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laiz [17]
3 years ago
10

Which Finance jobs can someone pursue with only a high school diploma? Check all that apply.

Engineering
2 answers:
Zolol [24]3 years ago
3 0

Answer:

Teller, Loan Officer, and Tax Preparer

Explanation:

marusya05 [52]3 years ago
3 0

Answer:

A, tax preparer and D, teller.

Explanation:

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As a means of preventing ice formation on the wings of a small, private aircraft, it is proposed that electric resistance heatin
DIA [1.3K]

Answer:

Average heat flux=3729.82 W/m^{2}

Explanation:

7 0
3 years ago
An air-standard Carnot cycle is executed in a closed system between the temperature limits of 350 and 1200 K. The pressures befo
SVEN [57.7K]

This question is incomplete, the complete question is;

An air-standard Carnot cycle is executed in a closed system between the temperature limits of 350 and 1200 K. The pressures before and after the isothermal compression are 150 and 300 kPa, respectively. If the net work output per cycle is 0.5 kJ, determine (a) the maximum pressure in the cycle, (b) the heat transfer to air, and (c) the mass of air. Assume variable specific heats for air.

Answer:

a) the maximum pressure in the cycle is 30.01 Mpa

b) the heat transfer to air is 0.7058 KJ

c) mass of Air is 0.002957 kg

Explanation:

Given the data in the question;

We find the relative pressure of air at 1200 K (T1) and 350 K ( T4)

so from the "ideal gas properties of air table"

Pr1 = 238

Pr4 = 2.379

we know that Pressure P1 is only maximum at the beginning of the expansion process,

so

now we express the relative pressure and pressure relation for the process 4-1

P1 = (Pr2/Pr4)P4

so we substitute

P1 = (238/2.379)300 kPa

P1 = 30012.6 kPa = 30.01 Mpa

Therefore the maximum pressure in the cycle is 30.01 Mpa

b)

the Thermal heat efficiency of the Carnot cycle is expressed as;

ηth = 1 - (TL/TH)

we substitute

ηth = 1 - (350K/1200K)

ηth = 1 - 0.2916

ηth = 0.7084

now we find the heat transferred

Qin = W_net.out / ηth

given that the net work output per cycle is 0.5 kJ

we substitute

Qin = 0.5 / 0.7084

Qin = 0.7058 KJ

Therefore, the heat transfer to air is 0.7058 KJ

c)

first lets express the change in entropy for process 3 - 4

S4 - S3 = (S°4 - S°3) - R.In(P4/P3)

S4 - S3 = - (0.287 kJ/Kg.K) In(300/150)kPa

= -0.1989 Kj/Kg.K = S1 - S2

so that; S2 - S1 = 0.1989 Kj/Kg.K

Next we find the net work output per unit mass for the Carnot cycle

W"_netout = (S2 - S1)(TH - TL)

we substitute

W"_netout = ( 0.1989 Kj/Kg.K )( 1200 - 350)K

= 169.065 kJ/kg

Finally we find the mass

mass m = W_ net.out /  W"_netout

we substitute

m = 0.5 / 169.065

m = 0.002957 kg

Therefore, mass of Air is 0.002957 kg

5 0
3 years ago
Which of the following terms is defined as small bumps and slashes within a fluid power system?
-BARSIC- [3]

Answer:

friction

Explanation:

''.''

7 0
3 years ago
Read 2 more answers
Does jupiter have a permanent storm thats looks like a large red spot?
alekssr [168]
Yes it does have a large spot
8 0
3 years ago
A thin plastic membrane is used to separate helium from a gas stream. Under steady-state conditions the concentration of helium
Juli2301 [7.4K]

Answer:

N_A=1.5*10^-8 kmol/s.m^2

Explanation:

<u>KNOWN: </u>

Molar concentration of helium at the inner and outer surfaces of a plastic membrane. Diffusion coefficient and membrane thickness.  

<u>FIND:</u>

Molar diffusion flux.  

<u>ASSUMPTIONS:</u>

(1) Steady-state conditions, (2) One-dimensional diffusion in a plane wall, (3) Stationary medium, (4) Uniform C = C_A + C_B.  

<u>ANALYSIS:</u> The molar flux may be obtained from

 N_A=D_AB/L(C_A,1-C_A,2)

       =10^-9 m^2/s/0.001 m(0.02-0.005)kmol/m^3

N_A=1.5*10^-8 kmol/s.m^2

<u>COMMENTS:</u> The mass flux is:

n_A,x=M_a*N_A,x

n_A,x=6*10^-8 kg/s m^2

5 0
3 years ago
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