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dusya [7]
3 years ago
11

I need to solve for d

Engineering
2 answers:
mrs_skeptik [129]3 years ago
6 0

Answer:

it's not included

Explanation:

plz exact ur explain

KonstantinChe [14]3 years ago
4 0

Answer:

si amor

Explanation:

Hoiykñjdnlklbutrk

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Only an outer panel is being replaced. Technician A says that removing the spot welds by drilling through both panels allows the
Angelina_Jolie [31]

Answer:

6e66363636633747747363637737373737337374

5 0
3 years ago
A closed, rigid tank is lled with a gas modeled as an ideal gas, initially at 27°C and a gage pressure of 300 kPa. The gas is he
sergejj [24]

Answer:

T₂ =93.77  °C

Explanation:

Initial temperature ,T₁ =27°C= 273 +27 = 300 K

We know that

Absolute pressure = Gauge pressure + Atmospheric pressure

Initial pressure ,P₁ = 300+1=301 kPa

Final pressure  ,P₂= 367+1 = 368  kPa

Lets take  temperature=T₂

We know that ,If the volume of the gas is constant ,then we can say that

\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}

{T_2}=T_1\times \dfrac{P_2}{P_1}

Now by putting the values in the above equation we get

{T_2}=300\times \dfrac{368}{301}\ K

The temperature in  °C

T₂ = 366.77 - 273  °C

T₂ =93.77  °C

8 0
3 years ago
A well is located in a 20.1-m thick confined aquifer with a conductivity of 14.9 m/day and a storativity of 0.0051. If the well
ahrayia [7]

Answer:

S = 5.7209 M

Explanation:

Given data:

B = 20.1 m

conductivity ( K ) = 14.9 m/day

Storativity  ( s ) = 0.0051

1 gpm = 5.451 m^3/day

calculate the Transmissibility ( T ) = K * B

                                                       = 14.9 * 20.1 = 299.5  m^2/day

Note :

t = 1

U = ( r^2* S ) / (4*T*<em> t </em>)

  = ( 7^2 * 0.0051 ) / ( 4 * 299.5 * 1 ) = 2.0859 * 10^-4

Applying the thesis method

W(u) = -0.5772 - In(U)

       = 7.9

next we calculate the pumping rate from well ( Q ) in m^3/day

= 500 * 5.451 m^3 /day

= 2725.5 m^3 /day

Finally calculate the drawdown at a distance of 7.0 m form the well after 1 day of pumping

S = \frac{Q}{4\pi T} * W (u)

 where : Q = 2725.5

               T = 299.5

               W(u)  = 7.9

substitute the given values into equation above

S = 5.7209 M

4 0
3 years ago
Zachary finished an internship as a Software Quality Assurance Engineer. For which job is he best qualified?
astra-53 [7]

Answer:

B. A software development firm needs someone to find and fix bugs on multiple computer platforms.

Explanation:

A software quality assurance engineer is someone who monitors every phase of the software development process so as to ensure design quality, making sure that the software adheres to the standards set by the development company. Finding bugs would make this intern a amazing bug finder

6 0
3 years ago
Read 2 more answers
Consider a Carnot heat pump cycle executed in a steady-flow system in the saturated mixture region using R-134a flowing at a rat
attashe74 [19]

Answer:

7.15

Explanation:

Firstly, the COP of such heat pump must be measured that is,

              COP_{HP}=\frac{T_H}{T_H-T_L}

Therefore, the temperature relationship, T_H=1.15\;T_L

Then, we should apply the values in the COP.

                           =\frac{1.15\;T_L}{1.15-1}

                           =7.67

The number of heat rejected by the heat pump must then be calculated.

                   Q_H=COP_{HP}\times W_{nst}

                          =7.67\times5=38.35

We must then calculate the refrigerant mass flow rate.

                   m=0.264\;kg/s

                   q_H=\frac{Q_H}{m}

                         =\frac{38.35}{0.264}=145.27

The h_g value is 145.27 and therefore the hot reservoir temperature is 64° C.

The pressure at 64 ° C is thus 1849.36 kPa by interpolation.

And, the lowest reservoir temperature must be calculated.

                   T_L=\frac{T_H}{1.15}

                        =\frac{64+273}{1.15}=293.04

                        =19.89\°C

the lowest reservoir temperature = 258.703  kpa                    

So, the pressure ratio should be = 7.15

8 0
2 years ago
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