Copper spheres of 20-mm diameter are quenched by being dropped into a tank of water that is maintained at 280 K . The spheres may be assumed to reach the terminal velocity on impact and to drop freely through the water. Estimate the terminal velocity by equating the drag and gravitational forces acting on the sphere. What is the approximate height of the water tank needed to cool the spheres from an initial temperature of 360 K to a center temperature of 320 K?

Answer and Explanation:

the answer is attached below

Answer:

Average access time for a hard disk = 11.38 ms

Explanation:

See attached pictures for step by step explanation.

Answer:

0.34

Explanation:

See the attached picture.

Answer: V = 208514.156 ft/hr

Explanation:

we will begin by giving a step by step order of answering.

given that

A = area available for convection which will be same for both cases

h (still) = heat transfer coefficient in still air

h(blowing) = heat transfer coefficient in blowing air.

Therefore,

h(still) A [40-(-30)] = h(blowing) A (40-15)

canceling out we have

h(still) (70) = h(blowing) (25)

where h(still) = 4 Btu/hr.ft².°F

4 × 70 =  h(blowing) (25)

h (blowing) = 11.2 Btu/hr.ft².°F

Also, we have that NUD = 0.0239 ReD 0805

h (blowing) D / k = 0.0239 (ρVD/μ)˄0.805

where from our data,

D = diameter = 1 ft

ρ = density = 0.0845 lbm/ft

μ = viscosity = 3.996×10-2 bm/ft hr

K = 0.0134 Btu/hr.ft²°F

So from

h (blowing) D / k = 0.0239 (ρVD/μ)˄0.805

we have;

11.2 × 1 / K =  0.0239 (0.0845×V×1 / 3.996×10-2 )˄0.805

where K = 0.0134

V = 208514.156 ft/hr

cheers i hope this helps

Answer:

(εtrue/εengr) = (1.3481/2.85) = 0.473

This shows that the engineering strain is truly a bit far off the true strain.

Explanation:

εengr

Engineering strain is a measure of how much a material deforms under a particular load. It is the amount of deformation in the direction of the applied force divided by the initial length of the material.

ε(engineering) = ΔL/L₀

Lf = final length = 3.85 L₀

L₀ = original length = L₀

ΔL = Lf - L₀ = 3.85 L₀ - L₀ = 2.85 L₀

ε(engineering) = ΔL/L₀ = (2.85L₀)/L₀ = 2.85

εtrue

True Strain measures instantaneous deformation. It is obtained mathematically by integrating strain over small time periods and Running them up. Hence,

ε(true) = In (Lf/L₀)

Lf = 3.85L₀

L₀ = L₀

ε(true) = In (Lf/L₀) = In (3.85L₀/L₀) = In 3.85 = 1.3481

(εtrue/εengr) = (1.3481/2.85) = 0.473

Answer:

Explanation:if you stretch the hose more tightly the speed of the pulse will reduce..

Answer:

Explanation:

The detailed calculation and appropriate equation with substitution is as shown in the attached file.

Answer and Explanation:

The answer is attached below

Answer:

note:

solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment

The amount of work produced by 1 kg of steam and 1 kg of R–134a would be the same in a closed system at the given conditions.

In a closed system, the amount of work produced depends on the change in internal energy of the system. The change in internal energy is given by the formula ΔU = Q - W, where Q is the heat added to the system and W is the work done by the system. Since both 1 kg of steam and 1 kg of R–134a are at the same temperature and pressure, their internal energy changes would be the same for the same amount of heat added. Therefore, the amount of work produced by both substances would also be the same in a closed system.

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Answer:

See image attached.

Explanation:

This device features priority encoding of the inputs

to ensure that only the highest order data line is en-

coded. Nine input lines are encoded to a four line

BCD output. The implied decimal zero condition re-

quires no input condition as zero is encoded when

all nine datalinesare athigh logic level. Alldata input

and outputs are active at the low logic level. All in-

puts are equipped with protection circuits against

static discharge and transient excess voltage.

A 10-to-4 encoder is requested to be designed for mapping 1-out-of-10 input code to a modified BCD output, where the digits 8 and 9 are encoded as 'E' and 'F'. The implementation involves the use of digital logic gates and could also reference quantum gates depending on the context of the course.

