nL;

possible value of the coefficient of
friction, we assume that the rod has
been pressed very strongly and,
hence, the force mg car be neglected.
Then
Ncoso:IsincrCotO( ' --r--I.
u2 \12
We've obtained the answer, but
some comments are in order. If J >
2R, we can't insert the rod into the
dome. If, on the other hand, the rod
is too small, then the corresponding
value forp will be too large and will
make no sense. Finally, rubber is a
substance with very complicated
properties; its i_nteraction with a
solid surface is not adequately described
by the laws of dry friction
(but that's another story).
P1 18
The conduction of heat from the
inner shell of the thermos (that is,
from the tea) is performed by molecules
of gas between the inner and
outer shells. That's why the space
between the shells was pumped out:
to prevent the gas that remains from
forming a continuous heat-conducting
medium-that is, to make the
molecules move between the shells
without colliding with one another.
Let's assumethat after a collision
with a shell, a molecule on average
has an energy proportional to the
temperature of the shell. Thus, at
the outer shell the kinetic energy of
the translational movement of a
molecule is K, - %kTr, where { is
room temperature; at the inner shell
the corresponding value is K, -
"/zkT, where 7, is the temperature
of the tea. Although this temperature
changes with time, the range of
its variation is not large: AT = 363 K
- 343 K = 20 K. So for the purpose of
our estimates we can assume thatKl
- %kT"where 7. = 353 K is the average
temperature of the hot shell.
Consequently, after a coliision
with the hot shell, each molecule
"takes" energy from the tea that is
equal to
LK - %kff"- T,l.
The number of molecules colliding
with the hot shell per unit time is
proportional to f nvs, where n is
the density of the gas molecules and
v is the average magnitude of the velocity
projected perpendicular to the
wall. The number of molecules
that collide with the wall during
time At is
N - fnvSLt.
These molecules take from the tea
the energy
AE : NAK - sfnvSk(T"- T,lLt.
This energy is equal to the change
AU in the intemal energy of the tea
in the period At-that is, AE : LU.
Insofar as
LU: McLT,
where AT is the change in the temperature
of the tea, then
sfnvSk(T"- T,lLt - MILT.
From this we get
4McLT
Af-:
ȧnvsk(4 -4)
Taking into accountthat v - $fli,
where p is the molar mass of the gas,
and n: PlkT, (which follows from
the Clapeyron-Clausius equation),
we finally obtain
Lt-
aMcLf
rE\
BP(4 - 4)sJR
Substituting numerical data lM :
1 kg, A7:}OIK,T,=293K,P: lPa,
p.= 29 . 10-3 kg/mol, and so on)
results in
Lt-3.104s-gh.
P1 19
At an arbitrary point A at a distance
R from the wire, the velocity
of the particle is directed at a small
angle cr with the x-axis, so
Figure 7
where v, is the vertical projection of
the velcicity and v, : TXlmis the
horizontal projection.
Let's write down Newton's second
law for the y-component (fig.7l:
where
v
A= -, v
Fydt = mdvr,
F, = gEc'ssy = e),cosy
ZneoR
The small time period dt canbe derived
from the formula v,= dxf dt:
,dx
uL -
v x
Rdv
Y COSTI,T
xr
During this time the vertical projection
changes by
F. e),,
dv..
o =
Jdt = ----"!' 4*.
m 2neomv,
The total projection of the velocity
on the y-axis is composed of small
increments:
'
rl2
-^,' ' Zeomv*
Thus, the angle we're seeking is
d=L
v, e?" e)"
_
=
v, 2eomtl 4eoK
P120
Two cases are possible. (1) The
lens is convergent. Drawing the
0lJAitIUil/AitstItR$, ililllT$ & $0t l|Tt0its 5l