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Temka [501]
3 years ago
10

Please help i will give brainilest

Engineering
2 answers:
Eduardwww [97]3 years ago
8 0

Answer:

I hot u

Explanation:

it makes things function better and more fluently than a manual version of the certain object

Troyanec [42]3 years ago
3 0

Answer:

l m a o

Explanationl

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A polymeric extruder is turned on and immediately begins producing a product at a rate of 10 kg/min. An operator realizes 20 min
hodyreva [135]

Answer:

The plot of the function production rate m(t) (in kg/min) against time t (in min) is attached to this answer.

The production rate function M(t) is:

m(t)=[H(t)\cdot10+H(t-20)\cdot5-H(t-80)\cdot14+H(t-81)\cdot9]kg/min (1)

The Laplace transform of this function is:

\displaystyle m(s)=[\frac{10+5e^{-20s}-14e^{-80s}+9e^{-81s}}{s}]kg/min    (2)

Explanation:

The function of the production rate can be considered as constant functions by parts in the domain of time. To make it a continuous function, we can use the function Heaviside (as seen in equation (1)). To join all the constant functions, we consider at which time the step for each one of them appears and sum each function multiply by the function Heaviside.

For the Laplace transform we use the following rules:

\mathcal{L}[f(x)+g(x)]=\mathcal{L}[f(x)]+\mathcal{L}[g(x)]=F(s)+G(s)    (3)

\mathcal{L}[aH(x-b)]=\displaystyle\frac{ae^{-bs}}{s}    (4)

4 0
4 years ago
In a 1D compression test with double drainage, the pore pressure readings are practically zero after 8 minutes for a clay sample
frosja888 [35]

Answer:

The duration of the consolidation process for the same clay is 32 min

Explanation:

for clay 1:

t1=0

H1=thickness=2 cm

for the clay 2:

t2=?

H2=2 cm

The time factor is equal to:

T=(\frac{Cv}{d^{2} })t

where Cv is the coefficient of consolidation

(\frac{Cvt}{d^{2} })_{1}=  (\frac{Cvt}{d^{2} })_{2}

if Cv is constant, we have:

(\frac{t1}{(\frac{H1}{2}) ^{2} })_{1}=(\frac{t2}{H2^{2} })_{2}\\\frac{0}{(\frac{2}{2})^{2})  }=\frac{t2}{2^{2} }

Clearing t2:

t2=32 min

3 0
3 years ago
How is me depressed or am me not? (but with a yoda voice)
Murrr4er [49]

Answer:

but the way is the way but the WAY is not the way

Explanation:

(yoda voice)

5 0
3 years ago
Read 2 more answers
A 0.35-ft3 well-insulated rigid can initially contains refrigerant-134a at 90 psia and 30°F. Now a crack develops in the can, an
nydimaria [60]

Answer:

m = 1.37 lbm

Explanation:

We are given that;

P1 = 90 psia

T1 = 30°F

From the table i attached, at T = 30°F, entropy, s1 = sf = 0.04752 Btu/lbm.R

We are also given;

P2 = 20 psia.

At this, s2 = s1 = 0.04752 Btu/lbm.R

From the table i attached, sf at 20 psia is; sf = 0.02605 Btu/lbm.R and sfg = 0.19962 Btu/lbm.R

Now, formula for quality of steam at Pressure P2 is;

X2 = (s2 - sf)/sfg

Plugging in the relevant values to obtain;

X2 = (0.04752 - 0.02605)/0.19962

X2 = 0.1076

Now, v2 = vf + x2•vfg

From the table i attached, at 20 psia, vf = 0.01182, vg =2.27772 and vfg = vg - vf = 2.27772 - 0.01182 = 2.2659 ft³/lbm

Thus,

v2 = 0.01182 + 0.1076*2.2659 = 0.2556 ft³/lbm

Now, let's find mass of the refrigerant from, m = V/v2

m = 0.35/0.2556 = 1.37 lbm

7 0
3 years ago
The human eye, as well as the light-sensitive chemicals on color photographic film, respond differently to light sources with di
jeka57 [31]

Answer:

a) at T = 5800 k  

  band emission = 0.2261

at T = 2900 k

  band emission = 0.0442

b) daylight (d) = 0.50 μm

    Incandescent ( i ) =  1 μm

Explanation:

To Calculate the band emission fractions we will apply the Wien's displacement Law

The ban emission fraction in spectral range λ1 to λ2 at a blackbody temperature T can be expressed as

F ( λ1 - λ2, T ) = F( 0 ----> λ2,T) - F( 0 ----> λ1,T )

<em>Values are gotten from the table named: blackbody radiati</em>on functions

<u>a) Calculate the band emission fractions for the visible region</u>

at T = 5800 k  

  band emission = 0.2261

at T = 2900 k

  band emission = 0.0442

attached below is a detailed solution to the problem

<u>b)calculate wavelength corresponding to the maximum spectral intensity</u>

For daylight ( d ) = 2898 μm *k / 5800 k  = 0.50 μm

For Incandescent ( i ) = 2898 μm *k / 2900 k = 1 μm

3 0
3 years ago
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