Answer:

**Answer:**

- branch 1: i = 25.440∠-32.005°; pf = 0.848 lagging
- branch 2: i = 21.466∠63.435°; pf = 0.447 leading
- total: i = 31.693∠10.392° leading; pf = 0.984 leading

**Explanation:**

To calculate the currents in the parallel branches, we need to know the impedance of each branch. That will be the sum of the resistance and reactance.

The inductive reactance is ...

The capacitive reactance is ...

__Branch 1__

The impedance of branch 1 is ...

Z1 = 8 +j4.99984 Ω

so the current is ...

** I1 =** V/Z = 240/(8 +j4.99984) ≈ **25.440∠-32.005°**

The **power factor** is cos(-32.005°) ≈ **0.848 (lagging)**

__Branch 2__

The impedance of branch 2 is ...

Z2 = 5 -j10 Ω

so the current is ...

** I2 = **240/(5 +j10) ≈ **21.466∠63.435°**

The **power factor** is cos(63.436°) ≈ **0.447 (leading)**

__Total current__

The total current is the sum of the branch currents. A suitable calculator can add these vectors without first converting them to rectangular form.

It = I1 +I2 = (21.573 -j13.483) +(9.6 +j19.2)

** It ≈** 31.173 +j5.717 ≈ **31.693∠10.392°**

The **power factor** for the circuit is cos(10.392°) ≈ **0.984 (leading)**

__

The phasor diagram of the currents is attached.

_____

*Additional comment*

Given two vectors, their sum can be computed several ways. One way to compute the sum is to use the Law of Cosines. In this application, the angle between the vectors is the supplement of the difference of the vector angles: 84.560°.

Required information NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. The gas-turbine cycle of a combined gas–steam power plant has a pressure ratio of 8. Air enters the compressor at 290 K and the turbine at 1400 K. The combustion gases leaving the gas turbine are used to heat the steam at 15 MPa to 450°C in a heat exchanger. The combustion gases leave the heat exchanger at 247°C. Steam expands in a high-pressure turbine to a pressure of 3 MPa and is reheated in the combustion chamber to 500°C before it expands in a low-pressure turbine to 10 kPa. The mass flow rate of steam is 17 kg/s. Assume isentropic efficiencies of 100 percent for the pump, 85 percent for the compressor, and 90 percent for the gas and steam turbines.Determine the rate of total heat input.

Q/For the circuit showm bellow:a) find the mathematical expression for the transient behavior of ve and ic after closing the switeh,b) sketch vc and ic.

An environmental engineer is considering three methods for disposing of a nonhazardous chemical sludge: land application, fluidized-bed incineration, and private disposal contract. The estimates for each method are below. (a) Determine which has the least cost on the basis of an annual worth comparison at 10% per year. (b) Determine the equivalent present worth value of each alternative using its AW value

Assume that in orthogonal cutting the rake angle is 15° and the coefficient of friction is 0.15. a. Determine the percentage change in chip thickness when the coefficient of friction is doubled. Justify your answer. b. Determine the percentage change in chip thickness when the rake angle is increased to 25o . Justify your answer.

A Coca Cola can with diameter 62 mm and wall thickness 300 um has an internal pressure of 100 kPa. Calculate the principal stresses at a point on the cylindrical surface of the can far from its ends.

Q/For the circuit showm bellow:a) find the mathematical expression for the transient behavior of ve and ic after closing the switeh,b) sketch vc and ic.

An environmental engineer is considering three methods for disposing of a nonhazardous chemical sludge: land application, fluidized-bed incineration, and private disposal contract. The estimates for each method are below. (a) Determine which has the least cost on the basis of an annual worth comparison at 10% per year. (b) Determine the equivalent present worth value of each alternative using its AW value

Assume that in orthogonal cutting the rake angle is 15° and the coefficient of friction is 0.15. a. Determine the percentage change in chip thickness when the coefficient of friction is doubled. Justify your answer. b. Determine the percentage change in chip thickness when the rake angle is increased to 25o . Justify your answer.

A Coca Cola can with diameter 62 mm and wall thickness 300 um has an internal pressure of 100 kPa. Calculate the principal stresses at a point on the cylindrical surface of the can far from its ends.

