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xenn [34]
2 years ago
8

Why do engineers need to be particularly aware of the impact of hubris?

Engineering
1 answer:
miv72 [106K]2 years ago
7 0

Answer:

A) Their creations change society.

You might be interested in
A six-lane divided highway (three lanes in each direction) is on rolling terrain with two access points per mile and has 10- ft
tatuchka [14]

Answer

given,

6 lanes divided highway 3 lanes in each direction

rolling terrain

lane width = 10'

shoulder on right = 5'

PHF = 0.9

shoulder on the left direction = 3'

peak hour volume = 3500 veh/hr

large truck = 7 %

tractor trailer = 3 %

speed = 55 mi/h

LOS is determined based on V p

10' lane weight ;  f_{Lw}=6.6 mi/h

5' on right   ;    f_{Lc} = 0.4 mi/hr

3' on left   ;      no adjustment

3 lanes in each direction    f n = 3 mi/h

v_p =\dfrac{V}{f_{HV}\times N\times f_p\times PHF}

f_{HV}=\dfrac{1}{1+P_T(E_T-1)+P_R(E_R-1)}

f_{HV}=\dfrac{1}{1+0.08(2.5-1)+0.02(2-1)}

          = 0.877

v_p =\dfrac{3500}{0.877\times 3\times 0.95\times0.9}

       = 1,555 veh/hr/lane

FFS = BFFS - F_{Lw}-F_{Lc}-F_{N}-F_{ID}

      = (55 + 5) - 6.6 - 0.4 -3 -0

      = 50 mi/h

D = \dfrac{V_P}{s}

D = \dfrac{1555}{55} =28.27

level of service is D using speed flow curves and LOS for basic free moving of vehicle

5 0
3 years ago
If 200 amperes flow from the positive terminal of a battery and operate the starter motor, how many amperes will flow back to th
garri49 [273]

Answer:

200 Amperes WB =<em> </em>200J

Explanation:

200 Amperes will flow back to the negative terminal because both the positive terminal and negative terminal contain the equivalent amount of current that will flow and it works in the starter motor.

To learn more about it, refer

to brainly.com/question/6561461

#SPJ4

5 0
1 year 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
Which of the following is NOT an ASE certification? Select one:
stiks02 [169]

The option that is not an ASE certification is . A/C and Refrigerants handling certification (609).

<h3>What is ASE certification?</h3>

The term ASE is known to be a body that tends to promotes excellence in regards to vehicle repair, service as well as parts distribution.

Note that in the world today more than a quarter of  million of people are known to possess ASE certifications.

Since ASE Certified professionals work in in all areas of the transportation industry. one can say that The option that is not an ASE certification is. A/C and Refrigerants handling certification (609).

Learn more about ASE certification from

brainly.com/question/5533417

#SPJ1

8 0
1 year ago
To compute the energy used by a motor, multiply the power that it draws by the time of operation. Con- sider a motor that draws
ehidna [41]

Answer:

E=52000Hp.h

E=38724920Wh

E=1.028x10^11 ftlb

Explanation:

To solve this problem you must multiply the engine power by the time factor expressed in h / year, to find this value you must perform the conventional unit conversion procedure.

Finally, when you have the result Hp h / year you convert it to Ftlb and Wh

E=(12.5hp)(\frac{16h}{day} )(\frac{5 days}{week} )(\frac{52week}{year} )\\

E=52000Hp.h

E=52000Hp.h(\frac{744.71Wh}{Hp.h} )\\

E=38724920Wh

E=52000Hph(\frac{1977378.4  ft lb}{1Hph}

E=1.028x10^11 ftlb

3 0
3 years ago
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