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nirvana33 [79]
3 years ago
5

Please help I need by today !! What is the purpose of a portfolio?

Engineering
1 answer:
svetoff [14.1K]3 years ago
6 0

Answer:

To document your work and projects.

Explanation:

I hope I got the right meaning. :)

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For some transformation having kinetics that obey the Avrami equation, the parameter n is known to have a value of 2. If, after
kotegsom [21]

This question is incomplete, the complete question is;

For some transformation having kinetics that obey the Avrami equation, the parameter n is known to have a value of 2. If, after 100 s, the reaction is 40% complete, how long (total time in seconds) will it take the transformation to go to 95% completion

y = 1 - exp( -ktⁿ )

Answer: the time required for 95% transformation is 242.17 s

Explanation:

First, we calculate the value of k which is the dependent variable in Avrami equation

y = 1 - exp( -ktⁿ )

exp( -ktⁿ ) = 1 - y

-ktⁿ = In( 1 - y )

k = - In( 1 - y ) / tⁿ

now given that; n = 2, y = 40% = 0.40, and t = 100 s

we substitute

k = - In( 1 - 0.40 ) / 100²

k = - In(0.60) / 10000

k = 0.5108 / 10000

k = 0.00005108 ≈ 5.108 × 10⁻⁵

Now calculate the time required for 95% transformation

tⁿ = - In( 1 - y ) / k

t = [- In( 1 - y ) / k ]^1/n

n = 2, y = 95% = 0.95 and k = 5.108 × 10⁻⁵

we substitute our values

t = [- In( 1 - 0.95 ) / 5.108 × 10⁻⁵ ]^1/2

t = [2.9957 / 5.108 × 10⁻⁵]^1/2

t = [ 58647.22 ]^1/2

t = 242.17 s

Therefore the time required for 95% transformation is 242.17 s

8 0
3 years ago
2.
den301095 [7]
Jae pain seems the most off
4 0
3 years ago
**Please Help. ASAP**
natima [27]

Answer:

The answer is below

Explanation:

1)

\frac{v-u}{a} =t\\\\Making \ v\ the \ subject\ of\ formula:\\\\First \ cross-multiply:\\\\v-u=at\\\\add\ u\ to \ both\ sides:\\\\v-u+u=at+u\\\\v=u+at

2)

\frac{y-x^2}{x}=3z\\ \\Making\ y\ the\ subject\ of\ formula:\\\\First \ cross \ multiply:\\\\y-x^2=3xz\\\\y=3xz+x^2\\\\y=x(x+3z)

3)

x+xy=y\\\\Making\ x\ the\ subject\ of\ formula:\\\\x(1+y)=y\\\\Divide\ through\ by\ 1+y\\\\\frac{x(1+y)}{1+y} =\frac{y}{1+y} \\\\x=\frac{y}{1+y}

4)

x+y=xy\\\\Making\ x\ the\ subject\ of\ formula:\\\\Subtract\ x\ from \ both\ sides:\\\\x+y-x=xy-x\\\\y=xy-x\\\\y=x(y-1)\\\\Divide\ through\ by \ y-1\\\\\frac{y}{y-1} =\frac{x(y-1)}{y-1}\\ \\x=\frac{y}{y-1}

5)

x=y+xy\\\\Making\ x\ the\ subject\ of\ formula:\\\\Subtract\ xy\ from \ both\ sides:\\\\x-xy=y+xy-xy\\\\x-xy=y\\\\x(1-y)=y\\\\Divide\ through\ by \ 1-y\\\\\frac{x(1-y)}{1-y} =\frac{y}{1-y}\\ \\x=\frac{y}{1-y}

6)

E=\frac{1}{2}mv^2-\frac{1}{2}mu^2\\  \\Making\ u\ the\ subject \ of\ formula:\\\\Multiply \ through\ by \ 2\\\\2E=mv^2-mu^2\\\\mu^2=mv^2-2E\\\\Divide\ through\ by\ m:\\\\u^2=\frac{mv^2-2E}{m}\\ \\Take\ square\ root\ of \ both\ sides:\\\\u=\sqrt{\frac{mv^2-2E}{m}}

7)

\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\\  \\Making\ y\ the\ subject \ of\ formula:\\\\\frac{x^2}{a^2}-1=\frac{y^2}{b^2}\\\\Multiply\ through\ by\ b^2\\\\b^2(\frac{x^2}{a^2} -1)=y^2\\\\Take\ square\ root\ of\ both\ sides:\\\\y=\sqrt{b^2(\frac{x^2}{a^2} -1)}

8)

ay^2=x^3\\\\Make\ y\ the\ subject\ of\ formula:\\\\Divide\ through\ by\ a:\\\\y^2=\frac{x^3}{a}\\ \\Take\ square\ root\ of\ both\ sides:\\\\y=\sqrt{\frac{x^3}{a}} \\

4 0
3 years ago
Both model building codes and NFPA __________ can be used to determine the type of construction used in a building.
Alex777 [14]

Answer:

Both model building codes and NFPA 220 can be used to determine the type of construction used in a building.

8 0
2 years ago
An object of irregular shape has a characteristic length of � = 1.00 [�] and is maintained at a uniform surface temperature of �
Pani-rosa [81]

The correct question;

An object of irregular shape has a characteristic length of L = 1 m and is maintained at a uniform surface temperature of Ts = 400 K. When placed in atmospheric air at a temperature of Tinfinity = 300 K and moving with a velocity of V = 100 m/s, the average heat flux from the surface to the air is 20,000 W/m² If a second object of the same shape, but with a characteristic length of L = 5 m, is maintained at a surface temperature of Ts = 400 K and is placed in atmospheric air at Too = 300 K, what will the value of the average convection coefficient be if the air velocity is V = 20 m/s?

Answer:

h'_2 = 40 W/K.m²

Explanation:

We are given;

L1 = 1m

L2 = 5m

T_s = 400 K

T_(∞) = 300 K

V = 100 m/s

q = 20,000 W/m²

Both objects have the same shape and density and thus their reynolds number will be the same.

So,

Re_L1 = Re_L2

Thus, V1•L1/v1 = V2•L2/v2

Hence,

(h'_1•L1)/k1 = (h'_2•L2)/k2

Where h'_1 and h'_2 are convection coefficients

Since k1 = k2, thus, we now have;

h'_2 = (h'_1(L1/L2)) = [q/(T_s - T_(∞))]• (L1/L2)

Thus,

h'_2 = [20,000/(400 - 300)]•(1/5)

h'_2 = 40 W/K.m²

5 0
3 years ago
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