What is a rarefraction?

what is the mass of vertical column of air that exerts pressure of 1 atm over a10m by 10m square surface

Oksana_A [137]

Answer:

$m=1.01\times 10^6\ kg$

Explanation:

Given that,

Pressure, P = 1 atm = 101325 Pa

Area of the square surface, A = 10² = 100 m²

We need to find the mass of vertical column of air. We know that, pressure is equal to the force acting per unit area. So,

$P=\dfrac{mg}{A}\\\\m=\dfrac{PA}{g}\\\\m=\dfrac{101325\times 10^2}{10}\\\\m=1.01\times 10^6\ kg$

So, the required mass of the vertical column of air is $1.01\times 10^6\ kg$ .

4 0

3 years ago

Find the rms speed of the molecules of a sample of n2 (diatomic nitrogen) gas at a temperature of 31.5°c.

Artemon [7]

The rms speed can be calculated using the following rule:
rms = sqrt ((3RT) / (M)) where:
R is the gas constant = 8.314 J/mol-K
T is the temperature = 31.5 + 273 = 304.5 degrees kelvin
M is the molar mass = 2*14 = 28 grams = 0.028 kg

Substitute with the givens to get the rms speed as follows:
rms speed = sqrt [(3*8.314*304.5) / (0.028)] = 520.811 m/sec

8 0

3 years ago

In young’s double slit experiment, the measured fringe width is 0.5 mm for a Sodium light of 589 nm at a distance of 1.5 m. A br

Levart [38]

Answer:

(A). The order of the bright fringe is 6.

(B). The width of the bright fringe is 3.33 μm.

Explanation:

Given that,

Fringe width d = 0.5 mm

Wavelength = 589 nm

Distance of screen and slit D = 1.5 m

Distance of bright fringe y = 1 cm

(A) We need to calculate the order of the bright fringe

Using formula of wavelength

$\lambda=\dfrac{dy}{mD}$

$m=\dfrac{d y}{\lambda D}$

Put the value into the formula

$m=\dfrac{1\times10^{-2}\times0.5\times10^{-3}}{589\times10^{-9}\times1.5}$

$m=5.65 = 6$

(B). We need to calculate the width of the bright fringe

Using formula of width of fringe

$\beta=\dfrac{yd}{D}$

Put the value in to the formula

$\beta=\dfrac{1\times10^{-2}\times0.5\times10^{-3}}{1.5}$

$\beta=3.33\times10^{-6}\ m$

$\beta=3.33\ \mu m$

Hence, (A). The order of the bright fringe is 6.

(B). The width of the bright fringe is 3.33 μm.

3 0

2 years ago

What crop is least likely to do well when the temperatures are very hot?

torisob [31]

a. Sweet corn and possibly d. okra.

3 0

3 years ago

Problem 1: Spherical mirrorConsider a spherical mirror of radius 2 m, and rays which go parallel to the optic axis. What is thep

SIZIF [17.4K]

Answer:

1) iii i= 1m, 2) iii and iv, 3) i = f₂ (L-f₁) / (L - (f₁ + f₂))

Explanation:

Problem 1

For this problem we use two equations the equations of the focal distance in mirrors

f = r / 2

f = 2/2

f = 1 m

The builder's equation

1 / f = 1 / o + 1 / i

Where f is the focal length, "o and i" are the distance to the object and the image respectively.

For a ray to arrive parallel to the surface it must come from infinity, whereby o = ∞ and 1 / o = 0

1 / f = 0 + 1 / i

i = f

i = 1 m

The image is formed at the focal point

The correct answer is iii

Problem 2

For this problem we have two possibilities the lens is convergent or divergent, in both cases the back face (R₂) must be flat

Case 1 Flat lens - convex (convergent)

R₂ = infinity

R₁ > 0

Cas2 Flat-concave (divergent) lens

R₂ = infinity

R₁ <0

Why the correct answers are iii and iv

Problem 3

For a thick lens the rays parallel to the first surface fall in their focal length (f₁), this is the exit point for the second surface whereby the distance to the object is o = L –f₁, let's apply the constructor equation to this second surface

1 / f₂ = 1 / (L-f₁) + 1 / i

1 / i = 1 / f₂ - 1 / (L-f₁)

1 / i = (L-f₁-f₂) / f₂ (L-f₁)

i = f₂ (L-f₁) / (L - (f₁ + f₂))

This is the image of the rays that enter parallel to the first surface

6 0

3 years ago