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kolbaska11 [484]
3 years ago
7

The author states that chemical engineering is one of the most difficult and complex aspects of engineering. Why do you think th

is is the case?
Why is stoichiometry so important for chemical engineering?
Do you think consumers would notice if process control standards were not met? Explain using an example.
Imagine that you are a chemical engineer who has been given the task of working with a museum curator to create a display case for a rare painting that must be preserved in a controlled environment, meaning elements such as temperature, humidity, exposure to light, etc. must be constantly regulated. What kind of control system would you choose and why?
According to the unit, part of workplace safety is “cultivating a culture of caution.” Explain what this means and how it could be accomplished.
Engineering
2 answers:
marishachu [46]3 years ago
8 0

Answer:

hope this helps

Explanation:

answers:

1. Chemical engineering is most difficult because it's a mix of physics, chemistry and math

2. Stoichiometry is so important because it shows how materials react, interact and play off each other

3. Yes I think consumers would notice if process control standards were not met. for example medicines, when people take Tylenol or cold pills, if the amount of time it took to kick it becomes longer, people will become aware that the product is not consistent and reliable.

4. i have no idea sorry :(

5. This is explaining how there are rules and regulations to make the workplace safe. it can be accomplished by following those rules and regulations

chrisss2 years ago
0 0

i dont know

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Gina is about to use a fire extinguisher on a small fire. What factor determines the
Effectus [21]

Answer:

Each portable fire extinguisher should only be used for its specific type of fire. Class A extinguishers could cause a Class B fire to spread or electrocution in a Class C fire. A Class B extinguisher could fail to completely extinguish a Class A fire, causing the flame to re-ignite later.

7 0
3 years ago
Some connecting rods have ____ to help lubricate the cylinder wall or piston pin.
Ket [755]

Answer:

some connecting rods have spit holes

4 0
4 years ago
Read 2 more answers
Assume that in orthogonal cutting the rake angle is 15° and the coefficient of friction is 0.15. a. Determine the percentage cha
alexira [117]

Answer:

Δr=20.45 %

Explanation:

Given that

Rake angle α =  15°

coefficient of friction ,μ = 0.15

The friction angle β

tanβ = μ

tanβ = 0.15

β=8.83°

2φ +  β - α  = 90°

φ=Shear angle

2φ + 8.833° - 15° = 90°

φ = 48.08°

Chip thickness r given as

r=\dfrac{tan\phi}{cos\alpha +sin\alpha\ tan\phi}

r=\dfrac{tan48.08^{\circ}}{cos15^{\circ} +sin15^{\circ}\ tan48.08^{\circ}}

r=0.88

New coefficient of friction ,μ'  = 0.3

tanβ' = μ'

tanβ' = 0.3

β'=16.69°

2φ' +  β' - α  = 90°

φ'=Shear angle

2φ' + 16.69° - 25° = 90°

φ' = 49.15°

Chip thickness r' given as

r'=\dfrac{tan\phi'}{cos\alpha +sin\alpha\ tan\phi'}

r'=\dfrac{tan44.15^{\circ}}{cos49.15^{\circ} +sin49.15^{\circ}\ tan44.15^{\circ}}

r'=0.70

Percentage change

\Delta r=\dfrac{r-r'}{r}\times 100

\Delta r=\dfrac{0.88-0.70}{0.88}\times 100

Δr=20.45 %

8 0
3 years ago
Suppose a consumer advocacy group would like to conduct a survey to find the proportion p of consumers who bought the newest gen
Anvisha [2.4K]

Answer:

a) Sample size = 1691

b) 95% Confidence Interval = (0.3696, 0.4304)

Explanation:

(a) How large a sample n should they take to estimate p with 2% margin of error and 90% confidence?

