The following fluids (air, H, N) at 350K and atmospheric pressure flow at velocity of 5 m/s over a 2 m long flat plate. The order of magnitude of the drag force from lowest to highest is a. air-H-N b. air-N-H Oc. H-N-air Od. H-air-N Oe. N-air-H Of. N-H-air

Branch circuits shall be sized in accordance with the maximum permitted.

The sizing of branch circuits ensure that they can safely and effectively carry the electrical load that will be placed on them. The National Electrical Code specifies the maximum ampacit  or current-carrying capacity of branch circuits based on the size and type of wire being used.

So, it important to size branch circuits correctly to avoid overloading the circuit which can result in overheating, fires, and other hazards. The contractors and other professionals responsible for installing and maintaining electrical systems should be familiar with the NEC requirements for branch circuit sizing and ensure that all installations comply with these standards.

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Answer:

Total time taken = 0.769 hour

Explanation:

using the velocity method

for sheet flow ;

Tt = [tex]\frac{0.007(nl)^{0.8} }{(Pl)^{5}s^{0.4} }[/tex]  

Tt = travel time

n = manning CaH

Pl = 25years

L = how length ( ft )

s = slope

For Location ( 1 )

s = 0.045

L = 1000 ft

n = 0.06 ( from manning's coefficient table )

Tt1 = 0.128 hour

For Location ( 2 )

s = 2.5 %

L= 750

n = 0.13

Tt2 = 0.239 hour

For Location ( 3 )

s = 1.5%

L = 500 ft

n = 0.15

Tt3 = 0.237  hour

For Location (4)

s = 0.5 %

L = 250 ft

n = 0.011

Tt4 = 0.165 hour

hence the Total time taken = Tt1 + Tt2 + Tt3 + Tt4

                                              = 0.128 + 0.239 + 0.237 + 0.165 = 0.769 hour

The equivalent force and a couple of moments acting at point O are:

F={i+j−4k}kN

M={-9i+17j+17k} N*m

To replace the given forces with an equivalent force and a couple of moments, we need to find the resultant force and the line of action. The resultant force can be calculated by adding the given forces vectorially. F1 + F2 = (-4i + 4j - 2k) + (5i - 3j - 2k) = i + j - 4k kN. To find the line of action, we can choose any point on the line of action of one of the forces, and then apply the conditions of equilibrium.

Let's choose point O as the reference point. The moment about O due to F1 is -40 + 40 + 20 = 0 Nm. The moment about O due to F2 is 50 - 30 - 20 = 0 Nm. Therefore, the line of action of the resultant force passes through point O. Now we need to find the couple moment that produces the same effect as the given forces.

We can choose any point on the line of action of the resultant force as the reference point. Let's choose point O again. The couple moment is given by the cross product of the position vector from O to the reference point with the resultant force. M = r x F = (0i + 0j + 0k) x (i + j - 4k) kN = -9i + 17j + 17k Nm. Therefore, the equivalent force and couple moment acting at point O are F={i+j−4k}kN and M={-9i+17j+17k} Nm, respectively.

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Answer:

a.

y[n] = x[n] x[n-1]  x[n+1]

(i) Memory-less - It is not memory-less because the given system is depend on past or future values.

(ii) Causal - It is non-casual because the present value of output depend on the future value of input.

(iii) Invertible - It is invertible and the inverse of the given system is [tex]\frac{1}{x[n] . x[n-1] x[n+1]}[/tex]

(iv) Stable - It is stable because for all the bounded input, output is bounded.

(v) Time invariant - It is not time invariant because the system is multiplying with a time varying function.

b.

y[n] = cos(x[n])

(i) Memory-less - It is memory-less because the given system is not depend on past or future values.

(ii) Causal - It is casual because the present value of output does not depend on the future value of input.

(iii) Invertible - It is not invertible because two or more than two input values can generate same output values .

For example - for x[n] = 0 , y[n] = cos(0) = 1

                       for x[n] = 2[tex]\pi[/tex] , y[n] = cos(2[tex]\pi[/tex]) = 1

(iv) Stable - It is stable because for all the bounded input, output is bounded.

