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3 years ago

BHE:FLI:JPM next sequence

Engineering

1 answer:

Natasha_Volkova [10]3 years ago

0 0

Answer:

NTQ

Explanation:

The given sequence is

BHE : FLI : JPM

If is clear that, alphabets on first places are B, F, J. Difference between their place vales is 4.

B+4=F,F+4=J ; so alphabet on first place of next term of sequence is J+4=N.

Similarly, alphabets on second places are H, L, P. Difference between their place vales is 4.

H+4=L,L+4=P ; so alphabet on second place of next term of sequence is P+4=T.

Alphabets on third places are E, I, M. Difference between their place vales is 4.

E+4=I,I+4=M ; so alphabet on third place of next term of sequence is M+4=Q.

Therefore, the next term is NTQ.

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Vedmedyk [2.9K]

The answer to your question should be 630, if you have a roof the size of 40 feet, by 35 feet.

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4 years ago

The pressure distribution over a section of a two-dimensional wing at 4 degrees of incidence may be approximated as follows: Upp

Aliun [14]

Answer:

The lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

Explanation:

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$Cp_u=-0.8 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.8*0.6+0.4*0.1$

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$Cp_l=-0.4 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.4*0.6+0.4*0.1$

The difference of the Cp over the airfoil is given as

$\Delta Cp=Cp_l-Cp_u\\\Delta Cp=-0.4*0.6+0.4*0.1-(-0.8*0.6-0.4*0.1)\\\Delta Cp=-0.4*0.6+0.4*0.1+0.8*0.6+0.4*0.1\\\Delta Cp=0.4*0.6+0.4*0.2\\\Delta Cp=0.32$

Now the Lift Coefficient is given as

$C_L=\Delta C_p cos(\alpha_i)\\C_L=0.32\times cos(4*\frac{\pi}{180})\\C_L=0.3192$

Now the coefficient of moment about the leading edge is given as

$C_M=-0.3*0.4*0.6-(0.6+\dfrac{0.4}{3})*0.2*0.4\\C_M=-0.1306$

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5 0

3 years ago

Select the correct answer.

Genrish500 [490]

A is the correct answer

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3 years ago

Serratia marcescens bacteria are used for the production of threonine. The maximum specific oxygen uptake rate of S. marcescens

Sati [7]

Answer:

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Explanation:

Data

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The maximum specific oxygen uptake rate = 4 mmol O2 / g h.

Concentration of oxygen = 0.5 × 10^-3 kg/ m^3

**The maximum cell density = 50 g/l

___________________

The calculated maximum cell concentration:

xmax= kLa · CAL*/ qo

CAL* is the solubility of oxygen in the broth and qo is the specific oxygen uptake rate

Replacing the data given

xmax= ( 0.17/s ) · (8 × 10^-3 kg / m^3) / 4 mmol O2 / g h

4 mmol O2 / g h to kg O2/ g s

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= 3.56 x 10^-3 kg O2/ g s

So then,

xmax= ( 0.17/s ) · (8 × 10^-3 kg / m^3) / 3.56 x 10^-3 o kg O2/ g s

xmax= 3. 8 x 10^4 g/ m^3 = 38 g/l

_____________________

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3 years ago

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garik1379 [7]

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