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3 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 .

1 answer:

Grace [21]3 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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Complete question

The complete question is shown on the first uploaded image

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Explanation:

From the question we are told that

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Now this length seen by the observer can be mathematically represented as

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Here t is the actual length of the major axis of of the Empire's symbol, (an ellipse)

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So

$b = nb [\sqrt{1 - \frac{v^2}{c^2} } ]$

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11. A vector M is 15.0 cm long and makes an angle of 20° CCW from x axis and another vector N is 8.0 cm long and makes an angle

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Answer:

The magnitude of the resultant vector is 22.66 cm and it has a direction of 29.33°

Explanation:

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For M vector you obtain:

$M=M_x\hat{i}+M_y\hat{j}\\\\M=15.0cm\ cos(20\°)\hat{i}+15.0cm\ sin(20\°)\hat{j}\\\\M=14.09cm\ \hat{i}+5.13\ \hat{j}$

For N vector:

$N=N_x\hat{i}+N_y\hat{j}\\\\N=8.0cm\ cos(40\°)\hat{i}+8.0cm\ sin(40\°)\hat{j}\\\\N=6.12cm\ \hat{i}+5.142\ \hat{j}$

The resultant vector is the sum of the components of M and N:

$F=(M_x+N_x)\hat{i}+(M_y+N_y)\hat{j}\\\\F=(14.09+6.12)cm\ \hat{i}+(5.13+5.142)cm\ \hat{j}\\\\F=20.21cm\ \hat{i}+10.27cm\ \hat{j}$

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$|F|=\sqrt{(20.21)^2+(10.27)^2}cm=22.66cm$

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2 years ago

A dolphin in an aquatic show jumps straight up out of the water at a velocity of 15.0 m/s. (a) List the knowns in this problem.

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Answer:

a)

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b) 7.71 m

c) 3.06 s

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Since acceleration is constant we can use the equation for uniformly accelerated movement:

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0 = Vy0 + a * t

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The doplhin jumps and falls back into the water, when it falls again it position will be 0 again. So we can equate the position to zero to find how long it was in the air knowing that it started the jump at t = 0s.

0 = Y0 + Vy0 * t + 1/2 * a * t^2

0 = 0 + 15 * t - 1/2 * 9.81 t^2

0 = 15 * t - 4.9 * t^2

0 = t * (15 - 4.9 * t)

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15 = 4.9 * t2

t2 = 3.06 s This is the moment when it falls again.

3.06 - 0 = 3.06 s

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2 years ago

When a 10 V battery is connected to a resistor, 5 A of current flows through the resistor. What is the resistor's value?

SashulF [63]

Given data

*The value of battery voltage is V = 10 V

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2 years ago