(v) Time invariant - It is time invariant because the system is not multiplying with a time varying function.

To calculate the variance of this project activity's estimated duration, we can use the formula:

Variance = [(b-a)/6]^2

where a is the optimistic time estimate, b is the pessimistic time estimate, and m is the most likely time estimate.

In this case, the optimistic time estimate (a) is 8, the pessimistic time estimate (b) is 27, and the most likely time estimate (m) is 16.

So, plugging these values into the formula:

Variance = [(27-8)/6]^2
Variance = 3.08

Therefore, the variance of this project activity's estimated duration is 3.08 (rounded to 2 decimal places).

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The wet unit weight is 106.5 lb/ft3, the dry unit weight is 97.3 lb/ft3, the porosity is 35.9%, and the degree of saturation is 23.3%.

To solve this problem, we need to use the following equations:

Void ratio (e) = Volume of voids (Vv) / Volume of solids (Vs)

Porosity (n) = Vv / Vt, where Vt is the total volume of the soil sample (Vt = Vv + Vs)

Degree of saturation (Sr) = (Vw / Vv) x 100, where Vw is the volume of water in the soil sample

Dry unit weight ([tex]γd[/tex]) = (Gs / (1 + e)) x [tex]γw[/tex], where Gs is the specific gravity of the soil and [tex]γw[/tex] is the unit weight of water (62.4 lb/ft3)

Wet unit weight [tex](γw[/tex]) = [tex]γd[/tex] + (w x [tex]γw[/tex]), where w is the water content of the soil sample

Given data:

Void ratio (e) = 0.56

Water content (w) = 15%

Specific gravity of soil (Gs) = 2.64

First, we need to calculate the dry unit weight:

[tex]γd[/tex] = (Gs / (1 + e)) x [tex]γw[/tex]

[tex]γd[/tex] = (2.64 / (1 + 0.56)) x 62.4

[tex]γd[/tex]= 97.3 lb/ft3

Next, we can calculate the wet unit weight:

[tex]γw[/tex] = [tex]γd[/tex] + (w x [tex]γw[/tex])

[tex]γw[/tex] = 97.3 + (0.15 x 62.4)

[tex]γw[/tex] = 106.5 lb/ft3

Now we can calculate the porosity:

n = Vv / Vt

n = e / (1 + e)

n = 0.56 / (1 + 0.56)

n = 0.359 or 35.9%

Finally, we can calculate the degree of saturation:

Sr = (Vw / Vv) x 100

Sr = (0.15 x Vt) / Vv

Sr = (0.15 x (Vv + Vs)) / Vv

Sr = (0.15 / (1 - n)) x 100

Sr = (0.15 / (1 - 0.359)) x 100

Sr = 23.3%

Therefore, the wet unit weight is 106.5 lb/ft3, the dry unit weight is 97.3 lb/ft3, the porosity is 35.9%, and the degree of saturation is 23.3%.

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The statement that is invalid is: "The final state is reachable from all other states."

This statement is incorrect because the final state is a state in which the object-oriented software has completed its execution, and it should not have any outbound transitions. Therefore, it should not be possible to reach the final state from any other state. The final state should only have inbound transitions from other states. On the other hand, the other statements are valid. The initial state should only have outbound transitions because it is the starting point of the object-oriented software. There should be an event and an action associated with a state transition to specify what triggers the transition and what happens during the transition. Every state should be reachable from any state to ensure that all possible paths through the software can be tested.