**Answer:**

The maximum load (in N) that may be applied to a specimen with this cross sectional area is

**Explanation:**

We know that the stress at which plastic deformation begins is 267 MPa.

We are going to assume that the stress is homogeneously distributed. As a definition, we know that stress (P) is force (F) over area(A), as follows:

(equation 1)

We need to find the maximum load (or force) that may be applied before plastic deformation. And the problem says that the maximum stress before plastic deformation is 267 MPa. So, we need to find the value of load when P= 267 MPa.

Before apply the equation, we need to convert the units of area in . So,

And then, from the equation 1,

**Answer:**

The code is attached.

**Explanation:**

I created a string **s** including 6 colors with spaces in between. Then I converted the string into a list **x** by using **split()** method. I used three different methods for removing elements from the list. These methods are **remove(), pop() and del**.

Then I used methods **append(), insert() and extend() **for adding elements to the list.

Finally I converted list into a string using **join() **and adding space in between the elements of the list.

from the MARIE instruction set architecture. Then write the corresponding signal sequence to perform these micro-operations and to reset the clock cycle counter.

You may refer to the provided "MARIE Architecture and Instruction Set" file in the Front Matter folder.

**Answer:**

So these are the RTL representation:

MAR<----X

MBR<-----AC

M[MAR]<------MBR

Control signal sequence are:

P3T0:MAR<----X

P2P3 P4T1:MBR<-----AC

P0P1 P3T2:M[MAR]<------MBR

**Explanation:**

STORE X instruction is used for storing the value of AC to the memory address pointed by X. This operation can be done by using the Register Transfers at System Level and this can be represented by using a notation called Register Transfer Language RTL. Let us see what are the register transfer operations happening at the system level.

1. First of all the address X has to be tranfered on to the Memory Address Register MAR.

MAR<----X

2. Next we have to tranfer the contents of AC into the Memory Buffer Register MBR

MBR<-----AC

3. Store the MBR into memory where MAR points to.

M[MAR]<------MBR

So these are the RTL representation:

MAR<----X

MBR<-----AC

M[MAR]<------MBR

Control signal sequence are:

P3T0:MAR<----X

P2P3 P4T1:MBR<-----AC

P0P1 P3T2:M[MAR]<------MBR

(B) Because rotating shafts are used in engineering applications

(C) We don't care - simply an academic exercise.

(D) Because it can determine G and inform us of a materials ability to resist shear deformation

**Answer:**

(A) Because the angle of twist of a material is often used to predict its shear toughness

**Explanation:**

In engineering, torsion is the solicitation that occurs when a moment is applied on the longitudinal axis of a construction element or mechanical prism, such as axes or, in general, elements where one dimension predominates over the other two, although it is possible to find it in diverse situations.

The torsion is characterized geometrically because any curve parallel to the axis of the piece is no longer contained in the plane initially formed by the two curves. Instead, a curve parallel to the axis is twisted around it.

The general study of torsion is complicated because under that type of solicitation the cross section of a piece in general is characterized by two phenomena:

1- Tangential tensions appear parallel to the cross section.

2- When the previous tensions are not properly distributed, which always happens unless the section has circular symmetry, sectional warps appear that make the deformed cross sections not flat.

**Answer:**

**Explanation:**

The specific heat for watermelon above freezing point is . The heat liberated by the watermelon to cool down to 8°C is:

The heat absorbed by the household refrigerator is:

Time needed to cool the watermelons is:

**Answer:**

import java.util.Scanner;

public class InputExample {

public static void main(String[] args) {

Scanner scnr = new Scanner(System.in);

int birthMonth;

int birthYear;

birthMonth = scnr.nextInt();

birthYear = scnr.nextInt();

System.out.println(birthMonth+"/"+birthYear);

}

}

**import java**.util.Scanner;

public class InputExample {

public static void main(**String**[] args) {

Scanner scnr = new Scanner(System.in);

int birthMonth;

int birthYear;

// Get input values for **birthMonth **and birthYear from the user

birthMonth = scnr.nextInt();

birthYear = scnr.nextInt();

// Output the month, a slash, and the year

System.out.println(birthMonth + "/" + birthYear);

}

}

When the program is tested with inputs 1 2000, the **output **will be:

**1/2000**

And when tested with **inputs **5 1950, the output will be:

**5/1950**

Know more about **java program:**

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