PHYSICS COLLEGE

At what partial pressure are argon atoms expected to have a free travel of approximately 5 µm, if the gas is at a temperature of 400 K? The cross section of collision, σ, or Argon is 0.28 nm2Ar molar mass is 39.9 g/mole

Answers

Answer 1

Answer:

Answer:

2790 Pa

Explanation:

Given wavelength λ= 5μm

temperature T= 400 K

cross section of collision σ= 0.28 nm^2

molar mass = 39.9 g/mole

pressure = $P= (RT)/(√(2)N_A\sigma\lambda )$

putting values we get

= $(8.314*400)/(√(2)*6.022*10^(23)*0.28*10^(-18)*5*10^(-6) )$

⇒P = 2790 J/m^3

the partial pressure are argon atoms expected= 2790 Pa

Related Questions

Which best describes a reference frame?
. Set the applied force to Force necessary to Keep the box Moving without accelerating. Restart the animation. Just before the box hits the wall, stop the animation. What can you tell me about relative magnitudes of the frictional force and the applied force
5. (Serway 9th ed., 7-3) In 1990, Walter Arfeuille of Belgium lifted a 281.5-kg object through a distance of 17.1 cm using only his teeth. (a) How much work was done on the object by Arfeuille in this lift, assuming the object was lifted at constant speed? (b) What total force was exerted on Arfeuille’s teeth during the lift? (Ans. (a) 472 J; (b) 2.76 kN)
How does the mass of an object affect its motion through the air?
(a) How many kilograms of water must evaporate from a 60.0-kg woman to lower their body temperature by 0.750ºC?

Define reflection of sound?

Answers

The reflection of sound is the movement of sound waves bouncing off of a surface and back into another direction, hope this helped :)

An electron enters a region of uniform electric field with an initial velocity of 50 km/s in the same direction as the electric field, which has magnitude E = 50 N/C, (a) what is the speed of the electron 1.5 ns after entering this region? (b) How far does the electron travel during the 1.5 ns interval?

Answers

Answer:

(a). The speed of the electron is $3.68*10^(4)\ m/s$

(b). The distance traveled by the electron is $4.53*10^(-5)\ m$

Explanation:

Given that,

Initial velocity = 50 km/s

Electric field = 50 N/C

Time = 1.5 ns

(a). We need to calculate the speed of the electron 1.5 n s after entering this region

Using newton's second law

$F = ma$ .....(I)

Using formula of electric force

$F = qE$ .....(II)

from equation (I) and (II)

$-qE= ma$

$a = (-qE)/(m)$

(a). We need to calculate the speed of the electron

Using equation of motion

$v = u+at$

Put the value of a in the equation of motion

$v = 50*10^(3)-(1.6*10^(-19)*50)/(9.1*10^(-31))*1.5*10^(-9)$

$v=36813.18\ m/s$

$v =3.68*10^(4)\ m/s$

(b). We need to calculate the distance traveled by the electron

Using formula of distance

$s = ut+(1)/(2)at^2$

Put the value in the equation

$s = 3.68*10^(4)*1.5*10^(-9)-(1)/(2)*(1.6*10^(-19)*50)/(9.1*10^(-31))*(1.5*10^(-9))^2$

$s=0.0000453\ m$

$s=4.53*10^(-5)\ m$

Hence, (a). The speed of the electron is $3.68*10^(4)\ m/s$

(b). The distance traveled by the electron is $4.53*10^(-5)\ m$

Based on the measured force between objects that are 10 meters apart, how can you find the force between objects that are any distance apart ?

Answers

The force between objects that are any distance apart is expressed as $P'=(100P)/(r^2)$

According to the gravitational law, the force acting on an object is directly proportional to the product of their masses and inversely proportional to the square of their distance apart. Mathematically,

$P=(GMm)/(r^2)$

M and m are the masses

r is the distance between the masses

If the force between objects that are 10 meters apart, hence;

$P=(GMm)/(10^2)\nP=(GMm)/(100)\nGMm = 100P$

To find the force between objects that are any distance apart, we will use the same formula above to have;

$P'=(GMm)/(r^2)\n$

Substitute the result above into the expression to have:

$P'=(100P)/(r^2)$

Hence the force between objects that are any distance apart is expressed as $P'=(100P)/(r^2)$

Learn more on gravitational law here: brainly.com/question/11760568

Answer:

F' = 100 F/r²

Explanation:

The gravitational force of attraction between two objects is given by the Newton's Gravitational Formula. The Newton's Gravitational Formula is as follows:

F = Gm₁m₂/r²

where,

F = Force between objects

G = Universal Gravitational Constant

m₁ = mass of first object

m₂ = mass of second object

r = distance between objects = 10 m

Therefore,

F = Gm₁m₂/10²

Gm₁m₂ = 100F --------------------- equation (1)

Now, we consider these objects at any distance r apart. So, the force becomes:

F' = Gm₁m₂/r²

using equation (1), we get:

F' = 100 F/r²

So, if the force (F) between objects 10 m apart is known, we can find it at any distance from the above formula.