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Answer:

A PULLY

Explanation:

HAD THIS ONE THAT IS THE CORRECT ANWSER

Answer:

The answer is B. a pulley

Explanation:

I hope I answered your question:)

a. Sketching P-v and T-s diagrams for the given power cycle:

In the P-v diagram, process 1-2 is an isentropic compression where the volume decreases and pressure increases. Processes 2-3 is a constant pressure heat addition where the volume increases and pressure remains constant. Process 3-1 is a constant volume heat rejection where the volume remains constant and pressure decreases.  In the T-s diagram, process 1-2 is an isentropic compression where the entropy decreases. Process 2-3 is a constant pressure heat addition where the entropy increases. Process 3-1 is a constant volume heat rejection where the entropy remains constant.

b. Calculation of heat and work interactions for each process, in kJ/kg:

Process 1-2: Isentropic compression

w12 = m*Cv*(T1-T2)/(1-k)

q12 = w12 + m*R*(T1-T2)/(1-k)

Process 2-3: Constant pressure heat addition

q23 = m*Cp*(T3-T2)

w23 = q23 - m*R*(T3-T2)

Process 3-1: Constant volume heat rejection

q31 = m*Cv*(T1-T4)

w31 = q31 - m*R*(T1-T4)

c. Calculation of the cycle thermal efficiency:

eta = (w12 + w23 - w31)/(q23)

d. Expression for the cycle thermal efficiency as a function of the compression ratio r and ratio of specific heats k:

eta = 1 - (1/r^((k-1)/k))*(T1/T3-1)

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To determine the maximum height that the block moves up the inclined plane, we can consider the conservation of mechanical energy.

The correct expression for the maximum height can be given by:

h = (v^2 * sin^2θ) / (2 * g)

Where:

h is the maximum height reached by the block

v is the initial velocity of the block along the inclined plane

θ is the angle of inclination of the plane

g is the acceleration due to gravity

This expression is derived from the conservation of mechanical energy, considering the initial kinetic energy and the potential energy at the maximum height.

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Answer: the engineers are performing a tensile test

Explanation:

i took the test and got it right

In Java, when a class is derived from another class, a relationship is formed between them. This relationship is known as the inheritance relationship, where the derived class inherits properties and methods from the parent class.

When you create a new object of the child class, the constructor of the parent class is automatically called before the constructor of the child class is executed. This is because the child class needs to initialize all the properties that it inherited from the parent class.

Now, if you experiment with a simple derivation relationship between two classes and put println statements in constructors of both the parent and child classes, but do not explicitly call the constructor of the parent in the child classes, you will see that the parent class constructor is still called before the child class constructor. This is because it is done implicitly by Java.

Here is a sample program that demonstrates this:

```
class Parent {
 public Parent() {
   System.out.println("Parent constructor called");
 }
}

class Child extends Parent {
 public Child() {
   System.out.println("Child constructor called");
 }
}

public class Main {
 public static void main(String[] args) {
   Child childObj = new Child();
 }
}
```

If you run this program, you will see the output as:

```
Parent constructor called
Child constructor called
```

Now, if you change the child's constructor to explicitly call the constructor of the parent, you will see that the output remains the same. However, the parent constructor is called explicitly this time.

Here is the modified sample program:

```
class Parent {
 public Parent() {
   System.out.println("Parent constructor called");
 }
}

class Child extends Parent {
 public Child() {
   super();
   System.out.println("Child constructor called");
 }
}

public class Main {
 public static void main(String[] args) {
   Child childObj = new Child();
 }
}
```

If you run this program, you will still see the output as:

```
Parent constructor called
Child constructor called
```

But this time, the parent constructor is called explicitly using the `super()` keyword inside the child's constructor.

I hope this helps you understand the inheritance relationship and how constructors work in Java. Let me know if you have any more questions.

