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3 years ago

As an engineer which types of ethical issues or problem you can face in industrial environment.

Engineering

1 answer:

Anarel [89]3 years ago

8 0

Explanation:

Answer ⬇️

Social and ethical issues in engineering, ethical principles of engineering, professional code of ethics, some specific social problems in engineering practice: privacy and data protection, corruption, user orientation, digital divide, human rights, access to basic services.

➡️dhruv73143(⌐■-■)

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The permeability coefficient of a type of small gas molecule in a polymer is dependent on absolute temperature according to the

jok3333 [9.3K]

Answer:

Answer for the question is explained in the attachment.

Explanation:

Diffusion flux at 350K for water in polystyrene is equal to

2.8 * 10^-7 .

Download pdf

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3 years ago

Read 2 more answers

The Environmental Protection Agency (EPA) has standards and regulations that says that the lead level in soil cannot exceed the

DENIUS [597]

Answer:

See below

Explanation:

Check One-Sample T-Interval Conditions

Random Sample? √

Sample Size ≥30? √

Independent? √

Population Standard Deviation Unknown? √

One-Sample T-Interval Information

Formula --> $CI=\bar{x}\pm t^*(\frac{S_x}{\sqrt{n}})$
Sample Mean --> $\bar{x}=390.25$
Critical Value --> $t^*=2.0096$ (given $df=n-1=50-1=49$ degrees of freedom at a 95% confidence level)
Sample Size --> $n=50$
Sample Standard Deviation --> $S_x=30.5$

Problem 1

The critical t-value, as mentioned previously, would be $t^*=2.0096$ , making the 95% confidence interval equal to $CI=\bar{x}\pm t^*(\frac{S_x}{\sqrt{n}})=390.25\pm2.0096(\frac{30.5}{\sqrt{50}})\approx\{381.5819,398.9181\}$

This interval suggests that we are 95% confident that the true mean levels of lead in soil are between 381.5819 and 398.9181 parts per million (ppm), which satisfies the EPA's regulated maximum of 400 ppm.

3 0

2 years ago

An elastic cable is to be designed for bungee jumping from a tower 130 ft high. The specifications call for the cable to be 85 f

azamat

Answer:

a) The spring constant is 50 lb/ft

b) The man is 26.3 ft close to the ground.

Explanation:

Height of tower is 130 ft

Specification calls for a cable of length 85 ft

the maximum this length stretches is 100 ft when subjected to a load of 750 lb

The extension of the cable is calculated from the formula from Hooke's law

F = kx

where F is the load or force on the cable

k is the spring constant of the cable

x is the extension on the cable

a) The extension on the cable is

x = 100 ft - 85 ft = 15 ft

substituting into the formula above, we'll have

750 = k*15

k = 750/15 = 50 lb/ft

b) for a 185 lb man, jumping down will give an extension gotten as

F = kx

185 = 50*x

x = 185/50 = 3.7 ft

The total length of the cable will be extended to 100 ft + 3.7 ft = 103.7 ft

closeness to the ground = 130 ft - 103.7 ft = 26.3 ft

3 0

3 years ago

____accelerates the rusting process A lubricant B improper storage CMoisture DOil

Dmitriy789 [7]

Answer:

i would go with-B

Explanation:

because Exposure to the elements has resulted in the formation of the blue-green patina

6 0

4 years ago

Read 2 more answers

At a ski resort, water at 40 °F is pumped through a 3-in.-diameter, 2001-ft-long steel pipe from a pond at an elevation of 4286

deff fn [24]

Answer:

P = 24.38 hp

Explanation:

Given data:

diameter of steel pipe = 3 inc

length of pipe = 2001 ft

elevation of pond = 4286 ft

elevation of snow making machine = 4620 ft

Rate of water = 0.26 ft^3/s

Applying bernouli equation on bioth side

$\frac{P}{\rho} + \frac{v_1^2}{2g} + z_1 + h_p =\frac{P_2}{\rho} + \frac{v_2^2}{2g} + z_2 + \frac{flv^2}{2gD}$ .....1

where P_2 = 182 psi

P_1 = 0, v_1 = 0

$V =V_2 = \frac{Q}{\frac{\pi}{4} D^2}= \frac{0.26}{\frac{\pi}{4} \times (3/12)^2} = 5.30 ft/s$

calculation fro friction factor

from standard table we have

$\frac{\epsilon}{D} = \frac{0.00015}{(3/12)} = 6\times 10^{-4}$

$Re =\frac{VD}[\nu} = \frac{5.30 \times (3/12)}{1.66 \times 10^{-5}} = 7.96 \times 10^4$

so for calculated R and $\frac{\epsilon}{D}$ , F value = 0.0212

from equation 1

$h_p = \frac{P_2}{\rho} + Z_2 -Z_1 +(1 + f\frac{l}{D}) \frac{V^2}{2g}$

$h_p = \frac{182 \times 144/ lb /ft^2}{62.4} + 4620 - 4286 + (1 + 0.0212 \frac{2001}{(3/12)}) \frac{5.30}{2\times 32.2}$

h_p = 826.76ft so that

$P = \rho Q h_p = 62.4 \times 0.26 \times 826.76 = 13413.38 ft lb/s = 24.38 hp$

3 0

3 years ago