An Opportunity need to have an amount rolled up from a cutom object that i not in a mater-detail relationhip. How can thi be achieved?

Answer:

-2 and 6

Explanation:

Let "x" and "y" be 2 numbers.

The sum of the two numbers is 4. The mathematical expression is:

x + y = 4

y = 4 - x   [1]

The sum of their squares minus three times their product is 76. The mathematical expression is:

x² + y² - 3 x y = 76   [2]

If we substitute [1] in [2], we get:

x² + (4 - x)² - 3 x (4 - x) = 76

x² + 16 - 8 x + x² - 12 x + 3 x² = 76

5 x² - 20 x - 60 = 0

We apply the solving formula for second order equations and we get x₁ = 6 and x₂ = -2.

If we replace these x values in [1], we get:

y₁ = 4 - x₁ = 4 - 6 = -2

y₂ = 4 - x₂ = 4 - (-2) = 6

As a consequence, one of the numbers is 6 and the other is -2.

Answer: the engineers are performing a tensile test

Explanation:

i took the test and got it right

Answer:

d.

Explanation:

Answer:

Explanation:

A scenario is an actual sequence of interactions (i.e., an instance) describing one specific situation; a use case is a general sequence of interactions (i.e., a class) describing all possible scenarios associated with a situation. Scenarios are used as examples and for clarifying details with the client. Use cases are used as complete descriptions to specify a user task or a set of related system features.

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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Solution :

Finding the cohesion of the soil(c) using the relation:

[tex]$c = \frac{q_u}{2}$[/tex]

Here, [tex]$q_u$[/tex] is the unconfined compression strength of the soil;

[tex]$c = \frac{800}{2}$[/tex]

   = 400 psf

∴ The cohesion value is greater than 0

So the use of the angle of internal friction is 0

Referring to the table relation between bearing capacity factors and angle of internal friction.

For the angle of inter friction [tex]$0^\circ$[/tex]

    [tex]$N_c = 5.14$[/tex]

   [tex]$N_q = 1.0$[/tex]

   [tex]$N_r = 0$[/tex]

Therefore,

[tex]$q_{ult} = (400 \times 5.14 )+(110 \times 3 \times 1.0)+(0.5 \times 100 \times 13 \times 0)$[/tex]

     =  2386 psf

∴ Allowable bearing capacity [tex]$q_{a} = \frac{Q_{allow}}{A}$[/tex]

                                                     [tex]$=\frac{30}{B^2}$[/tex]

∴ [tex]$q_a = \frac{q_{ult}}{F.O.S}$[/tex]

  [tex]$\frac{30}{B^2} = \frac{2386}{3}$[/tex]

∴ B = 0.2 ft

Therefore, the dimension of the square footing is 0.2 ft x 0.2 ft

                                                                                 [tex]$=0.04 \ ft^2$[/tex]

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

Answer:

the answer. for this is Crankshaft

Solution :

Given :

[tex]$V_1 = 7 \ m/s$[/tex]

Operation time, [tex]$T_1$[/tex] = 3000 hours per year

[tex]$V_2 = 10 \ m/s$[/tex]

Operation time, [tex]$T_2$[/tex] = 2000 hours per year

The density, ρ = [tex]$1.25 \ kg/m^3$[/tex]

The wind blows steadily. So, the K.E. = [tex]$(0.5 \dot{m} V^2)$[/tex]

                                                             [tex]$= \dot{m} \times 0.5 V^2$[/tex]

The power generation is the time rate of the kinetic energy which can be calculated as follows:

Power = [tex]$\Delta \ \dot{K.E.} = \dot{m} \frac{V^2}{2}$[/tex]

Regarding that [tex]$\dot m \propto V$[/tex]. Then,

Power [tex]$ \propto V^3$[/tex] → Power = constant x [tex]$V^3$[/tex]

Since, [tex]$\rho_a$[/tex] is constant for both the sites and the area is the same as same winf turbine is used.

