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2 years ago

Based on the ELR, the stability classification of the atmosphere at the airport between the surface and 3000 m is

Engineering

1 answer:

Anna11 [10]2 years ago

6 0

Answer:

The answer is "conditionally unstable"

Explanation:

The conditional volatility is really a condition of uncertainty, which reflects on whether increasing air is polluted or not. It determines the rate of ambient delay, which has been between humid and dry adiabatic rates. In general, the environment is in an unilaterally unhealthy region.

Classification dependent on ELR:

Larger than 10 $\frac{^{\circ}C}{1000}$ m Around 10 and 6 $\frac{^{\circ}C}{1000}$ m or less 6 $\frac{^{\circ}C}{1000}$ m volatile implicitly unreliable Therefore ELR is implicitly unreliable 9 $\frac{^{\circ}C}{1000}$ m, that's why it is "conditionally unstable".

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Anon25 [30]

Answer:

Following are the solution to the given question:

Explanation:

The material is necessary to cook since frying is a speedy process for evaporation.

Drug A is now in the compressed fluid area, and the material would not boil if the pressure is chilled. Because the ship is solid, its substance A claim is false.

Unless the volume comprises of a drop in the heat, the B substance reaches a vapor pressure area and a wet region. That's the area that melting may occur. His claim for material B could therefore be true.

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2 years ago

Given the following MATLAB statement: ( 3 + 2 ) / 5 * 4 + 5 ^ 2 In what order will these operations be done?

Andrei [34K]

Answer:

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2 years ago

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2 years ago

Read 2 more answers

Consider the expansion of a gas at a constant temperature in a water-cooled piston-cylinder system. The constant temperature is

Leona [35]

Answer:

$Q_{in} = W_{out} = nRT ln (\frac{V_{2}}{V_{1}})$

Explanation:

According to the first thermodynamic law, the energy must be conserved so:

$dQ = dU - dW$

Where Q is the heat transmitted to the system, U is the internal energy and W is the work done by the system.

This equation can be solved by integration between an initial and a final state:

(1) $\int\limits^1_2 {} \, dQ = \int\limits^1_2 {} \, dU - \int\limits^1_2 {} \, dW$

As per work definition:

$dW = F*dr$

For pressure the force F equials the pressure multiplied by the area of the piston, and considering dx as the displacement:

$dW = PA*dx$

Here A*dx equals the differential volume of the piston, and considering that any increment in volume is a work done by the system, the sign is negative, so:

$dW = - P*dV$

So the third integral in equation (1) is:

$\int\limits^1_2 {- P} \, dV$

Considering the gas as ideal, the pressure can be calculated as $P = \frac{n*R*T}{V}$ , so:

$\int\limits^1_2 {- P} \, dV = \int\limits^1_2 {- \frac{n*R*T}{V}} \, dV$

In this particular case as the systems is closed and the temperature constant, n, R and T are constants:

$\int\limits^1_2 {- \frac{n*R*T}{V}} \, dV = -nRT \int\limits^1_2 {\frac{1}{V}} \, dV$

Replacion this and solving equation (1) between state 1 and 2:

$\int\limits^1_2 {} \, dQ = \int\limits^1_2 {} \, dU + nRT \int\limits^1_2 {\frac{1}{V}} \, dV$

$Q_{2} - Q_{1} = U_{2} - U_{1} + nRT(ln V_{2} - ln V_{1})$

$Q_{2} - Q_{1} = U_{2} - U_{1} + nRT ln \frac{V_{2}}{V_{1}}$

The internal energy depends only on the temperature of the gas, so there is no internal energy change $U_{2} - U_{1} = 0$ , so the heat exchanged to the system equals the work done by the system:

$Q_{in} = W_{out} = nRT ln (\frac{V_{2}}{V_{1}})$

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3 years ago

A water supply agency is planning to add two reservoirs to its system. Water will flow from Reservoir A to Reservoir B via a 10,

NikAS [45]

Attached is the solution to the above question.

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2 years ago