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MakcuM [25]
3 years ago
7

\o you sell them?” "Fivepence farthing for one—Twopence for two,” the Sheep replied. "Then two are cheaper than one?” Alice said

in a surprised tone, taking out her purse. "Only you MUST eat them both, if you buy two,” said the Sheep. "Then I'll have ONE, please,” said Alice, as she put the money down on the counter. For she thought to herself, "They mightn't be at all nice, you know.” The Sheep took the money, and put it away in a box: then she said "I never put things into people's hands—that would never do—you must get it for yourself.” And so saying, she went off to the other end of the shop, and set the egg upright on a shelf. "I wonder WHY it wouldn't do?” thought Alice, as she groped her way among the tables and chairs, for the shop was very dark towards the end. "The egg seems to get further away the more I walk towards it. Let me see, is this a chair? Why, it's got branches, I declare! How very odd to find trees growing here! And actually here's a little brook! Well, this is the very queerest shop I ever saw!” The tone of this passage is best described as unusual and strange. logical and orderly. realistic and ordinary. negative and critical.
Engineering
1 answer:
NemiM [27]3 years ago
3 0

Answer:

unusual and strange

Explanation:

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On July 23, 1983, Air Canada Flight 143 required 22,300 kg of jet fuel to fly from Montreal to Edmonton. The density of jet fuel
Natasha2012 [34]

Answer:

20, 083 L

Explanation:

The mistake was the result of not using units when converting the 7862 l to Kg. They used the density in pounds hence they multiplied by 1.77 Lb/L and obtained 13597 Lb not Kg as they assumed.

To obtain the amount needed to refuel they subtracted this quantity from the 22,300 Kg required for the trip again obtaining the wrong quantity of 8703 Kg and they converted this to liters by dividing the density to get 4916 L and then placed then 5000 L of fuel

The quantity required was

7862 L * 1.77 Lb/L = 13915.74 Lb (pounds not kilos)

then converting this pounds to Kg by multiplying by  0.454 Kg/L one gets

6173 Kg on board

Amount Required

( 22,300 -6173)  :  16127 Kg

16127 Kg/ 0.803 Kg/L =  20083 L

5 0
3 years ago
-Electronic control modules can easily evaluate the voltage and current levels of circuits to which they are connected and deter
erma4kov [3.2K]

Answer:

multiplexing

Explanation:

3 0
3 years ago
The underground cafe has an operating cash flow of $187,000 and a cash flow to creditors of $71,400 for the past year. During th
Serggg [28]

Answer:

cash flow to stockholders = $39,700

Explanation:

Operating cash flow = $187,000

cash flow to creditors = $71,400

Net working capital = $28,000

Net capital spending =  $47,900

Cash flow to stockholders = ?

CFF = operating cash flow - net working capital - net capital spending

CFF = $187,000 - $28,000 - $47,900 = $111,100

CFF = cash flow to creditors + cash flow to stockholders

cash flow to stockholders = CFF - cash flow to creditors

cash flow to stockholders = $111,100 - $71,400 = $39,700

Hence $39,700 is the amount of the cash flow to stockholders for the last year.

3 0
3 years ago
How does the "E" in STEM work with the other letters
KIM [24]

Answer:

if you are speaking of the acronym then Engineering uses science and mathematics to solve everyday problems in society

4 0
3 years ago
If we have silicon at 300K with 10 microns of p-type doping of 4.48*10^18/cc and 10 microns of n-type doping at 1000 times less
liq [111]

Answer:

The resistance is 24.9 Ω

Explanation:

The resistivity is equal to:

R=\frac{1}{N_{o}*u*V } =\frac{1}{4.48x10^{15}*1500*106x10^{-19}  } =0.93ohm*cm

The area is:

A = 60 * 60 = 3600 um² = 0.36x10⁻⁴cm²

w=\sqrt{\frac{2E(V_{o}-V) }{p}(\frac{1}{N_{A} }+\frac{1}{N_{D} })

If NA is greater, then, the term 1/NA can be neglected, thus the equation:

w=\sqrt{\frac{2E(V_{o}-V) }{p}(\frac{1}{N_{D} })

Where

V = 0.44 V

E = 11.68*8.85x10¹⁴ f/cm

V_{o} =\frac{KT}{p} ln(\frac{N_{A}*N_{D}}{n_{i}^{2}  } , if n_{i}=1.5x10^{10}cm^{-3}  \\V_{o}=0.02585ln(\frac{4.48x10^{18}*4.48x10^{15}  }{(1.5x10^{10})^{2}  } )=0.83V

w=\sqrt{\frac{2*11.68*8.85x10^{-14}*(0.83-0.44) }{1.6x10^{-19}*4.48x10^{15}  } } =3.35x10^{-5} cm=0.335um

The length is:

L = 10 - 0.335 = 9.665 um

The resistance is:

Re=\frac{pL}{A} =\frac{0.93*9.665x10^{-4} }{0.36x10^{-4} } =24.9ohm

7 0
3 years ago
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