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the "Parent" constructor was specifically called from the "Child" constructor. This is such that each constructor can either call a constructor in the superclass (called "super()") or a constructor in the same class (called "this()").

```
class Parent {
   public Parent() {
       System.out.println("Parent constructor called");
   }
}

class Child extends Parent {
   public Child() {
       System.out.println("Child constructor called");
   }
}

public class Main {
   public static void main(String[] args) {
       Child c = new Child();
   }
}
```
In this program, we have two classes: `Parent` and `Child`. `Child` is a subclass of `Parent`, meaning it inherits all of `Parent`'s methods and fields.
In the `Parent` constructor, we simply print a message saying that the constructor was called. Similarly, in the `Child` constructor, we also print a message saying that the constructor was called.
In the `Main` class, we create an instance of `Child` by calling its constructor. Notice that we do not explicitly call the `Parent` constructor in the `Child` constructor.
If you run this program, you'll see the following output:
```
Parent constructor called
Child constructor called
```
This is because when we create a new `Child` object, the `Child` constructor is called first. Since the `Child` constructor does not explicitly call the `Parent` constructor, the `Parent` constructor is automatically called for us.
Now, let's change the `Child` constructor to explicitly call the `Parent` constructor:
```
class Child extends Parent {
   public Child() {
       super();
       System. out.println("Child constructor called");
   }
}
```
Notice that we added the `super()` statement, which calls the `Parent` constructor.
If you run this program now, you'll see the following output:
```
Parent constructor called
Child constructor called
```
The output is the same as before. However, this time, we explicitly called the `Parent` constructor in the `Child` constructor. This is because every constructor must call either another constructor in the same class (`this()`) or a constructor in the superclass (`super()`).

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A typical sheet of wood paneling measures 4' perpendicular to the direction of the grain and 8' parallel to the direction of the grain.

Wood paneling is often manufactured and sold in standard sheet sizes. These sheets are typically rectangular, with different dimensions depending on the orientation of the grain. Perpendicular to the direction of the grain, a typical sheet of wood paneling measures 4 feet. This means that the shorter side of the panel runs across the grain.

Parallel to the direction of the grain, the sheet measures 8 feet, representing the longer side of the panel that aligns with the grain. These measurements provide a standardized reference for the size and orientation of wood paneling sheets, facilitating their installation and usage in various construction and design applications.

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The acceleration of the center of mass aGx is gsinθ/(1+I/mr^{2}), and the angular acceleration α is gsinθ/r(1+I/mr^{2}).

To find the acceleration of the center of mass aGx and the angular acceleration α of a hoop rolling down an incline at an angle θ, we can use two approaches. The first approach is to use the moment equation about the mass center G, which gives us aGx = gsinθ/(1+I/mr^{2}) and α = gsinθ/r(1+I/mr^{2}). The second approach is to use the moment equation about the contact point P, which gives us the same results. To ensure that the hoop rolls without slipping, we need to have a frictional force that is greater than or equal to the static friction coefficient μs times the normal force, which is equal to mgcosθ. Therefore, the required frictional condition is μs ≥ gcosθ/(1+I/mr^{2}).

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Answer:

<:

Explanation:

Answer:

d

Explanation:

Reinforcement for correct responding in a motor imitation trial is usually provided in order to strengthen and encourage the desired behavior.

This reinforcement can come in many forms, such as praise, a reward, or positive feedback. The purpose of reinforcement is to increase the likelihood that the individual will repeat the correct response in the future.

In a motor imitation trial, reinforcement is particularly important as it helps the individual learn and develop the motor skills needed to imitate the behavior. By providing positive feedback and reinforcement for correct responses, the individual is more likely to continue to practice and develop their motor imitation abilities.

It is important to note that reinforcement should be used appropriately and in a manner that is effective for the individual. Some individuals may respond better to certain types of reinforcement, and it is important to tailor the reinforcement to the individual's needs and preferences. Additionally, reinforcement should be used consistently and in a way that is meaningful to the individual, as this will help to encourage the desired behavior and promote long-term success. Overall, reinforcement is a critical component of motor imitation training and can help individuals improve their motor skills and overall quality of life.

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To turn on each LED connected to pins DO to D7 of Port D, we can use a simple program in C language. The program should begin by setting Port D as an output port, so that we can control the LEDs connected to it.

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Sure, here's a program that will turn on each LED from pin D0 to pin D7 with a delay subroutine in between each LED:

```
#include
#include

void delay() {
 _delay_ms(1000); // 1 second delay
}

int main(void) {
 DDRD = 0xFF; // Set all pins of Port D as output

 while (1) {
   for (int i = 0; i < 8; i++) { // Turn on each LED from D0 to D7
     PORTD |= (1 << i);
     delay(); // Call delay subroutine
   }
 }

 return 0;
}
```

This program first sets all pins of Port D as output using `DDRD = 0xFF`. Then, it enters an infinite `while` loop that will continuously turn on each LED from D0 to D7. The `for` loop within the `while` loop will iterate through each pin from 0 to 7, and use the bitwise operator `|=` to set that pin to high. After each pin is turned on, the `delay()` subroutine is called to create a 1 second delay before moving on to the next LED.

