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3 years ago

¿Por qué la lógica de proposiciones es conocida también como lógica de las proposiciones sin analizar?

Engineering

1 answer:

Ganezh [65]3 years ago

6 0

CHICKEN WING CHICKEN WING HOT DOG AND BALONEY CHICKEN AND MACORONI

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A spring-loaded piston-cylinder contains 1 kg of carbon dioxide. This system is heated from 104 kPa and 25 °C to 1,068 kPa and 3

labwork [276]

Answer:

Q = -68.859 kJ

Explanation:

given details

mass $co_2 = 1 kg$

initial pressure P_1 = 104 kPa

Temperature T_1 = 25 Degree C = 25+ 273 K = 298 K

final pressure P_2 = 1068 kPa

Temperature T_2 = 311 Degree C = 311+ 273 K = 584 K

we know that

molecular mass of $co_2 = 44$

R = 8.314/44 = 0.189 kJ/kg K

c_v = 0.657 kJ/kgK

from ideal gas equation

PV =mRT

$V_1 = \frac{m RT_1}{P_1}$

$=\frac{1*0.189*298}{104}$

$V_1 = 0.5415 m3$

$V_2 = \frac{m RT_2}{P_2}$

$=\frac{1*0.189*584}{1068}$

$V_1 = 0.1033 m3$

WORK DONE

$W =P_{avg}*{V_2-V_1}$

w = 586*(0.1033 -0.514)

W =256.76 kJ

INTERNAL ENERGY IS

$\Delta U = m *c_v*{V_2-V_1}$

$\Delta U = 1*0.657*(584-298)$

$\Delta U =187.902 kJ$

HEAT TRANSFER

$Q = \Delta U +W$

= 187.902 +(-256.46)

Q = -68.859 kJ

7 0

3 years ago

A cylindrical bar of metal having a diameter of 20.9 mm and a length of 205 mm is deformed elastically in tension with a force o

natita [175]

Answer:

a. 1.91 b. -8.13 mm

Explanation:

Modulus =stress/strain; calculating stress =F/A, hence determine the strain

Poisson's ratio =(change in diameter/diameter)/strain

8 0

3 years ago

Please read and answer each question carefully.

Klio2033 [76]

the answer is (c)

After the vehicle is involved in a car accident or fire

5 0

3 years ago

Wastewater flows into a _________ once it is released into a floor drain.

vodomira [7]

Answer:

Explanation:

5 0

3 years ago

A reservoir delivers water to a horizontal pipeline 39 long The first 15 m has a diameter of 50 mm, after which it suddenly beco

allsm [11]

Answer:

The difference of head in the level of reservoir is 0.23 m.

Explanation:

For pipe 1

$d_1=50 mm,f_1=0.0048$

For pipe 2

$d_2=75 mm,f_2=0.0058$

Q=2.8 l/s

$Q=2.8\times 10^{-3]$

We know that Q=AV

$Q=A_1V_1=A_2V_2$

$A_1=1.95\times 10^{-3}m^2$

$A_2=4.38\times 10^{-3} m^2$

$So V_2=0.63 m/s,V_1=1.43 m/s$

head loss (h)

$h=\dfrac{f_1L_1V_1^2}{2gd_1}+\dfrac{f_2L_2V_2^2}{2gd_2}+0.5\dfrac{V_1^2}{2g}$

Now putting the all values

$h=\dfrac{0.0048\times 15\times 1.43^2}{2\times 9.81\times 0.05}+\dfrac{0.0058\times 24\times 0.63^2}{2\times 9.81\times 0.075}+0.5\dfrac{1.43^2}{2\times 9.81}$

So h=0.23 m

So the difference of head in the level of reservoir is 0.23 m.

8 0

3 years ago