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2 years ago

What type of test can show a chemical engineer if a material remains in a system and accumulates or if it moves right through?

1 answer:

7 0

A chemical engineer can clearly see from this kind of test if a substance stays in a system and builds up or if it just passes through.

<h3>What is a chemical engineer?</h3>

Processes for manufacturing chemicals are created and designed by chemical engineers.
To solve issues involving the manufacture or usage of chemicals, fuel, medications, food, and many other goods, chemical engineers use the concepts of chemistry, biology, physics, and math.
A wide range of sectors, including petrochemicals and energy in general, polymers, sophisticated materials, microelectronics, pharmaceuticals, biotechnology, foods, paper, dyes, and fertilizers, have a significant demand for chemical engineers.
Chemical engineering is undoubtedly difficult because it requires a lot of physics and math, as well as a significant number of exams at the degree level.

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True/False

agasfer [191]

Answer: This is true.

Hope this helps! :)

Explanation:

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3 years ago

How can historical design elements be added to modern building plans?

satela [25.4K]

Answer:

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3 years ago

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Finally, identify at least three colleges, universities, or training programs that are suited to that career choice.

gtnhenbr [62]

Answer:

Any of the top colleges are good for engineering. For example MIT and Stanford consistently rank among the top 5 best engineering schools in the country. Georgia Tech, Harvard, and Princeton are also good.

Explanation:

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3 years ago

A particle is moving along a straight line with an initial velocity of 6 m/s when it is subjected to a deceleration of a- (-1.5v

Juli2301 [7.4K]

Answer:

s= 6.53 m

t=3.27 s

Explanation:

velocity = 6 m/s

deceleration = -1.5 $v^\frac{1}{2}$

$a=-1.5v^\frac{1}{2}\\v\frac{\mathrm{d} v}{\mathrm{d} s}=-1.5v^\frac{1}{2}\\-1.5ds=v^\frac{1}{2}dv\\\int\ {-1.5} \, ds= \int\ v^\frac{1}{2}dv\\-1.5s=\frac{2}{3}\times v^{\frac{3}{2} }$

now inserting value of v=6s we get distance(s)

s= 6.53 m ( distance cannot be negative)

now for time calculation we know that

$a=\frac{\mathrm{d}v }{\mathrm{d} t}$

$-1.5v^\frac{1}{2} =\frac{\mathrm{d}v }{\mathrm{d} t}\\-1.5dt=\frac{dv}{v^\frac{1}{2}} \\\int -1.5 dt=\int v^{-\frac{1}{2}}dt \\1.5t=2v^\frac{1}{2}\\t=\frac{4}{3}v^\frac{1}{2}$

putting value of v=6s

t=3.27 s (time cannot be negative)

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3 years ago

Show that -40 F is approximately equal to -40 C.

V125BC [204]

Answer: Please see below the explanation.

Explanation:

The Celsius (ºC) and Farenheit (ºF), are different temperature scales, but both have identifiable values for two important physical phenomena, water freezing point and water boiling point.

For º C, these are follows:

Freezing Point= 0º C Boiling Point: 100ºC

For º F, we have the following:

Freezing Point= 32º F Boiling Point: 212 ºF

Assuming that all steps (in each scale) between two following values are equal, we have 100 º C between both points in ºC, and 190º in ºF, so degrees in Farenheit are "shorter" than in Celsius, being the relationship, approximately the following:

1 º C = 9/5 ºF.

Taking into account that, by definition, 0ºC = 32º F, we can use the following equations in order to convert ºC to ºF, and viceversa:

ºC = (ºF-32) 5/9

ºF = 32 + 9/5ºC

Now, let's calculate -40ºF, in ºC:

ºC= (-40-32)* 5/9 = -40º C

Let's do the same, converting -40ºC to ºF:

ºF = 32 + (-40)*9/5 = 32 + (-72) = -40ºF.

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3 years ago