The student is asking to design a 10-to-4 encoder with a specific bit pattern for the numbers 8 and 9. This encoder takes a l-out-of-10 input code and produces a BCD-like output with special cases for inputs 8 ('E') and 9 ('F'). The design would necessitate the use of digital logic gates to map each of the 10 inputs to correspondent 4-bit BCD outputs, with the additional requirement to encode the inputs representing the decimal numbers 8 and 9 into the hexadecimal digits 'E' and 'F', respectively.

In terms of circuitry, the encoder might use a combination of logical gates such as AND, OR, and NOT to create the desired output for each input. For example, when the ninth input is active (representing the number 8), the output should be '1110', which signifies 'E' in hexadecimal notation. Similarly, an active tenth input (representing the number 9) should produce '1111', corresponding to 'F' in hexadecimal.

The encoding and decoding elements are described using terms such as CnNOT, CNOT, I (Identity), and Toffoli gates, which suggests a more sophisticated setup, possibly involving quantum computing principles, as these terms relate to quantum gates.

Answer:

[tex] T_C = 27+273.15 = 300.15 K[/tex]

[tex] T_H = 477+273.15 = 750.15 K[/tex]

And replacing in the Carnot efficiency we got:

[tex] e= 1- \frac{300.15}{750.15}= 0.59988 = 59.98 \%[/tex]

[tex] W_{max}= e* Q_H = 0.59988 * 65000 \frac{KJ}{min}= 38992.2 \frac{KJ}{min}[/tex]

Explanation:

For this case we can use the fact that the maximum thermal efficiency for a heat engine between two temperatures are given by the Carnot efficiency:

[tex] e = 1 -frac{T_C}{T_H}[/tex]

We have on this case after convert the temperatures in kelvin this:

[tex] T_C = 27+273.15 = 300.15 K[/tex]

[tex] T_H = 477+273.15 = 750.15 K[/tex]

And replacing in the Carnot efficiency we got:

[tex] e= 1- \frac{300.15}{750.15}= 0.59988 = 59.98 \%[/tex]

And the maximum power output on this case would be defined as:

[tex] W_{max}= e* Q_H = 0.59988 * 65000 \frac{KJ}{min}= 38992.2 \frac{KJ}{min}[/tex]

Where [tex] Q_H[/tex] represent the heat associated to the deposit with higher temperature.

Answer:

Explanation:

For each instance, the `MachineStatus` class should store: the label of a vending machine the total amount of beverages the vending machine was previously stocked with the total amount of beverages currently in stock in the vending machine the total income of the machine from the last time it was stocked until now.

Answer:True

Explanation:Rotor matching is a term used to describe the process through a brake rotor is aligned to the hub and bearing assembly to produce a smooth, flat, and straight friction surface.

Rotor matching is an essential process which is required to prevent swirling of a break when a vehicle is stopping, rotor matching is needed for efficient and effective break system performance as it is one of the main determinant factors for accidents arising from break failures.

Explanation of rewriting code from for loop to while loop in Python.

To rewrite the given code using a while loop instead of a for loop, you can iterate over the list indices and manually accumulate the total.

Here is the code:

By using this code snippet, the sum accumulated using the while loop will be stored in the variable sum2, which should equal the sum1 calculated using the for loop.

Answer:

Explanation: see attachment

Answer:

183.75 cubic inches.

Explanation:

The volume of the wood board is determine by means of this expression:

[tex]V = w \cdot h \cdot l[/tex]

By replacing variables:

[tex]V = (7.5 in) \cdot (14 in) \cdot (1.75 in)\\V = 183.75 in^{3}[/tex]

Answer:

Thermal Conductivity (K) = 33.33 W/m. ° C

The value of the convection heat transfer coefficient = 3 W/m².° C

Explanation:

The attached document file gives a detailed and clear explanation about the question.

Answer:

(a)

(i) pseudo code :-

current = i

// assuming parent of root is -1

while A[parent] < A[current] && parent != -1 do,

if A[parent] < A[current] // if current element is bigger than parent then shift it up

swap(A[current],A[parent])

current = parent

(ii) In heap we create a complete binary tree which has height of log(n). In shift up we will take maximum steps equal to the height of tree so number of comparison will be in term of O(log(n))

(b)

(i) There are two cases while comparing with grandparent. If grandparent is less than current node then surely parent node also will be less than current node so swap current node with parent and then swap parent node with grandparent.