The margin of error is given by

MoE = z \cdot \frac{\sqrt{p(1-p)} }{\sqrt{n} }  \\\\

Where z is the corresponding z-score for 90% confidence level

z = 1.645 (from z-table)

for p = 0.50 and 2% margin of error, the required sample size would be

n = \frac{1.645^{2} \cdot 0.50(1-0.50)}{0.02^{2}}  \\\\n = \frac{0.6765}{0.0004}  \\\\n = 1691\\

(b) The advocacy group took a random sample of 1000 consumers who recently purchased this mobile phone and found that 400 were happy with their purchase. Find  a 95% confidence interval for p.

The sample proportion is

p = 400/1000

p = 0.40

z = 1.96 (from z-table)

n = 1000

The confidence interval is given by

CI = p \pm z \cdot \sqrt{\frac{p(1-p)}{n} } \\\\CI = 0.40 \pm 1.96 \cdot \sqrt{\frac{0.40(1-0.40)}{1000} } \\\\CI = 0.40 \pm 1.96 \cdot 0.01549 \\\\CI = 0.40 \pm 0.0304 \\\\CI = 0.40 - 0.0304 \: and \: 0.40 + 0.0304\\\\CI = (0.3696 ,\:  0.4304)

Therefore, we are 95% confident that the proportion of consumers who bought the newest generation of mobile phone were happy with their purchase is within the range of (0.3696, 0.4304)

What is Confidence Interval?

The confidence interval represents an interval that we can guarantee that the target variable will be within this interval for a given confidence level.  

3 0
3 years ago
25°C is flowing in a covered irrigation ditch below ground. Every 100 ft, there is a vent line with a 1.0 in. inside diameter an
Gnesinka [82]

The total evaporation loss of water is 87.873 ×10^{-5}  lbm /day.

<u>Explanation:</u>

Assume A is the water and B is air.

A is diffusing to non diffusing B.

N_{A}=\frac{D_{A B} P}{R T Z P_{B/ M}}\left(P_{A_{1}} \ - P_{B_{2}}\right)

By the table 6.2 - 1 at 25°C, the diffusivity of air and water system is 0.260 \times 10^{-4} \mathrm{m}^{2} / \mathrm{s}.

Total pressure P = 1 atm = 101.325 KPa

P_{A_{1} } = 23.76 mm Hg

P_{A_{1} } = \frac{23.76}{760}

P_{A_{1} } = 0.03126 atm

P_{A_{1} } = 3.167 K Pa

When air surrounded is dry air, then PA_{2} = 0 mm Hg

R = 8.314 \frac{K P a \cdot m^{3}}{mol \cdot k}

P_{B/M} = \frac{P_{B_{1}}-P_{B_{2}}}{\ln \left(P_{B_{1}} / P_{B_{2}}\right)}

P_{B_{1}}=P_{T}-P_{A1_{}}

= 101.325 - 3.167

P_{B_{1} } = 98.158 K Pa

P_{B_{2}}=P_{T}-P_{A_{2}}

P_{B_{2} } = 101.325 - 0

P_{B_{2} } = 101.325 K Pa

P_{B / M}=\frac{98.158-101.325}{\ln (98.158 / 101 \cdot 325)}

P_{B/M} = 99.733 K Pa

Z = 1 ft = 0.3048 m

T = 298 K

N_{A} =  \frac{(0.206*10^{-4})(101.325)(3.167 - 0) }{(8.314) ( 298 )( 0.3048 )( 99.733 ) }

N_{A} = 0.11077 × 10^{-6} mol/ m^{2}.s

N_{A} = 0.11077 × 10^{-6} × 18 × (60×60×24)

N_{A} = 0.1723 lb_{m} / m^{2}.day

\tilde{N}_{A}=N_{A} \times Area

Area of individual pipe is

A=\frac{\pi}{4}(0.0254)^{2}

A = 0.00051 m^{2}

\bar{N}_{\boldsymbol{A}} = 0.1723 × 0.00051

\bar{N}_{\boldsymbol{A}} = 0.000087873 lbm/day

In 1000 ft length of ditch,there will be a 10 pipes. The amount of evaporation water is

= 10 × 0.000087873 = 0.00087873 lbm /day

The total evaporation loss of water is 87.873 ×10^{-5}  lbm /day.

5 0
3 years ago
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