For the first site,

Power, [tex]$P_1= \text{const.} \times V_1^3$[/tex]

            [tex]$P_1 = \text{const.} \times 343 \ W$[/tex]

For the second site,

Power, [tex]$P_2 = \text{const.} \times V_2^3 \ W$[/tex]

           [tex]$P_2 = \text{const.} \times 1000 \ W$[/tex]

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:)

Answer:

attached below

Explanation:

Given data:

Gaged flow = 52 m^3 / sec

Depth covering drainage area = 1450 km^2

Develop a 24-hour rainfall duration unit hydrograph for the watershed using observed hydorgraph

Runoff flow = gaged flow - base flows

                   = 52 - 52 = 0 m^3/sec

For 18 hours time duration

Direct runoff volume ( Vr ) = ∑ ( QΔt1 )

where Δt = 6

           ∑Q = 3666 m^3/sec

hence Vr = Δt * ∑Q  = 79185600 m^3

Next we will convert the Direct runoff volume to its equivalent depth covering the drainage area

= Vr / drainage area depth

= 79185600 / 1450000000 = 5.46 cm

Next we will find the unit hydrograph flows by applying the relation below

[tex]Q_{1.0cm} = Q_{5.46cm} (\frac{1.0cm}{5.46cm} )[/tex]

where 14m^3/sec = [tex]Q_{5.6cm}[/tex]

input value back to the above relation

[tex]Q_{1.0cm} = 2.57 m^3/sec[/tex]

Attached below is The remaining part of the solution  tabulated below

and A graph of the unit hydorgraph for the given watershed

Note :

Base  flow total = 1560

UH total = 671.30

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:

I dont know the answer either

Explanation:

Answer:

flux

Explanation:

Answer:

Burj Khalifa is located in dubai UAE at over 828m

Explanation:

828 metres

Answer:

The Burji Khalifa, known as the Burj Dubai prior to its inauguration in 2010, is a skyscraper in Dubai, United Arab Emirates. With a total hight of 829.8 m and a roof hight of 828 m, the Burji Khalifa has been the tallest structure and building in the world since its topping out in 2009.

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

Answer:

shrinkae limit = 20.35%

shrinkage ratio = 1.45

Explanation:

1. to get the shrinkage limit we would first calculate the moisture content w.

w = (37-28)/28

= 9/28

= 0.3214

then the formula for shrinkage limit is

[tex][w-\frac{(V-Vd}{wd} ]*100[/tex]

w = 0.3214

V = 19.3

Vd = 16

Wd = 28

when we put these values into the formula:

[tex][0.3214-\frac{(19.3-16)}{28} ]*100\\[/tex]

= 20.35%

2. the shrinkage limit = Wd/V

= 28/19.3

= 1.45

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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Answer:

<:

Explanation:

Answer:

d

Explanation:

scrity añao devid codicie

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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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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Answer:

u need to plan it out

Explanation:

u need to plan it out

Answer:

use the turn 1 degrees option and put a repeat loop on it

Explanation:

u can add sound in ur loop

Answer:

this is your answer. if mistake don't mind.

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.

Solution :

Assuming that the wire has an uniform temperature, the equivalent convective heat transfer coefficient is given as :

[tex]$h_T= \epsilon \sigma (T_s+T_{surr})(T_s^2 +T^2_{surr})$[/tex]

[tex]$h_T= 0.20 \times 5.67 \times 10^{-8} (1473+323)(1473^2 +323^2)$[/tex]

[tex]$h_T=46.3 \ W/m^2 .K$[/tex]

The total heat transfer coefficient will be :

[tex]$h_T=(250+46.3) \ W/m^2 .K$[/tex]

    [tex]$=296.3 \ W/m^2 .K$[/tex]

Now calculating the maximum volumetric heat generation :

[tex]$\dot {q}_{max}=\frac{2h_t}{r_0}(T_s-T_{\infty})$[/tex]

[tex]$\dot {q}_{max}=\frac{2\times 296.3}{0.0005}(1200-50)$[/tex]

        [tex]$= 1.362 \times 10^{9} \ W/m^3$[/tex]

The heat generation inside the wire is given as :

[tex]$\dot{q} = \frac{I^2R}{V}$[/tex]

Here, R is the resistance of the wire

         V is the volume of the wire

∴ [tex]$\dot{q} = \frac{I^2\left( \rho \times \frac{L}{A} \right)}{A \times L}$[/tex]

     [tex]$=\frac{I^2 \rho}{\left(\frac{\pi}{4}D^2 \right)}$[/tex]

where, ρ is the resistivity.

