4. Use voltage divider concepts to find the voltages indicated in the following circuits. You may want to use some of your results from problem 1. You may need to use the voltage divider equation more than once.

Answers

Answer 1

Answer:

Hello, because there is not a circuit I'll explain the voltage divider and make an exercise, this way you can solve the problem using the method described here.

Answer with explanation:

A voltage divider uses the voltage distribution among components to find a voltage in a specific element of the circuit. If we have a source V1 connected to impedances Z1 and Z2 in series, we can use a voltage divider to find the voltage across Z1 or Z2 base on their value and the input voltage.

VZ1 = V1*Z1/(Z1+Z2)

VZ2 = V1*Z2/(Z1+Z2)

In the image, to find the voltage Vo across R2 we apply the following equation: Vo = (V1*R2)/(R1+R2).

To solve the exercise in the other image, we need to apply a voltage divider twice:

In-circuit 1 we are asked to find the voltage VAB that falls on R2 and R3 (the same voltage for both resistances because are in parallel), to do so we use a voltage divider using V1, R1 and RT where RT is the equivalent resistance RT = R2//R3 + R4, therefore, for circuit two VAC = (V1*R1)/(R1+RT). After finding VAC we apply voltage divider again to find VAB, see circuit 3, to do so we apply VAB = (VAC*R2//R3)/(R2//R3 + R4) = (VAC*R2//R3)/(RT)

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At its current short-run level of production, a firm's average variable costs equal $20 per unit, and its average fixed costs equal $30 per unit. Its total costs at this production level equal $2,500. A. What is the firm's current output level? B. What are its total variable costs at this output level? C. What are its total fixed costs?

Answers

Answer and Explanation:

The computation is shown below

Total average cost + total variable cost = total cost

Let number of output be x

So,

Total fixed average cost = x × $30

Total variable cost = x × $15

Total cost = $2,500

Therefore,

$20 × x + $30 × x = $2,500

50 × x = $2,500

x = 50

Now the total variable cost is

= 50 × $20

= $1,000

And, the fixed cost is

= 50 × $30

= $1,500

Answer:

12 4

Explanation:

because the production average is variable

Which statement best describes how power and work are related?O A. Power is the ability to do more work with less force.
O B. Power is a measure of how quickly work is done.
O C. Power and work have the same unit of measurement
O D. Power is the amount of work needed to overcome friction.
Pls answer quick

Answers

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Make a copy of the pthreads_skeleton.cpp program and name it pthreads_p2.cpp Modify the main function to implement a loop that reads 10 integers from the console (user input) and stores these numbers in a one-dimensional (1D) array (this code will go right after the comment that says ""Add code to perform any needed initialization or to process user input""). You should use a global array for this.

Answers

Answer:

The solution code is as follows:

#include <iostream>
using namespace std;
int main()
{
int myArray [10] = {};
int i;
for( i = 0; i < 10; i++ )
{
cout <<"Enter an integer: ";
cin>> myArray[i];
}
}

Explanation:

Firstly, we initialize a 10-elements array, myArray (Line 7) with no values.

Next, we create a for-loop (Line 10). Within the loop, we prompt user to enter an integer and assign the input value to the current element of myArray (Line 12-13).

The dam cross section is an equilateral triangle, with a side length, L, of 50 m. Its width into the paper, b, is 100 m. The dam material has a specific gravity, SG, of 3.1. You may assume that the dam is loosely attached to the ground at its base, though there is significant friction to keep it from sliding.Is the weight of the dam sufficient to prevent it from tipping around its lower right corner?

Answers

Answer:

Explanation:

In an equilateral trinagle the center of mass is at 1/3 of the height and horizontally centered.

We can consider that the weigth applies a torque of T = W*b/2 on the right corner, being W the weight and b the base of the triangle.

The weigth depends on the size and specific gravity.

W = 1/2 * b * h * L * SG

Then

Teq = 1/2 * b * h * L * SG * b / 2

Teq = 1/4 * b^2 * h * L * SG

The water would apply a torque of elements of pressure integrated over the area and multiplied by the height at which they are apllied:

$T1 = \int\limits^h_0 {p(y) * sin(30) * L * (h-y)} \, dy$

The term sin(30) is because of the slope of the wall

The pressure of water is:

p(y) = SGw * (h - y)

Then:

$T1 = \int\limits^h_0 {SGw * (h-y) * sin(30) * L * (h-y)} \, dy$

$T1 = \int\limits^h_0 {SGw * sin(30) * L * (h-y)^2} \, dy$

$T1 = SGw * sin(30) * L * \int\limits^h_0 {(h-y)^2} \, dy$

$T1 = SGw * sin(30) * L * \int\limits^h_0 {h^2 - 2*h*y + y^2} \, dy$

T1 = SGw * sin(30) * L * (h^2*y - h*y^2 + 1/3*y^3)(evaluated between 0 and h)

T1 = SGw * sin(30) * L * (h^2*h - h*h^2 + 1/3*h^3)

T1 = SGw * sin(30) * L * (h^3 - h^3 + 1/3*h^3)

T1 = 1/3 * SGw * sin(30) * L * h^3

To remain stable the equilibrant torque (Teq) must be of larger magnitude than the water pressure torque (T1)

1/4 * b^2 * h * L * SG > 1/3 * SGw * sin(30) * L * h^3

In an equilateral triangle h = b * cos(30)

1/4 * b^3 * cos(30) * L * SG > 1/3 * SGw * sin(30) * L * b^3 * (cos(30))^3

SG > SGw * 4/3* sin(30) * (cos(30))^2

SG > 1/2 * SGw

For the dam to hold, it should have a specific gravity of at leas half the specific gravity of water.

This is avergae specific gravity, including holes.

can anyone help me with this please.i have the current and pf for branch 1 and 2 but cant figure out the total current, pf and angle. thanks

Answers

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 ...

$X_L=j\omega L=j2\pi fL=j100\pi\cdot 15.915*10^(-3)\approx j4.99984\,\Omega$

The capacitive reactance is ...

$X_C=(1)/(j\omega C)=(-j)/(100\pi\cdot 318.31*10^(-6)F)\approx -j10.00000\,\Omega$

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°.

A furnace is shaped like a long equilateral triangular duct where the width of each side is 2 m. Heat is supplied from the base surface, whose emissivity is ε1 = 0.8, at a rate of 800 W/m2 while the side surfaces, whose emissivities are 0.5, are maintained at 500 K. Neglecting the end effects, determine the temperature of the base surface. Can you treat this geometry as a two-surface enclosure?