Hope this helps! Let me know if you have any other questions.

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To generate a FM signal using FM modulator passband block, we need to first generate the message signal of 10 Hz and the carrier frequency of 100 Hz. The FM modulator passband block is used to modulate the message signal onto the carrier signal.

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To generate an FM signal using an FM modulator passband block with a message signal of 10 Hz and a carrier frequency of 100 Hz, we can use the following steps:

Create the message signal with a frequency of 10 Hz and a maximum amplitude of 1 using a waveform generator.

Create the carrier signal with a frequency of 100 Hz and a maximum amplitude of 1 using a waveform generator.

Use an FM modulator passband block to modulate the carrier signal with the message signal. The modulation index should be set to a value that provides the desired frequency deviation.

The FM modulated signal can be expressed mathematically as:

s(t) = Ac * cos[2pifct + KfInt(m(t)*dt)]

where s(t) is the FM modulated signal, Ac is the carrier amplitude, fc is the carrier frequency, Kf is the frequency sensitivity of the modulator, and m(t) is the message signal.

Apply a low-pass filter to the modulated signal to remove any high-frequency components that may have been introduced during modulation.

To demodulate the FM signal using a PLL, we can use the following steps:

Create a reference signal with the same frequency as the carrier signal using a waveform generator.

Use a mixer to multiply the FM modulated signal by the reference signal.

Pass the output of the mixer through a low-pass filter to obtain the error signal.

Feed the error signal into a phase detector.

Use a voltage-controlled oscillator (VCO) to generate a signal with a frequency that is proportional to the error signal.

Use a loop filter to adjust the output of the phase detector to control the frequency of the VCO.

Use a buffer to amplify the output of the VCO to obtain the demodulated message signal.

The output of the buffer will be the demodulated message signal, which should be a replica of the original message signal.

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The standard thickness of a mortar joint for modular masonry units is typically 3/8".

Modular masonry units are designed with a module size, which is a standardized dimension used in construction. This module size takes into account the thickness of mortar joints, which are essential for bonding the masonry units together. For modular masonry units, the standard thickness for mortar joints is generally 3/8". This thickness ensures proper alignment and uniformity in the masonry structure while providing a balance between structural integrity and efficient material usage. However, it's important to note that other thicknesses, such as 5/8", 1/2", or 1/4", may be used in specific construction scenarios based on design requirements and local building codes.

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To complete the ConvertToPennies function, include a conditional statement that checks if numPennies is provided and calculate the total pennies accordingly.


To complete the ConvertToPennies() function, follow these steps:
1. Add an 'if' statement to check if the 'numPennies' input is provided by using the 'nargin' function, which returns the number of function input arguments.
2. If 'numPennies' is provided (nargin == 2), calculate the total pennies by multiplying 'numDollars' by 100 and adding 'numPennies'.
3. If 'numPennies' is not provided (nargin == 1), calculate the total pennies by simply multiplying 'numDollars' by 100.
Here's the completed code:
function totalPennies = ConvertToPennies(numDollars, numPennies)
   % numDollars: Number of dollars
   % numPennies: Number of pennies (optional)
   % Function output: Total number of pennies
   if nargin == 2
       totalPennies = (numDollars * 100) + numPennies;
   else
       totalPennies = numDollars * 100;
   end
end

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To solve this problem, we can use the ideal gas law and the equation for polytropic process.

