Strength of Materials Math Worksheet Answers (doc)

Strength of Materials Math Worksheet Answers (pdf)

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Explanation:

Required

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The pseudocode and program makes use of a 1 dimensional array to accept input for the 10 numbers;

The largest of the 10 numbers is then saved in variable Largest and printed afterwards.

Pseudocode (Number lines are used for indentation to illustrate the program flow)

1. Start:

2. Declare Number as 1 dimensional array of 10 integers

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4. Do:

4.1 Display “Enter Number ”+(counter + 1)

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C# Program (Console)

Comments are used for explanatory purpose

using System;

namespace ConsoleApplication1

{

   class Program

   {

       static void Main(string[] args)

       {

           int[] Number = new int[10];  // Declare array of 10 elements

           //Accept Input

           int counter = 0;

           while(counter<10)

           {

               Console.WriteLine("Enter Number " + (counter + 1)+": ");

               string var = Console.ReadLine();

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               counter++;                  

           }

           //Initialize largest to first element of the array

           int Largest = Number[0];

           //Determine Largest

           for(int i=0;i<10;i++)

           {

               if(Largest < Number[i])

               {

                   Largest = Number[i];

               }

           }

           //Print Largest

           Console.WriteLine("The largest input is "+ Largest);

           Console.ReadLine();

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V125BC [204]

Answer:

a) E = \frac{K}{e*d}  b) i.

Explanation:

Assuming no other forces acting on the electron, as the field is pushing the electron back till it comes to rest momentarily, applying the work-energy theorem, the loss in kinetic energy, must be equal to the work done by the field on the electron.

The force that does work on the electron is the one due to the electric field E, and by definition of electric field, can be expressed as follows:

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This work must be equal to the change in Kinetic Energy:

ΔK = Kf -K₀ = 0 - K = -K (2)

From (1) and (2), and solving for E, we have:

E =\frac{K}{e*d}

b) As the electric field (by convention) has the direction that it would have a positive test charge, the electrons (being negative charges),when starting from rest),have the direction opposite to the field.

In this case, as the electron has an initial velocity, and the field is opposing to the electron movement, we conclude that the electric field is in the direction of the electron´s motion, due to the electron slows down, and then comes momentarily to an stop, before changing direction and move opposite to the field as it is its natural behavior.

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FrozenT [24]

The code to

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were written in JavaScript and are found in the attached images

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Here, a function was created that receives two arguments (n1  and n2), then returns the product (n1 * n2)

<h3>Declaring an Array</h3>

This last code segment creates an array using the new keyword. The new keyword is generally used in constructing objects.

In this case the object constructed is an array having three strings;

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See another solved JavaScript problem here brainly.com/question/23610566

7 0
3 years ago
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