- #1

sponsoredwalk

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- 5

book it employs physics concepts ergo I have come here to ask you nice people

*The 5th point is where I have a problem, the rest is just information for you.*

I am assuming a particle is in a cube (ideal situations here) & it traverses a

distance l with a momentum p=mv.

1. As it travels forward it will hit the cube wall and the resultant change in

momentum will be;

[tex]change \ in \ momentum \ = \ (mv) \ - \ (-mv) \ = \ 2mv[/tex]

2. The number of collisions per second will be;

[tex]No.\ of \ collisions \ per \ second \ = \ \frac{velocity}{distance} \ = \ \frac{v}{l}[/tex]

3. The Change in momentum per second will be;

[tex]Change \ in \ momentum \ per \ second = 2mv \frac{v}{l} \ = \ \frac{2mv^2}{l}[/tex]

4. I've assumed one particle in one direction. Assuming n particles with

a total change in momentum per second of;

[tex]total \ change \ in \ momentum \ per \ second \ = \ \frac{2m}{l} \ \displaystyle\sum_{i=1}^n v_i^2 [/tex]

5. Here is where I have the problem. The text says to do this next;

Define the root square mean velocity, [tex]\overline{v} [/tex] such that;

[tex]\overline{v}^2 \ = \ \left( \ \frac{1}{n} \ \sum_{i=1}^n v_i^2 \right) [/tex]

What does this mean? Is this some mathematical concept from probabilities? I haven't done probabilities in math in like 5 years & even then that was in school :yuck: . Where did the [(2m)/l] fraction go?

Using this concept the book goes on to show that;

[tex] The \ total \ change \ in \ momentum \ = \ [2mn \overline{v}^2 ] / l [/tex]

(that is a fraction with l in the denominator, all the rest in the numerator [latex issues] )

I can't understand this (or any of the further concepts on pressure etc...) without getting this concept.