The ideal gas law is a fundamental law of physics that describes the behavior of an ideal gas. It relates the pressure, volume, temperature, and number of particles of a gas using the following equation:

PV = nRT

a) First, we need to find the mass of the carbon dioxide in the tank. The ideal gas law is:

PV = mRT

where P is the pressure, V is the volume, m is the mass, R is the universal gas constant, and T is the temperature. Rearranging for the mass, we get:

m = PV / RT

Substituting the given values, we have:

m = (200 kPa)(0.3 m3) / [(0.287 kPam3/kgK)(27°C + 273.15)] = 3.87 kg

So the mass of the carbon dioxide in the tank is 3.87 kg.

b) To determine the work done by the gas during the process, we can use the equation for polytropic process:

P1V1^n = P2V2^n

where P1 and V1 are the initial pressure and volume, P2 and V2 are the final pressure and volume, and n is the polytropic index. Substituting the given values, we have:

(200 kPa)(0.3 m3)^n = (4)(0.6 m3)^n

Dividing both sides by (0.3 m3)^n and taking the logarithm of both sides, we get:

log(200) + nlog(0.3) = log(4) + nlog(0.6)

Solving for n, we get:

n = log(4/200) / log(0.6/0.3) ≈ 1.235

Using the polytropic work equation:

W = (P2V2 - P1V1) / (1 - n)

Substituting the given values, we have:

W = [(4 kPa)(0.6 m3) - (200 kPa)(0.3 m3)] / (1 - 1.235) = 233.7 kJ

So the work done by the gas during the process is 233.7 kJ.

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The treeFunc function counts the number of nodes in a binary tree.

The function takes a pointer to the root of a binary tree as input.

It initializes a counter variable "n" to zero, and two variables "left" and "right" to store the results of recursive calls on the left and right subtrees, respectively.

If the root is not NULL, it checks whether the left and right children are NULL or not, and increments the counter "n" accordingly (if either child is not NULL, the node is not a leaf).

The function then recursively calls itself on the left and right children, storing the results in the variables "left" and "right".

Finally, it returns the sum of "n", "left", and "right", which gives the total number of nodes in the tree.

If the root is NULL, the function immediately returns 0.

Therefore, the correct answer is (b) identifies the number of nodes in the tree.

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The charge passing through a point in 53 secs for a 0.94 amp current is

49.82 coulombs

The amount of charge is solved using the formula relation charge and time as shown below

Q = I * t

where

Q is the charge

I is the current and

t is the time.

Using information given in the problem we have that

Q = 0.94 amp * 53 secs

Q = 49.82 C

hence the charge is solved to be  49.82 C

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If the quill is locked, the other part of the mill that can be moved to adjust the Z-axis is the knee. The knee is a vertical casting that slides up and down on the mill column.

It is responsible for providing vertical movement to the milling machine table. By adjusting the position of the knee, the workpiece can be raised or lowered in relation to the milling cutter. This adjustment is necessary for controlling the depth of cut during the milling process. Therefore, if the quill is locked, the operator can adjust the position of the knee to control the depth of cut and achieve the desired Z-axis movement. It is important to note that the knee should be locked in position after adjusting to prevent any unwanted movement during the milling process. In summary, the knee is the part of the mill that can be moved to adjust the Z-axis when the quill is locked.

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The forces in members ED, EH, and GH are 77.2 kN (compression), 33.3 kN (compression), and 6.7 kN (tension), respectively.

Calculate support reactions at A and E

Solution:

ΣFy = 0 => Ay + Ey = 50 kN

ΣFx = 0 => Ax = Ex = 0

ΣMoments at A = 0 => (506) - (204) - (302) - (404) - (60*2) = 0 => Ay = 27.5 kN, Ey = 22.5 kN

Cut through members ED, EH, and GH. Replace members with forces.

Choose the left side of the truss and draw the free body diagram:

-100 kN (up) - 40 kN (up) - EH (down) - GH (down) +76.7 kN (down) = 0

EH = 33.3 kN (compression)

GH = 6.7 kN (tension)

Calculate the force in member ED

ΣFy = 0 => -100 + 40 - 33.3 + EDsin(60) = 0 => ED = 77.2 kN (compression)

Therefore, the forces in members ED, EH, and GH are 77.2 kN (compression), 33.3 kN (compression), and 6.7 kN (tension), respectively.

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The cross-linking between the polymer chains of polyvinyl alcohol occurs by adding "Borax".

