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trapecia [35]
3 years ago
9

What is the definition of a duty cycle?

Engineering
1 answer:
ira [324]3 years ago
5 0

Answer:

D=\frac{PW}{T}*100

Explanation:

In electrical terms, is the ratio of time in which a load or circuit is ON compared to the time in which the load or circuit is OFF.

The duty cycle or power cycle, is expressed as a percentage of the activation time. For example, a 70% duty cycle is a signal that 70% of the time is activated and the other 30% disabled. Its equation can be expressed as:

D=\frac{PW}{T}*100

Where:

D=Duty\hspace{3}Cycle

PW=Pulse\hspace{3}Active\hspace{3}Time

T=Period\hspace{3}of\hspace{3}the\hspace{3}Signal

Here is a picture that will help you understand these concepts.

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11. Ten bagels is what percentage of a dozen bagels?
IRINA_888 [86]

Answer:

the answer is d 83.3%

Explanation:

3 0
3 years ago
In this exercise, you will write a Point structure that represents a space in two-dimensional space. This Point should have both
Afina-wow [57]

Answer:

Check the explanation

Explanation:

Points to consider:

We need to take the input from the user

We need to find the manhatan distance and euclidian using the formula

(x1, y1) and (x2, y2) are the two points

Manhattan:

|x_1 - x_2| + |y_1 - y_2|

Euclidian Distance:

\sqrt{(x1 - yl)^2 + (x2 - y2)^2)}

Code

#include<stdio.h>

#include<math.h>

struct Point{

  int x, y;

};

int manhattan(Point A, Point B){

  return abs(A.x - B.x) + abs(A.y- B.y);

}

float euclidean(Point A, Point B){

  return sqrt(pow(A.x - B.x, 2) + pow(A.y - B.y, 2));

}

int main(){

  struct Point A, B;

  printf("Enter x and Y for first point: ");

  int x, y;

  scanf("%d%d", &x, &y);

  A.x = x;

  A.y = y;

  printf("Enter x and Y for second point: ");

  scanf("%d%d", &x, &y);

  B.x = x;

  B.y = y;

  printf("Manhattan Distance: %d\n", manhattan(A, B));

  printf("Euclidian Distance: %f\n", euclidean(A, B));

 

}

Sample output

8 0
3 years ago
A triangular plate with a base 5 ft and altitude 3 ft is submerged vertically in water. If the base is in the surface of water,
Scorpion4ik [409]

Answer:

Hydrostatic force = 41168 N

Explanation:

Complete question

A triangular plate with a base 5 ft and altitude 3 ft is submerged vertically in water  so that the top is 4 ft below the surface. If the base is in the surface of water, find the force against onr side of the plate. Express the hydrostatic force against one side of the plate as an integral and evaluate it. (Recall that the weight density of water is 62.5 lb/ft3.)

Let "x" be the side length submerged in water.

Then

w(x)/base = (4+3-x)/altitude

w(x)/5 = (4+3-x)/3

w(x) = 5* (7-x)/3

Hydrostatic force = 62.5 integration of  x * 4 * (10-x)/3 with limits from 4 to 7

HF = integration of 40x - 4x^2/3

HF = 20x^2 - 4x^3/9 with limit 4 to 7

HF = (20*7^2 - 4*7^(3/9))- (20*4^2 - 4*4^(3/9))

HF = 658.69 N *62.5 = 41168 N

4 0
3 years ago
Consider 2 hosts, Host A and Host B, transmitting a large file to a Server C over a bottleneck link with a rate of R kbps.
nevsk [136]

Answer:

i want coins sorry use a calculator or sum

Explanation:

kk

5 0
3 years ago
1. Consider a city of 10 square kilometers. A macro cellular system design divides the city up into square cells of 1 square kil
kakasveta [241]

Answer:

a) n = 1000\,users, b)\Delta t_{min} = \frac{1}{30}\,h, \Delta t_{max} = \frac{\sqrt{2} }{30}\,h, \Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h, c) n = 10000000\,users, \Delta t_{min} = \frac{1}{3000}\,h, \Delta t_{max} = \frac{\sqrt{2} }{3000}\,h, \Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h

Explanation:

a) The total number of users that can be accomodated in the system is:

n = \frac{10\,km^{2}}{1\,\frac{km^{2}}{cell} }\cdot (100\,\frac{users}{cell} )

n = 1000\,users

b) The length of the side of each cell is:

l = \sqrt{1\,km^{2}}

l = 1\,km

Minimum time for traversing a cell is:

\Delta t_{min} = \frac{l}{v}

\Delta t_{min} = \frac{1\,km}{30\,\frac{km}{h} }

\Delta t_{min} = \frac{1}{30}\,h

The maximum time for traversing a cell is:

\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}

\Delta t_{max} = \frac{\sqrt{2} }{30}\,h

The approximate time is giving by the average of minimum and maximum times:

\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}

\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h

c) The total number of users that can be accomodated in the system is:

n = \frac{10\times 10^{6}\,m^{2}}{100\,m^{2}}\cdot (100\,\frac{users}{cell} )

n = 10000000\,users

The length of each side of the cell is:

l = \sqrt{100\,m^{2}}

l = 10\,m

Minimum time for traversing a cell is:

\Delta t_{min} = \frac{l}{v}

\Delta t_{min} = \frac{0.01\,km}{30\,\frac{km}{h} }

\Delta t_{min} = \frac{1}{3000}\,h

The maximum time for traversing a cell is:

\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}

\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h

The approximate time is giving by the average of minimum and maximum times:

\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}

\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h

8 0
3 years ago
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