(04.02 LC)

Gaseous helium is in thermal equilibrium with liquid helium at 6.4 K. The mass of a helium atom is 6.65 × 10−27 kg and Boltzmann

chubhunter [2.5K]

Answer:

162.78 m/s is the most probable speed of a helium atom.

Explanation:

The most probable speed:

$v_{mp}=\sqrt{\frac{2K_bT}{m}}$

$K_b$ = Boltzmann’s constant = $1.38066\times 10^{-23} J/K$

T = temperature of the gas

m = mass of the gas particle.

Given, m = $6.65\times 10^{-27} kg$

T = 6.4 K

Substituting all the given values :

$v_{mp}=\sqrt{\frac{2\times 1.38066\times 10^{-23} J/K\times 6.4 K}{6.65\times 10^{-27} kg}}$

$v_{mp}=162.78 m/s$

162.78 m/s is the most probable speed of a helium atom.

4 0

3 years ago

Two thin concentric spherical shells of radii r1 and r2 (r1 < r2) contain uniform surface charge densities V1 and V2, respect

Lyrx [107]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

So,

a) 0 < r < r1 :

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

Hence, E = 0 for r < r1

b) r1 < r < r2:

Electric field =?

Let, us consider the Gaussian Surface,

E x 4 $\pi$ $r^{2}$ = $\frac{Q1}{E_{0} }$

So,

Rearranging the above equation to get Electric field, we will get:

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$

Multiply and divide by $r1^{2}$

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ x $\frac{r1^{2} }{r1^{2} }$

Rearranging the above equation, we will get Electric Field for r1 < r < r2:

E= (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ )

c) r > r2 :

Electric Field = ?

E x 4 $\pi$ $r^{2}$ = $\frac{Q1 + Q2}{E_{0} }$

Rearranging the above equation for E:

E = $\frac{Q1+Q2}{E_{0} . 4 \pi. r^{2} }$

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ + $\frac{Q2}{E_{0} . 4 \pi. r^{2} }$

As we know from above, that:

$\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ = (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ )

Then, Similarly,

$\frac{Q2}{E_{0} . 4 \pi. r^{2} }$ = (σ2 x $r2^{2}$ ) /( $E_{0}$ x $r^{2}$ )

So,

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ + $\frac{Q2}{E_{0} . 4 \pi. r^{2} }$

Replacing the above equations to get E:

E = (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ ) + (σ2 x $r2^{2}$ ) /( $E_{0}$ x $r^{2}$ )

Now, for

d) Under what conditions, E = 0, for r > r2?

For r > r2, E =0 if

σ1 x $r1^{2}$ = - σ2 x $r2^{2}$

4 0

3 years ago

A 4.0 kg ball is traveling at 3.0 m/s and strikes a wall. The ball bounces off the wall with a velocity of 4.0 m/s in the opposi

trasher [3.6K]

Answer:

280 N

Explanation:

Applying Newton's third second law of motion,

F = m(v-u)/t................... Equation 1

Where F = Magnitude of the average force on the ball during contact, v = final velocity of the ball, u = initial velocity of the ball, t = time of contact of the ball and the wall.

Note: Let the direction of the initial velocity of the ball be positive

Given: m = 4 kg, u = 3.0 m/s, v = -4.0 m/s (bounce off), t = 0.1 s

Substitute into equation 1

F = 4(-4-3)/0.1

F = 4(-7)/0.1

F = -28/0.1

F = -280 N.

Note: The negative sign tells that the force on the ball act in opposite direction to the initial motion of the ball

3 0

3 years ago

Which law states that the pressure and absolute temperature of a fixed quantity of gas are directly proportional under constant

Brrunno [24]

<u>Gay Lussac’s law</u> state that the pressure and absolute temperature of a fixed quantity of a gas are directly proportional under constant volume conditions.

<h2>Further Explanation </h2><h3>Gay-Lussac’s law </h3>

It states that at constant volume, the pressure of an ideal gas I directly proportional to its absolute temperature.
Thus, an increase in pressure of an ideal gas at constant volume will result to an increase in the absolute temperature.

<h3>Boyles’s law </h3>

This gas law states that the volume of a fixed mass of a gas is inversely proportional to its pressure at constant absolute temperature.
Therefore, when the volume of an ideal gas is increased at constant temperature then the pressure of the gas will also increase.

<h3>Charles’s law </h3>

It states that the volume of a fixed mass of a gas is directly proportional to absolute temperature at constant pressure.
Therefore, an increase in volume of an ideal gas causes a corresponding increase in its absolute temperature and vice versa while the pressure is held constant.

<h3>Dalton’s law </h3>

It is also known as the Dalton’s law of partial pressure. It states that the total pressure of a mixture of gases is always equivalent to the total sum of the partial pressures of individual component gases.
Partial pressure refers to the pressure of an individual gas if it occupies the same volume as the mixture of gases.

Keywords: Gas law, Gay-Lussac’s law, pressure, volume, absolute temperature, ideal gas

<h3>Learn more about: </h3>

Gay-Lussac’s law: brainly.com/question/2644981
Charles’s law: brainly.com/question/5016068
Boyles’s law: brainly.com/question/5016068
Dalton’s law: brainly.com/question/6491675

Level: High school

Subject: Chemistry

Topic: Gas laws

Sub-topic: Gay-Lussac’s law

8 0

3 years ago

Read 2 more answers

what do you think is the purpose of a periscope? Are there any other uses for periscopes besides for submarines?

spin [16.1K]

Military personnel also use periscopes in some gun turrets and in armoured vehicles. More complex periscopes using prisms or advanced fibre optics instead of mirrors and providing magnification operate on submarines and in various fields of science

3 0

3 years ago