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Vanyuwa [196]
2 years ago
13

Who do you contact to get your credit report?

Engineering
1 answer:
kvasek [131]2 years ago
3 0
Contact the main office of your building or class, or talk to the head of charge
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An electric water heater held at 120° F is kept in a 60°F room. When purchased, its insulation is equivalent to R-5. An owner pu
Trava [24]

Answer:

Total saving would be of 36.917 $\yr

Explanation:

Given Data:

T_{heater} = 120 Degree F

T_{room} = 60 Degree F

A = 30 ft^2

\eta = 100%

Heat loss before previous final value = \frac{A \Delta T}{R}

                                                              =\frac{30\times *(120-60)}{5}

                                                              = 360 Btu/hr

Heat loss after new value= \frac{30\times \times (120-60)}{15} = 120 Btu/hr

saving would be = 360 - 120 Btu/hr \times kw hr/ 3412 Btu\times 24 hr/day \times 365 day/year

                           = 616.1782 kw hr/yr

cost = 616.1782 \times 0.06$

         = 36.917 $\yr

6 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
Corrective maintenance _________________________.​
andrew-mc [135]

Answer:

b. ​diagnoses and corrects errors in an operational system

Explanation:

Corrective maintenance is one that is performed with the purpose of repairing faults or defects that occur in equipment and machinery.  As such, it is the most basic way of providing maintenance, as it simply involves repairing what has broken down. In this sense, corrective maintenance is a process that basically consists of locating and correcting faults or damages that are preventing the machine from performing its function in a normal way.

3 0
4 years ago
An inductor of ????=6.95 H with negligible resistance is placed in series with a ℰ=12.5 V battery, a ????=3.00 Ω resistor, and a
elena-14-01-66 [18.8K]

Answer:

a) I=0 b) 4.17V c) 0.354 A d) 14.5s

Explanation:

a) consider circuit in the attachment

i(t)= E/R (1- e^(-t/RL))

i(0)= 12.5/3×(1-e^(0/RL))

i(0)=0

b) at t⇒∞

i(∞)= 12.5/3× (1- e^(-∞/RL))

    = 4.17V

c) 1/RL= 1/(6.95×3)= 0.0479616

i(1.85) = 12.5/3 × (1- e^(-1.85×0.0479616)

         = 0.354A

d) I/2= I (1- e^(-t/RL))

t= - RL ln0.5

t= - 3×6.95 × (-0.693)

t= 14.5 s

8 0
4 years ago
5. If you designed a part with an overall width dimension of 1.250 inches and printed it on the same 3D printer using the same f
sleet_krkn [62]

Answer:

hu22 54=52222222222222

3 0
3 years ago
Read 2 more answers
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