According to the World Flying Disk Federation, the world distance record for a flying disk throw in the men’s 85-years-and-older category is held by Jack Roddick of Pennsylvania, who on July 13, 2007, at the age of 86, threw a flying disk for a distance of 54.0 m. If the flying disk was thrown horizontally with a speed of 13.0 m/s, how long did the flying disk remain aloft? (Jack Roddick was also a physics teacher! Read more about him at

Answers

Answer:

t = 4.15 seconds

Explanation:

It is given that,

Distance traveled by a flying disk, d = 54 m

The speed at which it was thrown, v = 13 m/s

We need to find the time for which the flying disk remain aloft. Let the distance is d. We know that, speed is equal to the distance covered divided by time. So,

$t=(d)/(v)\n\nt=(54\ m)/(13\ m/s)\n\nt=4.15\ s$

Hence, for 4.15 seconds the flying disk remain aloft.

A train station bell gives off a fundamental tone of 505 Hz as the train approaches the station at a speed of 27.6 m/s. If the speed of sound in air on that day is 339 m/s, what will be the apparent frequency of the bell to an observer riding the train

Answers

Answer:

Apparent frequency of the bell to the observer is 546.12 Hz

Explanation:

The frequency of train bell (frequency of source) = 505 Hz

The speed of train (observer) = 27.6 m/s

The speed of sound in the air is (velocity of sound) = 339 m/s

The apparent frequency of the bell to the observer is calculated as follows:

Apparent frequency of bell to the observer.

$= \text{frequency of source} * (Observer + velocity \ of \ sound )/( velocity \ of \ sound ) \n= 505 * (27.6 + 339)/(339) \n= 546.12 Hz$

A dielectric-filled parallel-plate capacitor has plate area A = 10.0 cm2 , plate separation d = 10.0 mm and dielectric constant k = 3.00. The capacitor is connected to a battery that creates a constant voltage V = 15.0 V. Use ϵ0 = 8.85×10⁻¹² C²/N⋅m².
The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy U2 of the capacitor at the moment when the capacitor is half-filled with the dielectric.

Answers

Answer:

U_eq = 1.99 * 10^(-10) J

Explanation:

Given:

Plate Area = 10 cm^2

d = 0.01 m

k_dielectric = 3

k_air = 1

V = 15 V

e_o = 8.85 * 10 ^-12 C^2 / N .m

Equations used:

U = 0.5 C*V^2 .... Eq 1

C = e_o * k*A /d .... Eq 2

U_i = 0.5 e_o * k_i*A_i*V^2 /d ... Eq 3

For plate to be half filled by di-electric and half filled by air A_1 = A_2 = 0.5 A:

U_electric = 0.5 e_o * k_1*A*V^2 /2*d

U_air = 0.5 e_o * k_2*A*V^2 /2*d

The total Energy is:

U_eq = U_electric + U_air

U_eq = 0.5 e_o * k_1*A*V^2 /2*d + 0.5 e_o * k_2*A*V^2 /2*d

U_eq = (k_1 + k_2) * e_o * A*V^2 / 4*d

Plug the given values:

U_eq = (3 + 1) * (8.82 * 10^ -12 )* (0.001)*15^2 / 4*0.01

U_eq = 1.99 * 10^(-10) J

Other Questions

What is the magnitude and direction of the electric field atradiaConsider a coaxial conducting cable consisting of a conductingrod of radius R1 inside of a thin-walled conducting shell of radius 2(both are infinite length). Suppose the inner rod hasradiusR1= 1.3 mm and outer shell has radiusR2= 10R1Ifthe net charge density on the center rod isq1= 3.4×10−12C/mand the outer shell isq2=−2q1,a.)What is the magnitude and direction of the electric field atradial distancer= 5R1from the center rod

Which two processes allow water to enter the atmosphere

A 30 W engine generates 3600 J of energy. How long did it run for?

A 100-N uniform ladder, 8.0 m long, rests against a smooth vertical wall. The coefficient of static friction between ladder and floor is 0.40. What minimum angle can the ladder make with the floor before it slips?