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4 years ago

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Consider a process in which a carbon-based fuel is combusted in the presence of 70% excess oxygen (assume that all of the oxygen

present come from an air feel stream). Based on the information given above, which of the following statements are true? Select all that apply.
a. Exactly 70% of the oxygen presence in the process is not consumed.

b. 70% of the fuel in the process undergoes complete combustion.

c. At least 41% of the oxygen present in the grocers is not consumed

d. There is 70% more air present in the process that would be necessary for the complete combustion of all the fuel

e. There is 70% more oxygen present in the process that would be necessary for the complete combustion of all of the fuel

1 answer:

yuradex [85]4 years ago

7 0

Answer:

<u>Option-(A):</u>

When a carbon-based fuel is burned(combustion) or we can say burned in the presence of 70% excess oxygen (assuming that all of the oxygen present come from an air feel stream), we can say that there is 70% more oxygen present in the process that would be necessary for the complete combustion of all of the fuel.

Oxygen,O₂ is the precursor for the process of combustion as there is the requirement of an optimum level of oxygen for the burning up of the materials.While, there is about enough level of oxygen is present inside the free air or we can say inside the atmosphere for the purpose of combustion or burning of the fuel or consumption of the resources present in the environment.

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When determining risk, it is necessary to estimate all routes of exposure in order to determine a total dose (or CDI). Recognizi

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Answer:

The following are the solution to this question:

Explanation:

The Formula for calculating CDI:

$\bold{CDI = \frac{C \times CR \times EF \times ED}{BW \times AT}}$

$_{where} \\ CDI = \text{Chronic daily Intake rate} (\frac{mg}{kg-day})} \\\\\text{C = concentration of Toluene}\\\\\text{CR = contact rate} \frac{L}{day}\\\\\text{EF = Exposure frequency} \frac{days}{year}\\\\\text{ED = Exposure duration (in years)} = 10 \ \ years\\\\\text{BW = Body weight (kg) = 70 kg for adult}\\\\ \text{AT = average period of exposure (days) }$

calculating the value of AT:

$= 365 \frac{days}{year} \times 70 \ year \\\\ = 25550 \ days$

calculating the value of Intake based drinking:

$C = 1 \ \frac{mg}{L}$

$CR = 2 \frac{L}{day}$ Considering that adult females eat 2 L of water a day,

EF = 350 $\frac{days}{year}$ for drink

calculating the CDI value:

$\to CDI = \frac{(1 \times 2 \times 350 \times 10)}{(70 \times 25550)}\\\\$

$= \frac{(2 \times 3500)}{(70 \times 25550)}\\\\ = \frac{(7000)}{(70 \times 25550)}\\\\ = \frac{(100)}{(25550)}\\\\=0.00391 \frac{mg}{ kg-day}$

Centered on inhalation, intake:

$C = \frac{1 \mu g} { m^3} \ \ \ or \ \ \ \ 0.001 \ \ \frac{mg}{m^3}\\\\CR = 20 \frac{m^3}{day}\\\\EF = 15 \frac{min}{day} \ \ or\ \ 5475 \frac{min}{yr} \ \ \ or \ \ 3.80 \frac{days}{year}\\$

calculating the value of CDI:

$\to CDI = \frac{(0.001 \times 20 \times 3.80 \times 10)}{(70 \times 25550)}$

$= \frac{(0.76)}{(1788500)}\\\\= 4.24 \times 10^{-7} \ \ \frac{mg}{kg-day}$

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3 years ago

Evaluate to three significant figures and using appropriate prefix: (354 mg)(45 km)/(0.0356 kN)

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Answer:

0.447 s²

Explanation:

First, convert to SI units.

(354 mg) (45 km) / (0.0356 kN)

(0.354 g) (45000 m) / (35.6 N)

One Newton is kg m/s²:

(0.354 g) (45000 m) / (35.6 kg m/s²)

(0.000354 kg) (45000 m) / (35.6 kg m/s²)

Simplify:

0.447 s²

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3 years ago

How is an orthographic drawing similar to or different from an isometric drawing?

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An isometrical drawing is a nearly 3d drawing showing the object's width and depth in a complete image, from each curved plane of the orthhographic view, the viewpoint is at a 45 degree angle. From an observations point of view, isometric differs, since all longitudes are true.

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A group of students launches a model rocket in the vertical direction. Based on tracking data, they determine that the altitude

Fofino [41]

Answer:

$u = 260.22m/s$

$S_{max} = 1141.07ft$

Explanation:

Given

$S_0 = 89.6ft$ --- Initial altitude

$S_{16.5} = 0ft$ -- Altitude after 16.5 seconds

$a = -g = -32.2ft/s^2$ --- Acceleration (It is negative because it is an upward movement i.e. against gravity)

Solving (a): Final Speed of the rocket

To do this, we make use of:

$S = ut + \frac{1}{2}at^2$

The final altitude after 16.5 seconds is represented as:

$S_{16.5} = S_0 + ut + \frac{1}{2}at^2$

Substitute the following values:

$S_0 = 89.6ft$ $S_{16.5} = 0ft$ $a = -g = -32.2ft/s^2$ and $t = 16.5$

So, we have:

$0 = 89.6 + u * 16.5 - \frac{1}{2} * 32.2 * 16.5^2$

$0 = 89.6 + u * 16.5 - \frac{1}{2} * 8766.45$

$0 = 89.6 + 16.5u- 4383.225$

Collect Like Terms

$16.5u = -89.6 +4383.225$

$16.5u = 4293.625$

Make u the subject

$u = \frac{4293.625}{16.5}$

$u = 260.21969697$

$u = 260.22m/s$

Solving (b): The maximum height attained

First, we calculate the time taken to attain the maximum height.

Using:

$v=u + at$

At the maximum height:

$v =0$ --- The final velocity

$u = 260.22m/s$

$a = -g = -32.2ft/s^2$

So, we have:

$0 = 260.22 - 32.2t$

Collect Like Terms

$32.2t = 260.22$

Make t the subject

$t = \frac{260.22}{ 32.2}$

$t = 8.08s$

The maximum height is then calculated as:

$S_{max} = S_0 + ut + \frac{1}{2}at^2$

This gives:

$S_{max} = 89.6 + 260.22 * 8.08 - \frac{1}{2} * 32.2 * 8.08^2$

$S_{max} = 89.6 + 260.22 * 8.08 - \frac{1}{2} * 2102.22$

$S_{max} = 89.6 + 260.22 * 8.08 - 1051.11$

$S_{max} = 1141.0676$

$S_{max} = 1141.07ft$

Hence, the maximum height is 1141.07ft

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