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Mashutka [201]
3 years ago
6

What is a voltage divider circuit and how do you calculate the voltage across one element in a series

Engineering
1 answer:
Rama09 [41]3 years ago
6 0
Sorry I don’t know myself
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Four race cars are traveling on a 2.5-mile tri-oval track. The four cars are traveling at constant speeds of 195 mi/h, 190 mi/h,
Snezhnost [94]

Answer:

Explanation:

1) The number of times, the car with the speed of  195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

1) The number of times, the car with the speed of  195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

=187.5 mph

2)  There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt in that condition will be calculated as

Vt = 195+190+185+180/4

  = 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

2)  There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt, in that condition will be calculated as

Vt = 195+190+185+180/4

  = 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

4 0
4 years ago
3.24 Program: Drawing a half arrow (Java) This program outputs a downwards facing arrow composed of a rectangle and a right tria
Lostsunrise [7]

Answer:

Here is the JAVA program:

import java.util.Scanner; // to get input from user

public class DrawHalfArrow{ // start of the class half arrow

public static void main(String[] args) { // starts of main() function body

    Scanner scnr = new Scanner(System.in); //reads input

int arrowBaseHeight = 0; // stores the height of arrow base

int arrowBaseWidth  = 0; // holds width of arrow base

int arrowHeadWidth = 0; // contains the width of arrow head

// prompts the user to enter arrow base height, width and arrow head width

System.out.println("Enter arrow base height: ");

arrowBaseHeight = scnr.nextInt(); // scans and reads the input as int

System.out.println("Enter arrow base width: ");

arrowBaseWidth = scnr.nextInt();

/* while loop to continue asking user for an arrow head width until the value entered is greater than the value of arrow base width */

while (arrowHeadWidth <= arrowBaseWidth) {

    System.out.println("Enter arrow head width: ");

    arrowHeadWidth = scnr.nextInt(); }

//start of the nested loop

//outer loop iterates a number of times equal to the height of the arrow base

 for (int i = 0; i < arrowBaseHeight; i++) {

//inner loop prints the stars asterisks

      for (int j = 0; j <arrowBaseWidth; j++) {

          System.out.print("*");        } //displays stars

          System.out.println();          }

//temporary variable to hold arrowhead width value

int k = arrowHeadWidth;

//outer loop to iterate no of times equal to the height of the arrow head

for (int i = 1; i <= arrowHeadWidth; i++)

{     for(int j = k; j > 0; j--)     {//inner loop to print stars

       System.out.print("*");    } //displays stars

   k = k - 1;

   System.out.println(); } } } // continues to add more asterisks for new line

Explanation:

The program asks to enter the height of the arrow base, width of the arrow base and the width of arrow head. When asking to enter the width of the arrow head, a condition is checked that the arrow head width arrowHeadWidth should be less than or equal to width of arrow base arrowBaseWidth. The while loop keeps iterating until the user enters the arrow head width larger than the value of arrow base width.

The loop is used to output an arrow base of height arrowBaseHeight. So point (1) is satisfied.

The nested loop is being used which as a whole outputs an arrow base of width arrowBaseWidth. The inner loop draws the stars and forms the base width of the arrow, and the outer loop iterates a number of times equal to the height of the arrow. So (2) is satisfied.

A temporary variable k is used to hold the original value of arrowHeadWidth so that it keeps safe when modification is done.

The last nested loop is used to output an arrow head of width arrowHeadWidth. The inner loop forms the arrow head and prints the stars needed to form an arrow head. So (3) is satisfied.

The value of temporary variable k is decreased by 1 so the next time it enters  the nested for loop it will be one asterisk lesser.

The screenshot of output is attached.

8 0
4 years ago
Engine oil flows through a 25‐mm‐diameter tube at a rate of 0.5 kg/s. The oil enters the tube at a temperature of 25°C, while th
Elodia [21]

Answer:

a) the log mean temperature difference (Approx. 64.5 deg C)

b) the rate of heat addition into the oil.

The above have been solved for in the below workings

Explanation:

3 0
4 years ago
4. Which of the following are used for vehicle and/or major component identification?
AleksAgata [21]

Option C

VIN (vehicle identification number) are used for vehicle and/or major component identification

<h3><u>Explanation:</u></h3>

American automobile producers have practiced a vehicle identification number (V.I.N.) to name and recognize motor vehicles. They serve to retain an indication of obstacles, ownership exchanges, and prevent crime. This practice also posted the superimposed major component parts and replacement parts to be noted for each of the classes of vehicles.

Depending on the transport major parts may be marked with the vehicle’s VIN. The component part labels are composed of the alike material as the Federal Safety Certification Labels and are also planned to self destruct if excluded.

4 0
4 years ago
Consider a solid round elastic bar with constant shear modulus, G, and cross-sectional area, A. The bar is built-in at both ends
Ierofanga [76]

Answer:

\t(x)_{max} =\dfrac{p\times L}{2\times \pi}

Explanation:

Given that

Shear modulus= G

Sectional area = A

Torsional load,

t(x) = p sin( \frac{2\pi}{ L} x)

For the maximum value of internal torque

\dfrac{dt(x)}{dx}=0

Therefore

\dfrac{dt(x)}{dx} = p cos( \frac{2\pi}{ L} x)\times \dfrac{2\pi}{L}\\ p cos( \frac{2\pi}{ L} x)\times \dfrac{2\pi}{L}=0\\cos( \frac{2\pi}{ L} x)=0\\ \dfrac{2\pi}{ L} x=\dfrac{\pi}{2}\\\\x=\dfrac{L}{4}

Thus the maximum internal torque will be at x= 0.25 L

t(x)_{max} = \int_{0}^{0.25L}p sin( \frac{2\pi}{ L} x)dx\\t(x)_{max} =\left [p\times \dfrac{-cos( \frac{2\pi}{ L} x)}{\frac{2\pi}{ L}}  \right ]_0^{0.25L}\\t(x)_{max} =\dfrac{p\times L}{2\times \pi}

6 0
3 years ago
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