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daser333 [38]
4 years ago
6

Calculate the mean free path of air molecules at a pressure of 7.00×10^−13 atm and a temperature of 303 K . (This pressure is re

adily attainable in the laboratory.) Model the air molecules as spheres with a radius of 2.00×10^−10 m .
Physics
1 answer:
Grace [21]4 years ago
4 0

Answer:

82.8986 km

Explanation:

Given:

Pressure = 7.00×10⁻¹³ atm

Since , 1 atm = 101325 Pa

So, Pressure = 7.00×10⁻¹³×101325 Pa = 7.09275×10⁻⁸ Pa

Radius = 2.00×10⁻¹⁰ m

Diameter = 4.00×10⁻¹⁰ m (2× Radius)

Temperature = 303 K

The expression for mean free path is:

\lambda (Mean\ free\ path)=\frac {K (Boltzmann\ Constant)\times Temperature}{\sqrt {2}\times \pi\times (Diameter)^2\times Pressure}

Boltzmann Constant = 1.38×10⁻²³ J/K

So,

\lambda (Mean\ free\ path)=\frac {1.38\times 10^{-23}\times 303}{\sqrt {2}\times \frac {22}{7}\times (4.00\times 10^{-10})^2\times 7.09275\times 10^{-8}}

<u>Mean free path = 82.8986×10³ m = 82.8986 km</u>

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The answers to your questions are as written below:

  • The objects that represents a negatively charged particle is : Image B
  • The object that represents a positively charged molecule is : Image A
  • The object that represents an uncharged molecule is : Image C
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<h3>Different types of charges molecules</h3>

A negatively charged molecule move inwards when placed in an electric field while positively charged molecule placed in a electric field will move outwards the electric field.

A neutral/uncharged molecule will remains still when placd in an elctric field due to the absence of charges.

Hence we can concude that the answers to your questions are as listed above.

Learn more about electric charges :brainly.com/question/857179

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2 years ago
Describing a Physical Change
babymother [125]

Answer:

physical change is the temporary change or riversible change here the physical properties r only changed

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3 years ago
What is the change in potential energy of a squirrel that jumps from a tree branch that is 4.5 meters off of the ground and land
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Answer:

16.3 joules...................

3 0
3 years ago
The minute hand of a wall clock measures 16 cm from its tip to the axis about which it rotates. The magnitude and angle of the d
olya-2409 [2.1K]

Answer:

Explanation:

Given

Minute hand length =16 cm

Time at a quarter after the hour to half past i.e. 1 hr 45 min

Angle covered by minute hand in 1 hr is 360 and in 45 minutes 270

|r|=\frac{3\times 2\pi r}{4}=75.408 cm

Angle =270^{\circ}

(c)For the next half hour

Effectively it has covered 2 revolution and a quarter

|r|=\frac{2\pi r}{4}=25.136 cm

angle turned =90^{\circ}

(f)Hour after that

After an hour it again comes back to its original position thus displacement is same =25.136

Angle turned will also be same i.e. 90 ^{\circ}

7 0
4 years ago
What is the speed of sound at 33 °C (m/s)? For a frequency of 5 kHz, how large do you expect the wavelength to be (m)?
Vedmedyk [2.9K]

Answer:

-The speed of sound at 33°C is 362.8 m/s.

-The wavelength at a frequency at 5 kHz is 0.07256 m .

Explanation:

let v = 343 m/s be the speed of sound.

let T be the temperature.

then the speed of sound V, at 33°C is given by:

V = v + 0.6×T

   = 343 + 0.6×33

   = 362.8 m/s

Therefore, the speed of sound at 33°C is 362.8 m/s.

the wavelength at a frequency of f = 5kHz = 5000 Hz is given by:

λ = V/f

  = (362.8)/(5000)

  = 0.07256 m

Therefore, the wavelength at a frequency at 5 kHz is 0.07256 m .

7 0
3 years ago
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