A solid surface with dimensions 2.5 mm ✕ 3.0 mm is exposed to argon gas at 90. Pa and 500 K. How many collisions do the Ar atoms make with this surface in 20. s?v

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Answer 1

A solid surface with dimensions 2.5 mm ✕ 3.0 mm is exposed to argon gas at 90. Pa and 500 K, the Ar atoms make 4.6128 collisions with the surface in 20 seconds.

We may utilise the idea of the kinetic theory of gases to determine how many collisions the Ar (argon) atoms have with the solid surface.

The expression for the quantity of surface collisions per unit of time is:

Collisions per unit time = (Number of particles per unit volume) × (Velocity) × (Area of the surface)

Number of particles per unit volume = (Pressure) / (Gas constant * Temperature)

Number of particles per unit volume = (Pressure) / (Gas constant * Temperature)

= (90) / (8.314 * 500 K)

= 0.02154 [tex]mol/m^3[/tex]

Number of particles in the given volume = (Number of particles per unit volume) × (Volume)

= (0.02154) × (7.5 × [tex]10^{(-6)[/tex])

= 1.6155 × [tex]10^{(-7)[/tex] mol (approximately)

Number of collisions = (Number of particles in the given volume) × (Collisions per unit time) × (Time)

= (1.6155 × [tex]10^{(-7)[/tex]) × (Number of particles per unit volume) × (Velocity) × (Area of the surface) × (Time)

Velocity = √((3 * k_B * T) / M_Ar)

Velocity = √((3 * 1.380649 × [tex]10^{(-23)[/tex] J/K * 500) / (39.95 × [tex]10^{(-3)[/tex] )

≈ 1,558.45 m/s

Number of collisions = (1.6155 × [tex]10^{(-7)[/tex]) × (0.02154) × (1,558.45 m/s) × (7.5 × [tex]10^{(-6)[/tex]) × (20)

≈ 4.6128 collisions

Therefore, the Ar atoms make approximately 4.6128 collisions with the surface in 20 seconds.

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Related Questions

please fast.
- 14. A 0.400 kg physics cart is moving with a velocity of 0.22 m/s. This cart collides inelastically with a second stationary cart and the two move off together with a velocity of 0.16 m/s. What was

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In an inelastic collision, two or more objects stick together and travel as one unit after the collision. The principle of conservation of momentum states that the total momentum of a closed system remains constant if no external forces act on the system, which is also true for an inelastic collision.

As a result, the momentum of the first cart is equal to the combined momentum of the two carts after the collision, since the collision is inelastic. The velocity of the two carts after the collision can be calculated using the conservation of momentum, as follows:0.400 kg x 0.22 m/s + 0 kg x 0 m/s = (0.400 kg + 0 kg) x 0.16 m/s0.088 Ns = 0.064 NsThe total momentum of the system is 0.064 Ns.

The two carts move together after the collision with a velocity of 0.16 m/s. The mass of the second cart is 0 kg, therefore, its initial momentum is 0 Ns. The momentum of the first cart is therefore equal to the total momentum of the system.

The initial momentum of the first cart can be calculated using the following formula:p = mv0.088 Ns = 0.400 kg x v Therefore, the initial velocity of the first cart is:v = p/mv = 0.088 Ns / 0.400 kgv = 0.22 m/s Hence, the initial velocity of the first cart is 0.22 m/s.

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The displacement of a wave traveling in the negative y-direction
is D(y,t)=(9.0cm)sin(45y+70t+π)D(y,t)=(9.0cm)sin⁡(45y+70t+π), where
y is in m and t is in s.
What is the frequency of this wave?
Wh

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The displacement of a wave traveling in the negative y-direction depends on the amplitude and frequency of the wave.

