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777dan777 [17]
3 years ago
12

Ludwig Boltzmann performed a simple, but powerful experiment to gather evidence concerning the velocity distribution of a sample

of gas particles. His experiment revealed that the velocities of gases:
Physics
1 answer:
sp2606 [1]3 years ago
5 0

Answer:

Ludwig Boltzmann was an Austrian Physicist and he performed a simple but powerful experiment to gather evidence concerning the velocity distribution of a sample of gas particles.

His experiment revealed that the way Gases are distributed in its normal form across a range of temperature is dependent on the molar mass amen temperature of the gas. Gases with high temperature move faster due to the high number of colliding particles when compared to those with low temperature. Gases with lower molar mass move faster than those with higher molar mass.

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Answer:

CO2 increases in the winter

Explanation:

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A scientist pours 0.120 L of solution A into an Erlenmeyer flask. She adds 2.345 L of solution B. How many significant figures a
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Light enters water from air at an angle of 25° with the normal, Θ1. If water has an index of refraction of 1.33, determine Θ2.
vladimir1956 [14]
By Snell's law:

η = sini / sinr.        i = 25,  η = 1.33

1.33 = sin25° / sinr

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Option A.

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4 0
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Read 2 more answers
The rocket is fired vertically and tracked by the radar station shown. When θ reaches 66°, other corresponding measurements give
Flauer [41]

Answer:

velocity = 1527.52 ft/s

Acceleration = 80.13 ft/s²

Explanation:

We are given;

Radius of rotation; r = 32,700 ft

Radial acceleration; a_r = r¨ = 85 ft/s²

Angular velocity; ω = θ˙˙ = 0.019 rad/s

Also, angle θ reaches 66°

So, velocity of the rocket for the given position will be;

v = rθ˙˙/cos θ

so, v = 32700 × 0.019/ cos 66

v = 1527.52 ft/s

Acceleration is given by the formula ;

a = a_r/sinθ

For the given position,

a_r = r¨ - r(θ˙˙)²

Thus,

a = (r¨ - r(θ˙˙)²)/sinθ

Plugging in the relevant values, we obtain;

a = (85 - 32700(0.019)²)/sin66

a = (85 - 11.8047)/0.9135

a = 80.13 ft/s²

4 0
2 years ago
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