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telo118 [61]
3 years ago
12

write down your own definition of Engineering, preferably in 4-5 sentences. Maximum of 150 words for your definition???.​

Engineering
1 answer:
ollegr [7]3 years ago
4 0

Answer:

A charge q1=7.0mc is located at the origin and a second charge q2=-5.0mc is located on the x axis, 0.3m the origin find the electric field at the point p which he's coordinates (0,0.40)m

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What is brainstorming and why is it important to create the perfect solution to a problem? giving brainliest to the first answer
Diano4ka-milaya [45]
Brainstorming allows people to think freely without judge, or fear to share there answer. Basically encourages people to open up to what they believe.
5 0
3 years ago
Unit for trigonometric functions is always "radian". 1. 10 points: Do NOT submit your MATLAB code for this problem (a) Given f(x
RoseWind [281]

Answer:

Below is the required code.

Explanation:

%% Newton Raphson Method

clear all;

clc;

x0=input('Initial guess:\n');

x=x0;

f=exp(-x)-sin(x)-0.2;

g=-exp(-x)-cos(x);

ep=10;

i=0;

cc=input('Condition of convergence:\n');

while ep>=cc

i=i+1;

temp=x;

x=x-(f/g);

f=exp(-x)-sin(x)-0.2;

g=-exp(-x)-cos(x);

ep=abs(x-temp);

fprintf('x = %6f and error = %6f at iteration = %2f \n',x,ep,i);

end

fprintf('The solution x = %6f \n',x);

%% End of MATLAB Program

Command Window:

(a) First Root:

Initial guess:

1.5

Condition of convergence:

0.01

x = -1.815662 and error = 3.315662 at iteration = 1.000000

x = -0.644115 and error = 1.171547 at iteration = 2.000000

x = 0.208270 and error = 0.852385 at iteration = 3.000000

x = 0.434602 and error = 0.226332 at iteration = 4.000000

x = 0.451631 and error = 0.017029 at iteration = 5.000000

x = 0.451732 and error = 0.000101 at iteration = 6.000000

The solution x = 0.451732

>>

Second Root:

Initial guess:

3.5

Condition of convergence:

0.01

x = 3.300299 and error = 0.199701 at iteration = 1.000000

x = 3.305650 and error = 0.005351 at iteration = 2.000000

The solution x = 3.305650

>>

(b) Guess x=0.5:

Initial guess:

0.5

Condition of convergence:

0.01

x = 0.450883 and error = 0.049117 at iteration = 1.000000

x = 0.451732 and error = 0.000849 at iteration = 2.000000

The solution x = 0.451732

>>

Guess x=1.75:

Initial guess:

1.75

Condition of convergence:

0.01

x = 227.641471 and error = 225.891471 at iteration = 1.000000

x = 218.000998 and error = 9.640473 at iteration = 2.000000

x = 215.771507 and error = 2.229491 at iteration = 3.000000

x = 217.692636 and error = 1.921130 at iteration = 4.000000

x = 216.703197 and error = 0.989439 at iteration = 5.000000

x = 216.970438 and error = 0.267241 at iteration = 6.000000

x = 216.971251 and error = 0.000813 at iteration = 7.000000

The solution x = 216.971251

>>

Guess x=3.0:

Initial guess:

3

Condition of convergence:

0.01

x = 3.309861 and error = 0.309861 at iteration = 1.000000

x = 3.305651 and error = 0.004210 at iteration = 2.000000

The solution x = 3.305651

>>

Guess x=4.7:

Initial guess:

4.7

Condition of convergence:

0.01

x = -1.916100 and error = 1.051861 at iteration = 240.000000

x = -0.748896 and error = 1.167204 at iteration = 241.000000

x = 0.162730 and error = 0.911626 at iteration = 242.000000

x = 0.428332 and error = 0.265602 at iteration = 243.000000

x = 0.451545 and error = 0.023212 at iteration = 244.000000

x = 0.451732 and error = 0.000187 at iteration = 245.000000

The solution x = 0.451732

>>

Explanation:

The two solutions are x =0.451732 and 3.305651 within the range 0 < x< 5.

The initial guess x = 1.75 fails to determine the solution as it's not in the range. So the solution turns to unstable with initial guess x = 1.75.

7 0
3 years ago
B. Is the “Loading Time” of any online application a functional or a non-functional requirement? Can the requirement engineers s
Oksi-84 [34.3K]

Answer:

non-functional requirement,

Yes they can.

The application loading time is determined by testing system under various scenarios

Explanation:

non-functional requirement are requirements needed to justify application behavior.

functional requirements are requirements needed to justify what the application will do.

The loading time can be stated with some accuracy level after testing the system.

4 0
2 years ago
Liquid flows with a free surface around a bend. The liquid is inviscid and incompressible, and the flow is steady and irrotation
lions [1.4K]

Answer:

9 cm

Explanation:

The liquid on the bend will be affected by two accelerations: gravity and centripetal force.

Gravity will be of 9.81 m/s^2 pointing down at all points.

The centripetal acceleration will be of

ac = v^2/r

Pointing to the center of the bend (perpendicular to gravity).

The velocity will depend on the radius

v = (1 m^2/s) / r

Replacing:

ac = (1/r)^2 / r

ac = (1 m^4/s^2) / r^3

If we set up a cylindrical reference system with origin at the center of the bend, the total acceleration will be

a = (-1/r^3 * i - 9.81 * j)

The surface of the liquid will be an equipotential surface, this means all points on the surface have the same potential energy.

The potential energy of the gravity field is:

pg = g * h

The potential energy of the centripetal force is:

pc = ac * r

Then the potential field is:

p = -1/r^2 * - 9.81*h

Points on the surface at r = 1 m and r = 3 m have the same potential.

-1/1^2 * - 9.81*h1 = -1/3^2 * - 9.81*h2

-1 - 9.81*h1 = -1/9 - 9.81*h2

-1 + 1/9 = 9.81 * (h1 - h2)

h1 - h2 = (-8/9) / 9.81

h2 - h1 = 0.09 m

The outer part will be 9 cm higher than the inner part.

3 0
3 years ago
A cone penetration test was carried out in normally consolidated sand, for which the results are summarized below: Depth (m) Con
Cerrena [4.2K]

Answer:

hello your question is incomplete attached below is the missing equation related to the question  

answer : 40.389° , 38.987° , 38° , 39.869° , 40.265°

Explanation:

<u>Determine the friction angle at each depth</u>

attached below is the detailed solution

To calculate the vertical stress = depth * unit weight of sand

also inverse of Tan = Tan^-1

also qc is in Mpa while σ0 is in kPa

Friction angle at each depth

2 meters = 40.389°

3.5 meters  = 38.987°

5 meters = 38.022°

6.5 meters = 39.869°

8 meters = 40.265°

6 0
3 years ago
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