If above condition is not true then we will check for parent node and if it is less than current node then swap these.

pseudo code :-

current = i

// assuming parent of root is -1

parent is parent node of current node

while A[parent] < A[current] && parent != -1 do,

if A[grandparent] < A[current] // if current element is bigger than parent then shift it up

swap(A[current],A[parent])

swap(A[grandparent],A[parent])

current = grandparent

else if A[parent] < A[current]

swap(A[parent],A[current])

current = parent

(ii) Here we are skipping the one level so max we can make our comparison half from last approach, that would be (height/2)

so order would be log(n)/2

(iii) C++ code :-

#include<bits/stdc++.h>

using namespace std;

// function to return index of parent node

int parent(int i)

{

if(i == 0)

return -1;

return (i-1)/2;

}

// function to return index of grandparent node

int grandparent(int i)

{

int p = parent(i);

if(p == -1)

return -1;

else

return parent(p);

}

void shift_up(int A[], int n, int ind)

{

int curr = ind-1; // because array is 0-indexed

while(parent(curr) != -1 && A[parent(curr)] < A[curr])

{

int g = grandparent(curr);

int p = parent(curr);

if(g != -1 && A[g] < A[curr])

{

swap(A[curr],A[p]);

swap(A[p],A[g]);

curr = g;

}

else if(A[p] < A[curr])

{

swap(A[p],A[curr]);

curr = p;

}

}

}

int main()

{

int n;

cout<<"enter the number of elements :-\n";

cin>>n;

int A[n];

cout<<"enter the elements of array :-\n";

for(int i=0;i<n;i++)

cin>>A[i];

int ind;

cout<<"enter the index value :- \n";

cin>>ind;

shift_up(A,n,ind);

cout<<"array after shift up :-\n";

for(int i=0;i<n;i++)

cout<<A[i]<<" ";

cout<<endl;

}

Explanation:

Answer:

The lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

Explanation:

The Upper Surface Cp is given as

[tex]Cp_u=-0.8 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.8*0.6+0.4*0.1[/tex]

The Lower Surface Cp is given as

[tex]Cp_l=-0.4 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.4*0.6+0.4*0.1[/tex]

The difference of the Cp over the airfoil is given as

[tex]\Delta Cp=Cp_l-Cp_u\\\Delta Cp=-0.4*0.6+0.4*0.1-(-0.8*0.6-0.4*0.1)\\\Delta Cp=-0.4*0.6+0.4*0.1+0.8*0.6+0.4*0.1\\\Delta Cp=0.4*0.6+0.4*0.2\\\Delta Cp=0.32[/tex]

Now the Lift Coefficient is given as

[tex]C_L=\Delta C_p cos(\alpha_i)\\C_L=0.32\times cos(4*\frac{\pi}{180})\\C_L=0.3192[/tex]

Now the coefficient of moment about the leading edge is given as

[tex]C_M=-0.3*0.4*0.6-(0.6+\dfrac{0.4}{3})*0.2*0.4\\C_M=-0.1306[/tex]

So the lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

The lift coefficient and the pitching moment coefficient about the leading edge due to lift are respectively; 0.3192 and -0.13

We are given;

Distance of upper surface from leading edge to percentage of chord = -0.8

Percentage of chord for both surfaces = 60% = 0.6

Rate of increase at trailing edge for both surfaces = +0.1

Distance of lower surface from leading edge to percentage of chord = -0.6

angle of incidence; α_i = 4° = 4π/180 rad

Let us first calculate the Cp constant for both the upper and lower surface.

Cp for upper surface is;

Cp_u = (-0.8 × 0.6) - 0.1∫¹₀.₆ dx

Solving this integral gives;

Cp_u =  (-0.8 × 0.6) - (0.1 × 0.4)

Cp_u =  -0.52

Cp for lower surface is;

Cp_l = (-0.4 × 0.6) + 0.1∫¹₀.₆ dx

Solving this integral gives;

Cp_l =  (-0.4 × 0.6) + (0.1 × 0.4)

Cp_l =  -0.2

Change in Cp across the foil is;

ΔCp = Cp_l - Cp_u

ΔCp = -0.2 - (-0.52)

ΔCp = 0.32

Formula for the lift coefficient is;

C_L = ΔCp * cosα_i

C_L = 0.32 * cos (4π/180)

C_L = 0.3192

Formula for the pitching moment coefficient is;

(-0.3 * 0.4 * 0.6) - ((0.6 + (0.4/3)) * 0.2 * 0.4)

C_m,p =  -0.072 - 0.059

C_mp ≈ -0.13

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Answer:

Explanation:using the product immediately in the next step

The student's question involves calculating the water level drop at the center of a rotating cylindrical container, which requires understanding of rotational motion and equilibrium in physics.