[tex]$I_{max}= \left(\frac{\dot{q}_{max}}{\rho} \right)^{1/2} \times \frac{\pi}{4}D^2$[/tex]

[tex]$I_{max}= \left(\frac{1.36 \times 10^9}{10^{-6}} \right)^{1/2} \times \frac{3.14}{4}(1 \times 10^{-3})^2$[/tex]

        = 28.96 A

Now considering the relation for the current flow through the finite potential difference.

[tex]$E=I_{max} \times R$[/tex]

[tex]$E=I_{max} \times \rho \times \frac{L}{A}$[/tex]

[tex]$L=\frac{AE}{I_{max} \ \rho}$[/tex]

[tex]$L=\frac{\frac{\pi}{4} \times (1 \times 10^{-3})^2 \times 110}{28.96 \times 10^{-6}}$[/tex]

   = 2.983 m

Now calculating the power rating of the heater:

[tex]$P= E \times I_{max}$[/tex]

[tex]$P= 110 \times 28.96}$[/tex]

   = 3185.6 W

   = 3.1856 kW

Answer:

Explanation:

From the given information:

The equation for applied stress can be expressed as:

[tex]\sigma_{app} = \dfrac{\tau_{CRSS}}{cos \phi \ cos \lambda}[/tex]

where;

[tex]\phi[/tex] = angle between the applied stress [100] and [111]

To determine the [tex]\phi[/tex] and [tex]\lambda[/tex] for the system

Using the equation:

[tex]\phi= cos^{-1}\Big [\dfrac{l_1l_2+m_1m_2+n_1n_2}{\sqrt{(l_1^2+m_1^2+n_1^2)(l_2^2+m_2^2+n_2^2)}}\Big][/tex]

for [100]

[tex]l_1 = 1, m_1 = 0, n_1 = 0[/tex]

for [111]

[tex]l_1 = 1 , m_1 = 1, n_1 = 1[/tex]

Thus;

[tex]\phi= cos^{-1}\Big [\dfrac{1*1+0*1+0*1}{\sqrt{(1^2+0^2+0^2)(1^2+1^2+1^2)}}\Big][/tex]

[tex]\phi= cos^{-1}\Big [\dfrac{1}{\sqrt{(3)}}\Big][/tex]

[tex]\phi= 54.74^0[/tex]

To determine  [tex]\lambda[/tex]  for [tex][1 \overline 1 0][/tex]

where;

for [100]

[tex]l_1 = 1, m_1 = 0, n_1 = 0[/tex]

for [tex][1 \overline 1 0][/tex]

[tex]l_1 = 1 , m_1 = -1, n_1 = 0[/tex]

Thus;

[tex]\lambda= cos^{-1}\Big [\dfrac{1*1+0*1+0*0}{\sqrt{(1^2+0^2+0^2)(1^2+(-1)^2+0^2)}}\Big][/tex]

[tex]\phi= cos^{-1}\Big [\dfrac{1}{\sqrt{(2)}}\Big][/tex]

[tex]\phi= 45^0[/tex]

Thus, the magnitude of the applied stress can be computed as:

[tex]\sigma_{app} = \dfrac{\tau_{CRSS}}{cos \phi \ cos \lambda }[/tex]

[tex]\sigma_{app} = \dfrac{2.00}{cos (54.74) \ cos (45) }[/tex]

[tex]\mathbf{\sigma_{app} =4.89 \ MPa}[/tex]