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galben [10]
3 years ago
5

When mining diamonds with a stone pick what will be the outcome

Engineering
2 answers:
allsm [11]3 years ago
6 0

Answer: Jump to Mining - Wooden: 59; Stone: 131; Iron: 250; Golden: 32; Diamond: 1561 ... For example, while stone can be mined with any pickaxe, gold ore must ...

‎

Explanation:

Soloha48 [4]3 years ago
4 0

Answer:

The diamond ore will break and you won't get any diamonds.

Explanation:

You might be interested in
You will create three classes, the first two being Student and LineAtOfficeHour. The instances of the first class defines a sing
White raven [17]

Answer:

Complete solution is given below:

Explanation:

//student class

class Student{

  private String firstname,lastname;

 

  //constructor

  Student(String first,String last){

      this.firstname=first;

      this.lastname=last;

  }

 

  //getters and setters

  public String getFirstname() {

      return firstname;

  }

  public void setFirstname(String firstname) {

      this.firstname = firstname;

  }

  public String getLastname() {

      return lastname;

  }

  public void setLastname(String lastname) {

      this.lastname = lastname;

  }

  //function to get the fullname of student

  public String fullName() {

      return this.firstname+" "+this.lastname;

  }

}

//class for line at office hour

class LineAtOfficeHour{

 

  private Student line[];

  private int N=0;

  private int back=0;

 

  //empty constructor

  LineAtOfficeHour() {

      line=new Student[5];

  }

  //parameterized constructor

  LineAtOfficeHour(Student st[]) {

      int i=0;

      line=new Student[5];

      while(i<st.length && i<5) {

          line[i]=st[i];

          i++;

      }

      this.N=i;

      this.back=i-1;

  }

  //function to check if line is empty or not

  public boolean isEmpty() {

      if(this.N==0)

          return true;

      else

          return false;

  }

  //function to check if line is full

  public boolean isFull() {

      if(this.N==5) {

          return true;

      }else

          return false;

  }

  ///function to get the size of line

  public int size() {

      return this.N;

  }

 

  //function to add a student to the line

  public void enterLine(Student s) {

      if(isFull())

          System.out.println("Line is full!!!!");

      else {

          line[++back]=s;

          this.N++;

      }

  }

  public Student seeTeacher() {

      Student result=null;

      if(this.N>=0) {

          result=line[0];

          int i=0;

          for(i=1;i<N;i++) {

              line[i-1]=line[i];

          }

          line[i-1]=null;

          this.N--;

          this.back--;

      }

     

     

      return result;

  }

  //function to print students in line

  public String whosInLine() {

      String result ="";

      for(int i=0;i<this.N;i++) {

          result+=line[i].fullName()+",";

      }

      return result;

  }

}

//driver method

public class TestLine {

  public static void main(String[] args) {

      LineAtOfficeHour list=new LineAtOfficeHour();

     

      if(list.isEmpty()) {

          System.out.println("Line is empty!!!!!!!!!");

      }

     

      Student s1[]=new Student[3];

      s1[0]=new Student("John","Smith");

      s1[1]=new Student("Sam","Zung");

      s1[2]=new Student("Peter","Louis");

      list=new LineAtOfficeHour(s1);

     

      if(list.isEmpty()) {

          System.out.println("Line is empty!!!!!!!!!");

      }else {

          System.out.println("Line is not empty.........");

      }

     

      System.out.println("Students in line: "+list.whosInLine());

     

      System.out.println("Student removed: "+list.seeTeacher().fullName());

     

      System.out.println("Students in line: "+list.whosInLine());

  }

}

6 0
4 years ago
A very large plate is placed equidistant between two vertical walls. The 10-mm spacing between the plate and each wall is filled
Vikentia [17]

Answer:

Force per unit plate area is 0.1344 N/m^{2}

Solution:

As per the question:

The spacing between each wall and the plate, d = 10 mm = 0.01 m

Absolute viscosity of the liquid, \mu =1.92\times 10^{- 3} Pa-s

Speed, v = 35 mm/s = 0.035 m/s

Now,

Suppose the drag force that exist between each wall and plate is F and F' respectively:

Net Drag Force = F' + F''

F = \tau A

where

\tau = shear stress

A = Cross - sectional Area

Therefore,

Net Drag Force, F = (\tau ' +\tau '')A

\frac{F}{A} = \tau ' +\tau ''

Also

F = \frac{\mu v}{d}

where

\mu = dynamic coefficient of viscosity

Pressure, P = \frac{F}{A}

Therefore,

\frac{F}{A} = \frac{\mu v}{d} + \frac{\mu v}{d} = 2\frac{\mu v}{d}

\frac{F}{A} = 2\frac{1.92\times 10^{- 3}\times 0.035}{0.010} = 0.01344 N/m^{2}

8 0
3 years ago
Isormophous phase diagram
shusha [124]

Answer:

Phase diagrams represent the relationship between temperature and the composition of phases present at equilibrium. An isomorphous system is one in which the solid has the same structure for all compositions. The phase diagram shown is the diagram for Cu-Ni, which is an isomorphous alloy system.

Hope it help you friend

6 0
3 years ago
A steam power plant operating on a simple ideal Rankine cycle maintains the boiler at 6000 kPa, the turbine inlet at 600 °C, and
ohaa [14]

Answer:

ηa=0.349

ηb=0.345

Explanation:

The enthalpy and entropy at state 3 are determined from the given pressure and temperature with data from table:

h_{3}=3658.8kJ/kg\\ s_{3}=7.1693kJ/kg

The quality at state 4 is determined from the condition  s_{4} =s_{3} and the entropies of the components at the condenser pressure taken from table:

 q_{4} =\frac{s_{4}-s_{liq50}  }{s_{evap,50} } \\=\frac{7.1693-1.0912}{6.5019}=0.935

The enthalpy at state 4 then is:  

h_{4} =h_{liq50} +q_{4} h_{evap,50}\\ (340.54+0.935*2304.7)kJ/kg\\=2495.43kJ/kg\\

Part A

In the case when the water is in a saturated liquid state at the entrance of the pump the enthalpy and specific volume are determined from A-5 for the given pressure:  

h_{1}=340.54kJ/kg\\ \alpha _{1}=0.00103m^3/kg

The enthalpy at state 2 is determined from an energy balance on the pump:

h_{2} =h_{1} +\alpha _{1}( P_{2}-P_{1}  )

    =346.67 kJ/kg

The thermal efficiency is then determined from the heat input and output in the cycle:  

=1-\frac{q_{out} }{q_{in} } \\=1-\frac{h_{4} -h_{1} }{h_{3} -h_{2}} \\=0.349

Part B  

In the case when the water is at a lower temperature than the saturation temperature at the condenser pressure we look into table and see the water is in a compressed liquid state. Then we take the enthalpy and specific volume for that temperature with data from  and the saturated liquid values:  

h_{1}=293.7kJ/kg\\ \alpha _{1}=0.001023m^3/kg

The enthalpy at state 2 is then determined from an energy balance on the pump:  

h_{2} =h_{1} +\alpha _{1}( P_{2}-P_{1}  )

    =299.79 kJ/kg  

The thermal efficiency in this case then is:  

=1-\frac{q_{out} }{q_{in} } \\=1-\frac{h_{4} -h_{1} }{h_{3} -h_{2}} \\=0.345

8 0
3 years ago
- Consider a 2024-T4 aluminum material with ultimate tensile strength of 70 ksi. In a given application, a component of this mat
Sliva [168]

Answer:

2.43 ksi

Explanation:

In this question, we are asked to calculate the value of σmax that would ensure that the component does not fall below 1,000,000 cycles having a 99.99% chance.

Please check attachment for complete solution and step by step explanation

6 0
4 years ago
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