The student is asking about the change in water level in a rotating cylindrical container due to the centrifugal force. In a rotating frame, the water surface forms a parabolic shape and the level at the center drops. To solve this problem, principles of rotational motion and physics are applied. The task is to calculate the drop in the center of the water surface within a cylindrical container rotating at a constant angular speed. This can be determined by setting up the equilibrium of forces acting on a particle of the water in the radial direction and using the relationship between the angular speed, radius, and acceleration in a rotating system.

Answer: 2.26x10^-4 v

Explanation:

Lenght of the selonoid = 24x10^-2m

Diameter of the selonoid = 2.3cm

The radius will then be = 1.15cm = 1.15x10^-2m

The area of the selonoid = ¶r^2 = 3.142 x (1.15x10^-2)^2 = 0.000415m^2.

Number of turns on selonoid N1 is 750

For the small center coil, number of turns N2 is 19.

There is a change in current dI/dt from 0 to 5.1 in 0.7s, dI/dt = (5.1-0)/0.7

dI/dt = 7.29A/s.

Induced EMF on selonoid due to magnetic Flux due to changing current in small coil is given as;

E = -M(dI/dt), where M is the mutual inductance of the coils.

but M = (u°AN1N2)/L, where u°= 4¶x10^-7,

A = area of selonoid,

L = Lenght of selonoid.

M = (4¶X10^-7X0.000415X750X19)/(24X10^-2)

M = 3.096X10^-5H

Induced EMF E = 3.096X10^-5 x 7.29

E = 2.26x10^-4V

(a) The RTL specification for an arithmetic shift right on an eight-bit cell can be described as follows:

1. Take the eight-bit input value and assign it to a variable, let's call it "input".

2. Create a temporary variable, let's call it "shifted".

3. Assign the most significant bit (MSB) of the "input" to the least significant bit (LSB) of the "shifted" variable.

4. Shift the remaining seven bits of the "input" one position to the right, discarding the least significant bit (LSB) and shifting in the sign bit to the most significant bit (MSB).

5. Repeat step 3 and step 4 seven more times to shift all the bits in the "input" variable to the right.

6. Output the final value of the "shifted" variable, which represents the result of the arithmetic shift right.

Here's an example to illustrate the RTL specification:

Input: 10111010

Shifted: 11011101 (after one shift)

Shifted: 11101110 (after two shifts)

Shifted: 11110111 (after three shifts)

Shifted: 11111011 (after four shifts)

Shifted: 11111101 (after five shifts)

Shifted: 11111110 (after six shifts)

Shifted: 11111111 (after seven shifts)

Shifted: 11111111 (after eight shifts)

(b) The RTL specification for an arithmetic shift left on an eight-bit cell can be described as follows:

1. Take the eight-bit input value and assign it to a variable, let's call it "input".

2. Create a temporary variable, let's call it "shifted".

3. Assign the least significant bit (LSB) of the "input" to the most significant bit (MSB) of the "shifted" variable.

4. Shift the remaining seven bits of the "input" one position to the left, discarding the most significant bit (MSB) and shifting in a zero to the least significant bit (LSB).

5. Repeat step 3 and step 4 seven more times to shift all the bits in the "input" variable to the left.

6. Output the final value of the "shifted" variable, which represents the result of the arithmetic shift left.

Here's an example to illustrate the RTL specification:

Input: 10111010

Shifted: 01110100 (after one shift)

Shifted: 11101000 (after two shifts)

Shifted: 11010000 (after three shifts)

Shifted: 10100000 (after four shifts)

Shifted: 01000000 (after five shifts)

Shifted: 10000000 (after six shifts)

Shifted: 00000000 (after seven shifts)

Shifted: 00000000 (after eight shifts)

Know more about RTL specifications,

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The Poisson's ratio is 0.36

Explanation:

Given-

Diameter = 10 mm = 10 X 10⁻³m

Force, F = 15000 N

Diameter reduction = 7 X 10⁻3 mm = 7 X 10⁻⁶ mm

The equation for Poisson's ratio:

ν = [tex]\frac{-E_{x} }{E_{z} }[/tex]

and

εₓ = Δd / d₀

εz = σ / E

    = F / AE

We know,

Area of circle = π (d₀/2)²

[tex]E_{z}[/tex]= F / π (d₀/2)² E

εz = 4F / πd₀²E

Therefore, Poisson's ratio is

ν = - Δd ÷ d / 4F ÷ πd₀²E

ν =  -d₀ΔDπE / 4F

ν = (10 X 10⁻³) (-7 X 10⁻⁶) (3.14) (100 X 10⁻⁹) / 4 (15000)

ν = 0.36

Therefore, the Poisson's ratio is 0.36

Answer:

The code is as below whereas the output is attached herewith

Explanation:

package brainly.priorityQueue;

class PriorityJobQueue {

   Job[] arr;

   int size;

   int count;

   PriorityJobQueue(int size){

       this.size = size;

       arr = new Job[size];

       count = 0;

   }

   // Function to insert an element into the priority queue

   void insert(Job value){

       if(count == size){

           System.out.println("Cannot insert the key");

           return;

       }

       arr[count++] = value;

       heapifyUpwards(count);

   }

   // Function to heapify an element upwards

   void heapifyUpwards(int x){

       if(x<=0)

           return;

       int par = (x-1)/2;

       Job temp;

       if(arr[x-1].getPriority() < arr[par].getPriority()){

           temp = arr[par];

           arr[par] = arr[x-1];

           arr[x-1] = temp;

           heapifyUpwards(par+1);

       }

   }

   // Function to extract the minimum value from the priority queue

   Job extractMin(){

       Job rvalue = null;

       try {

           rvalue = arr[0].clone();

       } catch (CloneNotSupportedException e) {

           e.printStackTrace();

       }

       arr[0].setPriority(Integer.MAX_VALUE);

       heapifyDownwards(0);

       return rvalue;

   }

   // Function to heapify an element downwards

   void heapifyDownwards(int index){

       if(index >=arr.length)

           return;

       Job temp;

       int min = index;

       int left,right;

       left = 2*index;

       right = left+1;

       if(left<arr.length && arr[index].getPriority() > arr[left].getPriority()){

           min =left;

       }

       if(right <arr.length && arr[min].getPriority() > arr[right].getPriority()){

           min = right;

       }

       if(min!=index) {

           temp = arr[min];

           arr[min] = arr[index];

           arr[index] = temp;

           heapifyDownwards(min);

       }

   }

   // Function to implement the heapsort using priority queue

   static void heapSort(Job[] array){

       PriorityJobQueue object = new PriorityJobQueue(array.length);

       int i;

       for(i=0; i<array.length; i++){

           object.insert(array[i]);

       }

       for(i=0; i<array.length; i++){

           array[i] = object.extractMin();

       }

   }

}

package brainly.priorityQueue;

import java.io.BufferedReader;

import java.io.IOException;

import java.io.InputStreamReader;

import java.util.Arrays;

public class PriorityJobQueueTest {

   // Function to read user input

   public static void main(String[] args) {

       BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

       int n;

       System.out.println("Enter the number of elements in the array");

       try{

           n = Integer.parseInt(br.readLine());

       }catch (IOException e){

           System.out.println("An error occurred");

           return;

       }

       System.out.println("Enter array elements");

       Job[] array = new Job[n];

       int i;

       for(i=0; i<array.length; i++){

           Job job  =new Job();

           try{

               job.setJobId(i);

               System.out.println("Element "+i +"priority:");

               job.setJobName("Name"+i);

               job.setSubmitterName("SubmitterName"+i);

               job.setPriority(Integer.parseInt(br.readLine()));

               array[i] = job;

           }catch (IOException e){

               System.out.println("An error occurred");

           }

       }

       System.out.println("The initial array is");

       System.out.println(Arrays.toString(array));

       PriorityJobQueue.heapSort(array);

       System.out.println("The sorted array is");

       System.out.println(Arrays.toString(array));

       Job[] readyQueue =new Job[4];

   }

}

Answer:

note:

solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment

Answer:

The part I called command in the first diagram has been renamed to opcode, or operation code. This is a set of bits (a number) that will tell the ALU which action to perform. I can get the LC-3 opcodes for ADD and NOT and ADD from the book, so I'm not too worried.

