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2 years ago

What did August Comte contribute to sociology including positivism

Engineering

1 answer:

Travka [436]2 years ago

4 0

Answer:

Auguste Comte was the first to develop the concept of "sociology." He defined sociology as a positive science. Positivism is the search for "invariant laws of the natural and social world." Comte identified three basic methods for discovering these invariant laws, observation, experimentation, and comparison.

Explanation:

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The emissivity of galvanized steel sheet, a common roofing material, is ε = 0.13 at temperatures around 300 K, while its absorpt

Step2247 [10]

Answer:

759.99W/m²

Explanation:

Question: If the temperature of the sheet is 77C,what is the incident solar radiation on aday with Tinf= Tsurr= 16°C?

Given

Energy Equation of the Gas

αs * Gs * A + h * A * (T inf - Tg) + εσA (Tsurr⁴- Tg⁴) = 0

Where σ= 5.67 *10^-8 W/m²K⁴ (Stefan-Boltzmann constant)

ε = 0.13 (Emisivity)

αs = 0.65 (Absorptivity for solar radiation)

h = 7W/m²K⁴

Tg = 77 + 273.15K = 350.15K

T inf = 16 + 273.15 = 288.15K

T surr= T inf = 288.15

Substitute the above values in the Gas Equation, we have

0.65 * Gs * A + 7 * A * (288.15 - 350.15) + 0.13 * 5.67 * 10^-8 * A * (288.15⁴ - 350.15⁴) = 0

0.65 * Gs * A = - 7 * A * (288.15 - 350.15) - 0.13 * 5.67 * 10^-8 * A * (288.15⁴ - 350.15⁴)

A cancels out, so we are left with

0.65 * Gs = - 7 * (288.15 - 350.15) - 0.13 * 5.67 * 10^-8 * (288.15⁴ - 350.15⁴)

0.65Gs = 434 - 0.7372 * 10^-8(−8,137,940,481.697)

0.65Gs = 434 + 0.7372 * 81.37940481697

0.65Gs = 493.992897231070284

Gs = 493.992897231070284/0.65

Gs = 759.9890726631850

Gs = 759.99W/m² ------- Approximated

3 0

2 years ago

A 400-m^3 storage tank is being constructed to hold LNG, liquefied natural gas, which may be assumed to be essentially pure meth

GuDViN [60]

Answer:

mass of LNG: 129501.3388 kg

quality: 0.005048662

Explanation:

Volume occupied by liquid:

400 m^3*0.9 = 360 m^3

Volume occupied by vapor

400 m^3*0.1 = 40 m^3

A figured with thermodynamic properties of saturated methane is attached. Notice that a liquid-gas mixture is present

For liquid phase specific volume (vf) at 150 K is 0.002794 m^3/kg and for vapor phase specific volume (vg) is 0.06118 m^3/kg

From specific volume definition:

vf = liquid volume/liquid mass

liquid mass = liquid volume/vf

liquid mass = 360 m^3/0.002794 m^3/kg

liquid mass = 128847.5304 kg

vg = vapor volume/vapor mass

vapor mass = liquid volume/vg

vapor mass = 40 m^3/0.06118 m^3/kg

vapor mass = 653.8084341 kg

total mass = 128847.5304 kg + 653.8084341 kg = 129501.3388 kg

Quality is defined as the ratio between vapor mass and total mass

quality = 653.8084341 kg/129501.3388 kg = 0.005048662

4 0

2 years ago

Water vapor at 6 MPa, 600 degrees C enters a turbine operating at steady state and expands to 10kPa. The mass flow rate is 2 kg/

kirill115 [55]

Answer:

Explanation:

Obtain the following properties at 6MPa and 600°C from the table "Superheated water".

$h_1=3658.8KL/Kg\\s_1=7.1693kJ/kg.k$

Obtain the following properties at 10kPa from the table "saturated water"

$h_{f2}=191.81KJ/Kg.K\\h_{fg2}=2392.1KJ/Kg\\s_{f2}=0.6492KJ/Kg.K\\s_{fg2}=7.4996KJ/Kg.K$

Calculate the enthalpy at exit of the turbine using the energy balance equation.

$\frac{dE}{dt}=Q-W+m(h_1-h_2)$

Since, the process is isentropic process $Q=0$

$0=0-W+m(h_1-h_2)\\h_2=h_1-\frac{W}{m}\\\\h_2=3658.8-\frac{2626}{2}\\\\=2345.8kJ/kg$

Use the isentropic relations:

$s_1=s_{2s}\\s_1=s_{f2}+x_{2s}s_{fg2}\\7.1693=6492+x_{2s}(7.4996)\\x_{2s}=87$

Calculate the enthalpy at isentropic state 2s.

$h_{2s}=h_{f2}+x_{2s}.h_{fg2}\\=191.81+0.87(2392.1)\\=2272.937kJ/kg$

a.)

Calculate the isentropic turbine efficiency.

$\eta_{turbine}=\frac{h_1-h_2}{h_1-h_{2s}}\\\\=\frac{3658.8-2345.8}{3658.8-2272.937}=0.947=94.7%$

b.)

Find the quality of the water at state 2

since $h_f$ at 10KPa < $h_2$ < $h_g$ at 10KPa

Therefore, state 2 is in two-phase region.

$h_2=h_{f2}+x_2(h_{fg2})\\2345.8=191.81+x_2(2392.1)\\x_2=0.9$

Calculate the entropy at state 2.

$s_2=s_{f2}+x_2.s_{fg2}\\=0.6492+0.9(7.4996)\\=7.398kJ/Kg.K$

Calculate the rate of entropy production.

$S=\frac{Q}{T}+m(s_2-s_1)$

since, Q = 0

$S=m(s_2-s_1)\\=2\frac{kg}{s}(7.398-7.1693)kJ/kg\\=0.4574kW/k$

6 0

3 years ago

All aspects of the Kirby-Bauer test are standardized to assure reliability. What might be the consequence of pouring the plates

EleoNora [17]

Answer:

it would affect the distance the antiantibodies diffuse from the disk

Explanation:

7 0

2 years ago

The following electrical characteristics have been determined for both intrinsic and p-type extrinsic gallium antimonide (GaSb)

xxTIMURxx [149]

Answer:

0.5m^2/Vs and 0.14m^2/Vs

Explanation:

To calculate the mobility of electron and mobility of hole for gallium antimonide we have,

$\sigma = n|e|\mu_e+p|e|\mu_h$ (S)

Where

e= charge of electron

n= number of electrons

p= number of holes

$\mu_e=$ mobility of electron

$\mu_h=$ mobility of holes

$\sigma =$ electrical conductivity

Making the substitution in (S)

Mobility of electron

$8.9*10^4=(8.7*10^{23}*(-1.602*10^{-19})*\mu_e)+(8.7*10^{23}*(-1.602*10^{-19})*\mu_h)$

$0.639=\mu_e+\mu_h$

Mobility of hole in (S)

$2.3*10^5 = (7.6*10^{22}*(-1.602*10^{-19})*\mu_e)+(1*10^{25}*(-1.602*10^{-19}*\mu_h))$

$0.1436 = 7.6*10^{-3}\mu_e+\mu_h$

Then, solving the equation:

$0.639=\mu_e+\mu_h$ (1)

$0.1436 = 7.6*10^{-3}\mu_e+\mu_h$ (2)

We have,

Mobility of electron $\mu_e = 0.5m^2/V.s$

Mobility of hole is $\mu_h = 0.14m^2/V.s$

6 0

2 years ago