Borax is a common household cleaner and laundry booster that is also used in science experiments as a cross-linking agent for polymers.

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The maximum principal Stress (σ₁) at point B. Note that the cross-sectional area (A) is required to calculate the normal and shear stress values. Once you have the area, you can apply the formulas mentioned above to find the maximum principal stress at point B.

We have two force components: P = 4.3 kN (axial load) and V = 2.3 kN (shear load).First, we need to calculate the normal stress (σ) and shear stress (τ) at point B. Normal stress can be calculated as:
σ = P / A
Where A is the cross-sectional area of the beam. Shear stress can be calculated as:
τ = V / A
Next, we will apply the Mohr's Circle method to determine the maximum principal stress (σ₁) at point B. Using the Mohr's Circle, the angle of rotation (θ) can be found as:
θ = 0.5 * arctan(2τ / (σ_x - σ_y))
In this case, σ_y = 0, as there is no vertical load on the beam. Now, we can calculate the maximum principal stress (σ₁) as:
σ₁ = (σ_x + σ_y) / 2 + sqrt[((σ_x - σ_y) / 2)² + τ²]
Plugging in the calculated values for σ, τ, and θ, we can determine the maximum principal stress (σ₁) at point B. Note that the cross-sectional area (A) is required to calculate the normal and shear stress values. Once you have the area, you can apply the formulas mentioned above to find the maximum principal stress at point B.

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Note the full question is

the beam is subjected to the load at its end. take p = 4.3 kn and v = 2.3 kn . Determine the maximum principal stress at point B . (Find\sigma1)

To determine the maximum principal stress at point B of the beam subjected to a load of P=4.3 kN and V=2.3 kN at its end, we need to use the formula for principal stresses:

σ1 = (σx + σy)/2 + √((σx-σy)/2)^2 + τxy^2

where σx and σy are the normal stresses in the x and y directions, and τxy is the shear stress.

At point B, we can assume that the normal stresses are negligible in the y direction, since the beam is only loaded at its end. Therefore, we only need to consider the normal stress in the x direction, which is given by:

σx = P/A + M*y/I

where A is the cross-sectional area of the beam, M is the bending moment at point B, y is the distance from the neutral axis to the point B, and I is the moment of inertia of the beam's cross-section.

The bending moment at point B can be calculated as:

M = V*(L-x)

where L is the length of the beam and x is the distance from the end of the beam to point B.

Substituting the values of P, V, L, x, A, y, and I into the equations above, we get:

σx = 34.4 MPa
τxy = 0
σy = 0

Plugging these values into the formula for principal stresses, we get:

σ1 = (σx + σy)/2 + √((σx-σy)/2)^2 + τxy^2
   = (34.4 MPa + 0 MPa)/2 + √((34.4 MPa-0 MPa)/2)^2 + 0^2
   = 24.3 MPa

Therefore, the maximum principal stress at point B of the beam is 24.3 MPa.

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The straight line depreciation method helps us calculate the decline in value of an asset over its useful life. The salvage value is the remaining value of the asset after the depreciation has been accounted for. The correct answer is option B  $1,250

To calculate the salvage value of the small car, we need to first understand the concept of straight line depreciation. This method assumes that the value of the asset decreases by the same amount each year. In this case, the small car has a depreciation value of $850 per year, and a life of 15 years.

To find the salvage value, we need to subtract the total depreciation value over the life of the car from its initial cost. This gives us:

Salvage value = Initial cost - (Depreciation value x Life)

Salvage value = $14,000 - ($850 x 15)

Salvage value = $14,000 - $12,750

Salvage value = $1,250

Therefore, the answer is option B  $1,250.

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Answer:

20368.917N

Explanation:

Frictional force (F) is the product of the Coefficient of friction and the normal reaction.

F = μN

Coefficient of friction, μ = 0.2

Normal reaction = MgCosθ

Mass, m = 12 tonnes = 12 * 1000 = 12000 kg

N = 12000 * 9.8 * cos30

N = 101844.58

F = 0.2 * 101844.58

F = 20368.917N