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I am Lyosha [343]
3 years ago
14

How many types of engineering specialist are there?

Engineering
1 answer:
klasskru [66]3 years ago
3 0

There are now six major branches of engineering: Mechanical, Chemical, Civil, Electrical, Management, and Geotechnical, and literally hundreds of different subcategories of engineering under each branch.

You might be interested in
Consider the smoothie example given in class: Write down the production function of Tutti Frutti Smoothie example given in class
mart [117]

Answer:

Production Function : TFSq = f { ingredient 1 ...... ing. i }

Increasing/ Constant/  Increasing Returns to Scale :

Output change > / = / < Input change respectively

Explanation:

Production Function is the relationship between production inputs & outputs, given technology. It denotes the maximum output that can be generated with given inputs.

Tutti Frutti Smoothie [TFS] quantity = Function of {Ingredient1.....ingredient i}

Returns to Scale represents change in output when all inputs change in same proportion.

  • Constant Returns to Scale [CRS] : Output Change = All inputs change
  • Increasing Returns to Scale [IRS] : Output Change > All inputs change
  • Negative Returns to Scale [NRS] : Output Change < All inputs change

When all inputs (ingredients) change by same proportion i.e get twice 2X :- If output of Tutti Frutti Smoothie increases by > 2X  i.e 3X - IRS.  If it increases equal ie 2X - CRS. If it increases lesser i.e 1.5X - CRS.

6 0
4 years ago
Engineering stress and strain are calculated using the actual cross-sectional area and length of the specimen. a. True b. False?
Flauer [41]

Answer:

False

Explanation:

True stress and strain are what are calculated by diving applied load by actual cross sectional area and change in length by length of specimen respectively. However, for engineering stress and strain, we use original area and original length. Generally, engineering Stress is always greater than the corresponding True Stress because the actual area increases due to Poisson's effect.

3 0
4 years ago
IV. An annealed copper strip 9 inches wide and 2.2 inches thick, is rolled to its maximum possible draft in one pass. The follow
Irina-Kira [14]

Answer:

13.9357 horse power

Explanation:

Annealed copper

Given :

Width, b = 9 inches

Thickness, $h_0=2.2$ inches

K= 90,000 Psi

μ = 0.2, R = 14 inches, N = 150 rpm

For the maximum possible draft in one pass,

$\Delta h = H_0-h_f=\mu^2R$

     $=0.2^2 \times 14 = 0.56$ inches

$h_f = 2.2 - 0.56$

     = 1.64 inches

Roll strip contact length (L) = $\sqrt{R(h_0-h_f)}$

                                             $=\sqrt{14 \times 0.56}$

                                             = 2.8 inches

Absolute value of true strain, $\epsilon_T$

$\epsilon_T=\ln \left(\frac{2.2}{1.64}\right) = 0.2937$

Average true stress, $\overline{\gamma}=\frac{K\sum_f}{1+n}= 31305.56$ Psi

Roll force, $L \times b \times \overline{\gamma} = 2.8 \times 9 \times 31305.56$

                                 = 788,900 lb

For SI units,

Power = $\frac{2 \pi FLN}{60}$  

           $=\frac{2 \pi 788900\times 2.8\times 150}{60\times 44.25\times 12}$

           = 10399.81168 W

Horse power = 13.9357

6 0
3 years ago
Determine the real roots of f (x) = −0.6x2 + 2.4x + 5.5:(a) Graphically.(b) Using the quadratic formula.(c) Using three iteratio
zubka84 [21]

The three methods used to find the real roots of the function are,

graphically, the quadratic formula, and by iteration.

The correct vales are;

(a) Graphically, the roots obtained are; <u>x ≈ -1.629, and 5.629</u>

(b) Using the quadratic formula, the real roots of the given function are; <u>x ≈ -1.62589, x ≈ 5.62859</u>

(c) Using three iterations, we have; the bracket is x_l = <u>5.625</u>, and x_u =<u> 6.25</u>

Reasons:

The given function is presented as follows;

f(x) = -0.6·x² + 2.4·x + 5.5

(a) The graph of the function is plotted on MS Excel, with increments in the

x-values of 0.01, to obtain the approximation of the x-intercepts which are

the real roots as follows;

\begin{array}{|c|cc|}x&&f(x)\\-1.63&&-0.00614\\-1.62&&0.03736\\5.62&&0.03736 \\5.63&&-0.00614\end{array}\right]

Checking for the approximation of x-value of the intercept, we have;

x = -1.63 + \dfrac{0 - (-0.00614)}{0.0376 - (-0.00614)} \times (-1.62-(-1.63)) \approx -1.629

Therefore, based on the similarity of the values at the intercepts, the x-

values (real roots of the function) at the x-intercepts (y = 0) are;

<u>x ≈ -1.629, and 5.629</u>

(b) The real roots of the quadratic equation are found using the quadratic

formula as follows;

The quadratic formula for finding the roots of the quadratic equation

presented in the form f(x) = a·x² + b·x + c, is given as follows;

x = \mathbf{ \dfrac{-b\pm \sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}}

Comparison to the given function, f(x) = -0.6·x² + 2.4·x + 5.5, gives;

a = -0.6, b = 2.4, and c = 5.5

Therefore, we get;

x = \dfrac{-2.4\pm \sqrt{2.4^{2}-4\times (-0.6)\times 5.5}}{2\times (-0.6)} = \dfrac{-2.4\pm\sqrt{18.96} }{-1.2} = \dfrac{12 \pm\sqrt{474} }{6}

Which gives

The real roots are; <u>x ≈ -1.62859, and x ≈ 5.62859</u>

(c) The initial guesses are;

x_l = 5, and x_u = 10

The first iteration is therefore;

x_r = \dfrac{5 + 10}{2} = 7.5

Estimated \ error , \ \epsilon _a = \left|\dfrac{10- 5}{10 + 5} \right | \times 100\% = 33.33\%

True \ error, \ \epsilon _t = \left|\dfrac{5.62859 - 7.5}{5.62859} \right | \times 100\% = 33.25\%

f(5) × f(7.5) = 2.5 × (-10.25) = -25.625

The bracket is therefore; x_l = <u>5</u>, and x_u = <u>7.5</u>

Second iteration:

x_r = \dfrac{5 + 7.5}{2} = 6.25

Estimated \ error , \ \epsilon _a = \left|\dfrac{7.5- 5}{7.5+ 5} \right | \times 100\% = 20\%

True \ error, \ \epsilon _t = \mathbf{\left|\dfrac{5.62859 - 6.25}{5.62859} \right | \times 100\%} \approx 11.04\%

f(5) × f(6.25) = 2.5 × (-2.9375) = -7.34375

The bracket is therefore; x_l = <u>5</u>, and x_u = <u>6.25</u>

Third iteration

x_r = \dfrac{5 + 6.25}{2} = 5.625

Estimated \ error , \ \epsilon _a = \left|\dfrac{5.625- 5}{5.625+ 5} \right | \times 100\% = 5.88\%

True \ error, \ \epsilon _t = \mathbf{\left|\dfrac{5.62859 - 5.625}{5.62859} \right | \times 100\%} \approx 6.378 \times 10^{-2}\%

f(5) × f(5.625) = 2.5 × (0.015625) = 0.015625

Therefore, the bracket is x_l = 5.625, and x_u = 6.25

Learn more here:

brainly.com/question/14950153

6 0
3 years ago
Only an outer panel is being replaced. Technician A says that removing the spot welds by drilling through both panels allows the
Angelina_Jolie [31]

Answer:

6e66363636633747747363637737373737337374

5 0
3 years ago
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