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()
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. 1924. 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 meansquare speed is given by
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 Spring '08
 Any
 Physics

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