Note the #? comment by the switch above opcode. This means I'm not sure how many switches I will need. How many bits do I need to perform all the operations I want? The textbook will tell me.

Materials

Now I make a list of all the materials you have accumulated so far. This list is just an example; yours may be different.

Two 4-bit inputs

One 4-bit output

Two keypads for 4-bit input

Three 7-segment displays (2 for input, 1 for output)

A bunch of switches for opcode (could use a keypad, I guess, but switches are so much more geeky)

A bunch of lights too

The "is zero" LED

One button for clock

One button for reset

One switch for carry-in

Include logic to perform a SUB instruction. That is, subtract the second operand from the first (out = in1 - in2). All three values -- both inputs and the output -- must be two's complement numbers (negative numbers must be represented). Your design may work in one (8 points) or two (4 points) clock cycles.

Explanation:

You are allowed to use the Logisim built-in registers.

The clear input of the register should not be used (do not connect anything to

them).

Custom ALU

Use the provided subcircuit in the template to implement your ALU. You do not have to create additional subcircuits to do this. The ALU has a total of three inputs: First number, second number and select operation input. And one output: Result. The first and second number are used as input for the operations the ALU performs. The select input decides which operation result will be on the single output of the ALU. The ALU is supposed to calculate: NumberA OPERATION NumberB. Register 1 of the storage contains NumberA and Register 2 contains NumberB. The ALU must be able to compute signals with a 4-bit width. Make sure to add labels to all inputs and outputs.

The following operations should be performed for each select input combination (s1s0): • 00: Logic Bitwise XOR

• 01: Multiplication

• 10: Division

• 11: Addition Notes:

You can change the inputs bit width / data bits of any gate to more that 1-bit.

The Logic Bitwise XOR operation can be done with a single XOR gate.

You are also allowed to use the built-in arithmetic logic components and multi- plexer provided by Logisim.

If the result is larger than 4 bits, it will be truncated (only 4 LSB will be shown). This behavior is intended for this assignment. Also, negative results do not have to be considered.

Once you have implemented the ALU circuit, connect the wires in the main circuit properly and test all four operations of your ALU in combination with the storage component.

Answer:

a) the velocity of the implant immediately after impact is 20 m/s

b) the average resistance of the implant is 40000 N

Explanation:

a) The impulse momentum is:

mv1 + ∑Imp(1---->2) = mv2

According the exercise:

v1=0

∑Imp(1---->2) = F(t2-t1)

m=0.2 kg

Replacing:

[tex]0+F(t_{2} -t_{1} )=0.2v_{2}[/tex]

if F=2 kN and t2-t1=2x10^-3 s. Replacing

[tex]0+2x10^{-3} (2x10^{-3} )=0.2v_{2} \\v_{2} =\frac{4}{0.2} =20m/s[/tex]

b) Work and energy in the system is:

T2 - U(2----->3) = T3

where T2 and T3 are the kinetic energy and U(2----->3) is the work.

[tex]T_{2} =\frac{1}{2} mv_{2}^{2} \\T_{3} =0\\U_{2---3} =-F_{res} x[/tex]

Replacing:

[tex]\frac{1}{2} *0.2*20^{2} -F_{res} *0.001=0\\F_{res} =40000N[/tex]

Answer:

Explanation: see attachment

Comments on Question

Your questions is incomplete.

I'll provide a general answer to create a checkbox programmatically

Answer:

// Comments are used for explanatory purpose

// Function to create a checkbox programmatically

public class CreateCheckBox extends Application {

// launch the application

public void ClubObject(Stage chk)

{

// Title

chk.setTitle("CheckBox Title");

// Set Tile pane

TilePane tp = new TilePane();

// Add Labels

Label lbl = new Label("Creating a Checkbox Object");

// Create an array to hold 2 checkbox labels

String chklbl[] = { "Checkbox 1", "Checkbox 2" };

// Add checkbox labels

tp.getChildren().add(lbl);

// Add checkbox items

for (int i = 0; i < chklbl.length; i++) {

// Create checkbox object

CheckBox cbk = new CheckBox(chklbl[i]);

// add label

tp.getChildren().add(cbk);

// set IndeterMinate to true

cbk.setIndeterminate(true);

}

// create a scene to accommodate the title pane

Scene sc = new Scene(tp, 150, 200);

// set the scene

chk.setScene(sc);

chk.show();

}