The displacement of a wave traveling in the negative y-direction is a combination of factors. The first factor is the amplitude, which is the maximum distance that a particle moves from its rest position as a wave passes through it. The second factor is the frequency, which is the number of waves that pass a fixed point in a given amount of time. The displacement of a wave is given by the formula y = A sin(kx - ωt + ϕ), where A is the amplitude, k is the wave number, x is the position, ω is the angular frequency, t is the time, and ϕ is the phase constant. This formula shows that the displacement depends on the amplitude and frequency of the wave.

These variables have the same fundamental meaning for waves. In any case, it is useful to word the definitions in a more unambiguous manner that applies straightforwardly to waves: Amplitude is the distance between the wave's maximum displacement and its resting position. Frequency is the number of waves that pass by a particular point every second.

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Relative to the ground, a car has a velocity of 17.3 m/s, directed due north. Relative to this car, a truck has a velocity of 23.0 m/s, directed 52.0° north of east. What is the magnitude of the truc

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The

magnitude

of the truck's velocity

is approximately 22.783 m/s.

To solve this problem, we can break down the velocities into their x and y components.

The

car's velocity

is directed due north, so its

x-component is 0 m/s and its y-component is 17.3 m/s.

The truck's velocity is directed 52.0° north of east. To find its x and y components, we can use trigonometry. Let's define the

angle

measured counterclockwise from the positive x-axis.

The x-component of the truck's velocity can be found using the cosine function:

cos(52.0°) = adjacent / hypotenuse

cos(52.0°) = x-component / 23.0 m/s

Solving for the x-component:

x-component = 23.0 m/s * cos(52.0°)

x-component ≈ 14.832 m/s

The y-component of the truck's velocity can be found using the sine function:

sin(52.0°) = opposite / hypotenuse

sin(52.0°) = y-component / 23.0 m/s

Solving for the y-component:

y-component = 23.0 m/s * sin(52.0°)

y-component ≈ 17.284 m/s

Now, we can find the magnitude of the truck's velocity by using the

Pythagorean theorem

:

magnitude = √(x-component² + y-component²)

magnitude = √((14.832 m/s)² + (17.284 m/s)²)

magnitude ≈ √(220.01 + 298.436)

magnitude ≈ √518.446

magnitude ≈ 22.783 m/s

Therefore, the magnitude of the truck's

velocity

is approximately 22.783 m/s.

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magnitude

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What value below has 3 significant digits? a) 4.524(5) kev b) 1.48(4) Mev c) 58 counts d) 69.420 lols Q13: What is the correct count-rate limit of precision for an exactly 24 hour live time count with 4.00% dead time, a count rate of 40.89700 counts/second, and a Fano Factor of 0.1390000? a) 40.897(8) counts/sec b) 40.90(12) counts/sec c) 41.0(5) counts/sec d) 41(5) counts/sec e) Infinite Q14: What kind of detectors have the risk of a wall effect? a) Neutron gas detectors b) All gas detectors c) Neutron semiconductor detectors d) Gamma semiconductor detectors e) Geiger-Müller counters

Answers

The value below that has 3 significant digits is: c) 58 counts

In this value, the digits "5" and "8" are considered significant, and the trailing zero does not contribute to the significant figures. The value "58" has two significant digits.

Q13: The correct count-rate limit of precision for an exactly 24 hour live time count with 4.00% dead time, a count rate of 40.89700 counts/second, and a Fano Factor of 0.1390000 is:

b) 40.90(12) counts/sec

The value has 4 significant digits, and the uncertainty is indicated by the value in parentheses. The uncertainty is determined by the count rate's precision and the dead time effect.

Q14: The detectors that have the risk of a wall effect are:

c) Neutron semiconductor detectors

d) Gamma semiconductor detectors

The wall effect refers to the phenomenon where radiation interactions occur near the surface of a detector, leading to reduced sensitivity and accuracy. In the case of neutron and gamma semiconductor detectors, their thin semiconductor material can cause a significant portion of radiation interactions to occur close to the detector surface, resulting in the wall effect.

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