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Sloan [31]
3 years ago
5

I logged on today to work on my makeup work. A: True B: False

Engineering
2 answers:
Lelu [443]3 years ago
7 0
True I don’t really know.
____ [38]3 years ago
5 0

Answer:

True?

Explanation:

Is this a personal question about yourself?

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Consider a unidirectional continuous fiber-reinforced composite with epoxy as the matrix with 55% by volume fiber.i. Calculate t
ohaa [14]

Answer:

I)E= 40.95 GPa

II)E=5.29 GPa

Explanation:

I)

Given that

E₁ = 2.41 GPa  ,V₁=1-0.55 = 0.45

E₂ = 72.5 GPa   ,V₂=0.55

Longitudinal moduli  given as ;

E= E₁V₁+E₂V₂

E= 2.41 x 0.45 + 72.5 x 0.55 GPa

E= 40.95 GPa

II)

E₁ = 2.41 GPa  ,V₁=1-0.55 = 0.45

E₂ =230 GPa   ,V₂=0.55

Transverse moduli given as:

\dfrac{1}{E}=\dfrac{V_1}{E_1}+\dfrac{V_2}{E_2}

\dfrac{1}{E}=\dfrac{0.45}{2.41}+\dfrac{0.55}{230}

E=5.29 GPa

7 0
3 years ago
A heat engine receives 6 kW from a 250oC source and rejects heat at 30oC. Examine each of three cases with respect to the inequa
taurus [48]

Answer:

Explanation:

Given

T_h=250^{\circ}C\approx 523\ K

T_L=30^{\circ}C\approx 303\ K

Q_1=6 kW

From Clausius inequality

\oint \frac{dQ}{T}=0  =Reversible cycle

\oint \frac{dQ}{T}  =Irreversible cycle

\oint \frac{dQ}{T}>0  =Impossible

(a)For P_{out}=3 kW

Rejected heat Q_2=6-3=3\ kW

\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}

=\frac{6}{523}-\frac{3}{303}=1.57\times 10^{-3} kW/K

thus it is Impossible cycle

(b)P_{out}=2 kW

Q_2=6-2=4 kW

\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}

=\frac{6}{523}-\frac{4}{303}=-1.73\times 10^{-3} kW/K

Possible

(c)Carnot cycle

\frac{Q_2}{Q_1}=\frac{T_1}{T_2}

Q_2=3.47\ kW

\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}

=\frac{6}{523}-\frac{3.47}{303}=0

and maximum Work is obtained for reversible cycle when operate between same temperature limits

P_{out}=Q_1-Q_2=6-3.47=2.53\ kW

Thus it is possible

6 0
4 years ago
Explain why systems of measurement are defined and provide an example to support your answer.
kotegsom [21]

Answer:

Explanation:

Chicken wings

8 0
3 years ago
If the total energy change of an system during a process is 15.5 kJ, its change in kinetic energy is -3.5 kJ, and its potential
drek231 [11]

Answer:

The change in specific internal energy is 3.5 kj.

Explanation:

Step1

Given:

Total change in energy is 15.5 kj.

Change in kinetic energy is –3.5 kj.

Change in potential energy is 0 kj.

Mass is 5.4 kg.

Step2

Calculation:

Change in internal energy is calculated as follows:

\bigtriangleup E=\bigtriangleup KE+\bigtriangleup PE+\bigtriangleup U15.5=-3.5+0+\bigtriangleup U

\bigtriangleup U=19 kj.

Step3

Specific internal energy is calculated as follows:

\bigtriangleup u=\frac{\bigtriangleup U}{m}

\bigtriangleup u=\frac{19}{5.4}

\bigtriangleup u=3.5 kj/kg.

Thus, the change in specific internal energy is 3.5 kj/kg.

7 0
4 years ago
[ 3/8-28 UNC] According to the thread note, what does 3/8 mean?
lara [203]

Answer:

It diameter

Explanation:

That where the diameter always is.

4 0
3 years ago
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