ch19-p041

# ch19-p041 - 41(a The distribution function gives the...

This preview shows pages 1–2. Sign up to view the full content.

() 0 0 2 23 3 3 2 00 0 0 71 4 8. 33 9 v v aa av d v v v v v =− = = ³ Thus, 22 rms rms 0 0 0 0 11 4 1.31 1.31 . 69 v vv v v v =+ = ¡ = 41. (a) The distribution function gives the fraction of particles with speeds between v and v + dv , so its integral over all speeds is unity: ³ P ( v ) dv = 1. Evaluate the integral by calculating the area under the curve in Fig. 19-24. The area of the triangular portion is half the product of the base and altitude, or 1 0 2 av . The area of the rectangular portion is the product of the sides, or av 0 . Thus, 0 13 P v dv av av av = ³ , so 3 0 2 1 av = and av 0 = 2/3=0.67. (b) The average speed is given by avg . P v d v = ³ For the triangular portion of the distribution P ( v ) = av / v 0 , and the contribution of this portion is 0 2 0 0 2 , 9 v av vdv v v == = ³ where 2/3 v 0 was substituted for a . P ( v ) = a in the rectangular portion, and the contribution of this portion is 0 0 2 2 0 0 3 4. v v d v v v = = ³ Therefore, avg avg 0 0 0 0 2 1.22 1.22 9 v v v v = ¡ = . (c) The mean-square speed is given by

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

### Page1 / 2

ch19-p041 - 41(a The distribution function gives the...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online