A resistance of 30 ohms is placed in a circuit with a 90 volt battery. What current flows in the circuit?

How nany degrées is the included angle of General Purpose Acme threads? A. 60 B. 29 c. 14.5 D. 10

horrorfan [7]

Answer:

B.29

Explanation:

In general purpose acme thread:

Nominal depth of thread=0.5 $ltimes Pitch$

Included angle =29 degrees

Generally Acme thread are following types

1.General purpose(G) Acme

2.Centralizing(C) Acme

3. Stub Acme

Centralizing(C) Acme threads have tighter tolerance during manufacturing as compare to General purpose(G) Acme threads.

8 0

3 years ago

Air enters an adiabatic compressor through 0:5m2 opening and exhausts through a 0:2m2 opening. The inlet conditions of air are 2

gladu [14]

Answer:

i) 43.55 kg/s

ii) 40 m/s

iii) -199.32 KW

Explanation:

To resolve the above question we have to make some assumptions :

mass flow through the system is constant
The only interactions that are between the system and the surrounding are work and heat
The fluid is uniform

i) first we have to determine the mass flow rate of the air

M = $(\frac{P1}{RT1} )A1v1$

= $( \frac{200}{0.287*325} ) 0.5 * V1$ ---------- (1) hence M = 43.55 kg/s

ii) using this relationship : A1V1 = A2V2 hence V1 = (0.2/0.5) * 100 = 40m/s ( inlet velocity )

input this value into equation 1

iii) Next we will determine the power required to run compressor

attached below

power required = -199.32 KW ( this value indicates that there is power supplied )

8 0

3 years ago

Can anyone solve this for me?sequential circuit ,flipflop

andrezito [222]

Explanation:

We assume the T flip-flop changes state on the rising edge of the clock input.

The first stage is connected to the clock. The second stage clock is connected to the inverse of the Q output of the first stage, so that when the first stage Q makes a 1 to 0 transition, the second stage changes state.

6 0

2 years ago

A capillary tube is immersed vertically in a water container. Knowing that water starts to evaporate when the pressure drops bel

Anuta_ua [19.1K]

Answer:

$d=0.414\times 10^{-4}\ m$

Explanation:

Given that

P = 4 KPa

Contact angle = 6°

Surface tension = 1 N/m

Lets assume that atmospheric pressure = 100 KPa

Lets take that density of water = $1000\ kg/m^3$

So the capillarity rise h

$h=\dfrac{\Delta P}{\rho g}$

$h=\dfrac{100\times 1000-4\times 1000}{1000\times 10}$

h= 9.61 m

We know that for capillarity rise h

$h=\dfrac{2\sigma cos\theta }{r\rho g}$

$r=\dfrac{2\sigma cos\theta }{h\rho g}$

$r=\dfrac{2\times 1 cos4^{\circ} }{9.61\times 1000\times 10}$

$r=0.207\times 10^{-4}\ m$

$d=0.414\times 10^{-4}\ m$

3 0

3 years ago

Write a program that dynamically allocates an array large enough to hold a user-defined number of test scores. Once all the scor

Readme [11.4K]

Answer:

#include <iostream>

#include <iomanip>

using namespace std;

//Function prototypes

void arrSelectSort(double *, int);

double arrAvgScore(double *, int);

int main()

{

//Variables definition

double *TestScores,

total = 0.0,

average;

int numTest,

count;

//Enter the number of test scores you want to get to their average in ascending order

cout << "How many test scores do you wish to enter?";

cin >> numTest;

//Dynamically allocate an array large enough to hold that many scores

TestScores = new double[numTest];

//Get the test scores

cout << "Enter the test scores below.\n";

for (count = 0; count < numTest; count++)

{

//Display score

cout << "Test Score " << (count + 1) << ": ";

cin >> TestScores[count];

}

// Input validation. Only numbers between 0-100

while (numTest<0)

{

cout << "You must enter a scores that non-negative" << endl;

cout << "Please enter a non-negative interger between 0 and 100: ";

cin >> TestScores[count];

}

//Calculate the total test scores

for (count = 0; count < numTest; count++)

{

total += TestScores[count];

}

average = total / numTest;

//Dsiplay the results

cout << fixed << showpoint << setprecision(2);

cout << "The average of all the test score is " << average << endl;

//Free dynamically allocated memory

delete [] TestScores;

TestScores = 0; //make TestScores point to null

//Display the Test Scores in ascending order

cout << "The test scores, sorted in ascending order, are: \n";

system ("pause");

return 0;

}

//Ascending order selection sort

void arrSelectSort(double *arr, int size)

{

int startScan;

double minIndex;

double minElem;

for(startScan = 0; startScan < (size - 1); startScan++)

{

minIndex = startScan;

minElem = arr[startScan];

}

for(int index = startScan + 1; index < size; index++)

{

if (arr[index] < minElem)

{

minElem = arr[index];

minIndex = index;

}

void arrAvgScore (double *arr[], int size)

{

double total = 0;

int numTest;

for (int count = 0; count < numTest; count++)

{

total += numTest[count];

average = total / numTest;

}

6 0

4 years ago