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3 years ago

Dfine ' weight of moist air '

Physics

1 answer:

grandymaker [24]3 years ago

6 0

Answer:

First of all, “moist air” is air with a high water vapor content. Water vapor, the invisible, gaseous form of water, occurs in highly variable amounts in the atmosphere. Water is composed of a hydrogen atom and two oxygen atoms (H2O) and has a molecular weight of 18 grams per mole.

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A window in a skyscraper has a surface area of 3.50 m^2. Wind rushes by the outside of the window at 17.4 m/s, while inside the

Mariana [72]

The difference in the pressure between the inside and outside will be 369.36 N/m²

<h3>What is pressure?</h3>

The force applied perpendicular to the surface of an item per unit area across which that force is spread is known as pressure.

It is denoted by P. The pressure relative to the ambient pressure is known as gauge pressure.

The given data in the problem is;

dP is the change in the presure=?

Using Bernoulli's Theorem;

$\rm \rho\frac{V^2_{12}}{2} +P_1= \rho \frac{V^2_{22}}{2} +P_2 \\\\\ P_2-P_1=\rho \frac{v_2^2-v_1^2}{2} \\\\ P_2-P_1= 1.21 \times \frac{17.4^2-0}{2} \\\\ \triangle p=369.36 \ N/m^2$

Hence, the difference in the pressure between the inside and outside will be 369.36 N/m²

To learn more about the pressure refer to the link;

brainly.com/question/356585

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3 0

2 years ago

lesya692 [45]

Answer:

John Dalton

Explanation:

Dalton's atomic theory was the foundation for a new understanding of chemical structures. He proposed that matter was constituted by indivisible and indestructible particles "atoms." He theorized that all atoms of a particular substance were equal, and the atoms of different substances had atoms of different sizes and masses.

He also proposed that all compounds of elements were combinations of elements but in a very precise ratio.

7 0

3 years ago

Read 2 more answers

. A solenoid coil consists of a single layer of 250 circular turns of wire with each turn having a 0.02m radius. The axial lengt

Vinil7 [7]

Answer:

a) L = 3.29 10⁻⁴ H, b)U = 5.33 10⁻² J

Explanation:

a) The inductance is a solenoid this given carrier

L = $\frac{N \ \phi_B }{I}$

The magnetic field inside the solenoid is

B = μ₀ $\frac{N}{l} i$

hence the magnetic flux

Ф_B = B. A = μ₀ $\frac{N \ A}{l \ i}$

we substitute in the expression of inductance

L = N² μ₀ A /l

let's find the area of each turn

A = π r²

A = π 0.02²

A = 1.2566 10⁻³ m²

let's calculate

L = 250² 4π 10⁻⁷ 1.2566 10⁻² / 0.3

L = 3.29 10⁻⁴ H

b) The stored energy is

U = ½ L i²

let's calculate

U = ½ 3.29 10⁻⁴ 18²

U = 5.33 10⁻² J

5 0

3 years ago

How do boron-10 and boron-11 differ?

tensa zangetsu [6.8K]

Answer:

I think it has to do something with their ionizations... not entirely sure though.

Explanation:

3 0

3 years ago

How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe

Rus_ich [418]

Explanation:

(a) For an isothermal process, work done is represented as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

Putting the given values into the above formula as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

= $- 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})$

= $-12280.82 \times ln (0.09)$

= $-12280.82 \times -2.41$

= 29596.78 J

or, = 29.596 kJ (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

W = $\frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}$

= $\frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}$

= $\frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}$

= 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c) We know that for an isothermal process,

$P_{1}V_{1} = P_{2}V_{2}$

or, $P_{2} = \frac{P_{1}V_{1}}{V_{2}}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})$

= 11 atm

Hence, the required pressure is 11 atm.

(d) For adiabatic process,

$P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}$

or, $P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}$

= 28.7 atm

Therefore, required pressure is 28.7 atm.

6 0

4 years ago

Dfine ' weight of moist air '​

Dfine ' weight of moist air '