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geniusboy [140]
3 years ago
9

Not much of a question :/ dont answer

Engineering
1 answer:
Mamont248 [21]3 years ago
5 0

Answer: ok

Explanation: this is totally not an answer

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A local bank, with several branches in three cities, requests changes to its mortgage calculation software.

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Which federal agency issues the model food code?
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The FDA! (food and drug administration)

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Read 2 more answers
A saturated 1.5 ft3 clay sample has a natural water content of 25%, shrinkage limit (SL) of 12% and a specific gravity (GS) of 2
Svetllana [295]

79 f t^{3} is the volume of the sample when the water content is 10%.

<u>Explanation:</u>

Given Data:

V_{1}=100\ \mathrm{ft}^{3}

First has a natural water content of 25% = \frac{25}{100} = 0.25

Shrinkage limit, w_{1}=12 \%=\frac{12}{100}=0.12

G_{s}=2.70

We need to determine the volume of the sample when the water content is 10% (0.10). As we know,

V \propto[1+e]

\frac{V_{2}}{V_{1}}=\frac{1+e_{2}}{1+e_{1}}  ------> eq 1

e_{1}=\frac{w_{1} \times G_{s}}{S_{r}}

The above equation is at S_{r}=1,

e_{1}=w_{1} \times G_{s}

Applying the given values, we get

e_{1}=0.25 \times 2.70=0.675

Shrinkage limit is lowest water content

e_{2}=w_{2} \times G_{s}

Applying the given values, we get

e_{2}=0.12 \times 2.70=0.324

Applying the found values in eq 1, we get

\frac{V_{2}}{100}=\frac{1+0.324}{1+0.675}=\frac{1.324}{1.675}=0.7904

V_{2}=0.7904 \times 100=79\ \mathrm{ft}^{3}

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3 years ago
A machine used to shred cardboard boxes for composting has a first cost of $10,000, an AOC of $7000 per year, a 3-year life, and
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Annual Payment where F is accumulated sum of amount, n is number of years and i is annual rate of interest. The standard notation equation is in the image since i can’t type it-
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2 years ago
For some transformation having kinetics that obey the Avrami equation (Equation 10.17), the parameter n is known to have a value
OleMash [197]

Answer:

t = 25.10 sec

Explanation:

we know that Avrami equation

Y = 1 - e^{-kt^n}

here Y is percentage of completion  of reaction = 50%

t  is duration of reaction = 146 sec

so,

0.50 = 1 - e^{-k^146^2.1}

0.50 = e^{-k306.6}

taking natural log on both side

ln(0.5) = -k(306.6)

k = 2.26\times 10^{-3}

for 86 % completion

0.86 = 1 - e^{-2.26\times 10^{-3} \times t^{2.1}}

e^{-2.26\times 10^{-3} \times t^{2.1}} = 0.14

-2.26\times 10^{-3} \times t^{2.1} = ln(0.14)

t^{2.1} = 869.96

t = 25.10 sec

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3 years ago
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