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Ilia_Sergeevich [38]
4 years ago
13

When using a computer, you should not keep your:

Engineering
2 answers:
kykrilka [37]4 years ago
8 0
Your thighs should not be perpendicular and lower legs not parallel because you have either broke your knee caps backwards or your kneeling
Gre4nikov [31]4 years ago
3 0
thighs perpendicular to the floor
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Which of the following elements would make a good insulator?
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Glass

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The rear window of an automobile is defogged by passingwarm air at 40ºC over its inner surface, and the associatedconvection coe
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4 years ago
Under normal conditions, gauge pressure is ______ absolute pressure.
levacccp [35]

Answer:

Option B  Lower than

Explanation:

Gauge pressure is a relative measurement based on atmospheric pressure. Gauge pressure can be positive if it is above atmospheric pressure or it can also be negative it is below.  On another hand, absolute pressure is an actual pressure in a space and its value has always to be zero or above. Basically absolute pressure is zero if it is in a perfect vacuum. So the measurement of absolute pressure is gauge pressure + atmospheric pressure.  This is the reason in normal condition the gauge pressure = absolute pressure - atmospheric pressure and therefore is lower than absolute pressure

3 0
4 years ago
Serratia marcescens bacteria are used for the production of threonine. The maximum specific oxygen uptake rate of S. marcescens
Sati [7]

Answer:

The rate of cell metabolism is limited by mass transfer since the value of maximum cell concentration obtained (38 g/l) is lower than 50 g l-1, the value planed.

Explanation:

                                                     Data

<u>kLa</u> = 0.17/s

<u>Solubility of oxygen</u> =  8 × 10^-3 kg / m^3

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

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

<u>**The maximum cell density</u> = 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

4 \frac{mmol}{gh} \frac{1 gmol}{1000mmol}\frac{1h}{3600s}\frac{32 g}{gmol} \frac{1Kg}{1000g}

= 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

_____________________

5 0
3 years ago
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