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a_sh-v [17]
3 years ago
13

Alberta Einstein teaches a business class at Podunk University. To evaluate the students in this class, she has given three test

s. It is now the end of the semester and Alberta asks you to create a program that inputs each student’s test scores and outputs the average score for each student and the overall class average. (Hint: The outer loop should allow for Ms. Einstein to input all the students, one by one, and the inner loop should accept the three exam scores and compute the average for each student.)

Engineering
1 answer:
madreJ [45]3 years ago
7 0

Answer:

See explaination

Explanation:

Please kindly check attachment for step by step solution of the given problem represented as a flow chart.

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A rigid 10-L vessel initially contains a mixture of liquid and vapor water at 100 °C, with a quality factor of 0.123. The mixtur
masya89 [10]

Answer:

Q_{in} = 46.454\,kJ

Explanation:

The vessel is modelled after the First Law of Thermodynamics. Let suppose the inexistence of mass interaction at boundary between vessel and surroundings, changes in potential and kinectic energy are negligible and vessel is a rigid recipient.

Q_{in} = U_{2} - U_{1}

Properties of water at initial and final state are:

State 1 - (Liquid-Vapor Mixture)

P = 101.42\,kPa

T = 100\,^{\textdegree}C

\nu = 0.2066\,\frac{m^{3}}{kg}

u = 675.761\,\frac{kJ}{kg}

x = 0.123

State 2 - (Liquid-Vapor Mixture)

P = 476.16\,kPa

T = 150\,^{\textdegree}C

\nu = 0.2066\,\frac{m^{3}}{kg}

u = 1643.545\,\frac{kJ}{kg}

x = 0.525

The mass stored in the vessel is:

m = \frac{V}{\nu}

m = \frac{10\times 10^{-3}\,m^{3}}{0.2066\,\frac{m^{3}}{kg} }

m = 0.048\,kg

The heat transfer require to the process is:

Q_{in} = m\cdot (u_{2}-u_{1})

Q_{in} = (0.048\,kg)\cdot (1643.545\,\frac{kJ}{kg} - 675.761\,\frac{kJ}{kg} )

Q_{in} = 46.454\,kJ

3 0
2 years ago
A cubic transmission casing whose side length is 25cm receives an input from the engine at a rate of 350 hp. If the vehicle's ve
Musya8 [376]

Answer:

The surface temperature is 921.95°C .

Explanation:

Given:

   a=25 cm ,P=350 hp⇒P=260750 W

Power transmitted 0.95\times 260750W and remaining will lost in the form of heat.This heat transmitted to air by the convection.

 h=230\frac{W}{m^2-K},\eta =0.95

Actually heat will be transmit by the convection.

In convection Q=hA\Delta T

So P=\Delta T\times Q

0.05\times 260750=230\times0.25^2\(T-15)

T=921.95°C

So the surface temperature is 921.95°C .

6 0
3 years ago
A part made from annealed AISI 1018 steel undergoes a 20 percent cold-work operation. Obtain the yield strength and ultimate str
Charra [1.4K]

Answer:

yield strength before cold work = 370 MPa

yield strength after cold work = 437.87 MPa

ultimate strength before cold work = 440 MPa

ultimate strength after cold work = 550 MPa

Explanation:

given data

AISI 1018 steel

cold work factor W = 20% = 0.20

to find out

yield strength and ultimate strength before and after the cold-work operation

solution

we know the properties of AISI 1018 steel is

yield strength σy =  370 MPa

ultimate tensile strength σu = 440 MPa

strength coefficient K = 600 MPa

strain hardness n = 0.21

so true strain is here ∈ = ln\frac{1}{1-0.2} = 0.223

so

yield strength after cold is

yield strength = K \varepsilon ^n

yield strength =  600*0.223^{0.21)

yield strength after cold work = 437.87 MPa

and

ultimate strength after cold work is

ultimate strength = \frac{\sigma u}{1-W}

ultimate strength = \frac{440}{1-0.2}

ultimate strength after cold work = 550 MPa

8 0
3 years ago
Turn on your____
storchak [24]

Answer:

b

Explanation:

5 0
2 years ago
Read 2 more answers
What is the sun's degree angle in the sky in summer and in winter?
Vladimir [108]

Answer:

At the time of the summer or winter solstices, the Sun is 23.44° degrees above or below the horizon, respectively, irrespective of time of day.

Explanation:

6 0
1 year ago
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