, сечение полусферы , изменение массы за время m 10 0,2

10
1.
,
(
),
.
,
1
?
20
.
.
:
Sdx = dV , S −
ρSdx = dm,
dm
dt
dm = m0 Sdt
m0 Sdt = ρSdx
ρR 103 ⋅ 0, 2
=
= 1,11⋅104 ≈ 3, 09
t=
−3
m0 18 ⋅10
ρdx
dt =
,
m0
2.
«!
"
»,
" ,
" 1000 $
"
#
#
5 . %
10 . &
.
:
' II
$
:
F ∆t ( F − F )t1
=
−
m
m
m∆v m ( F − F )t1
F = ma1 =
=
− II
t2
mt2
"
∆v =
F t2 = Ft1 − F t1
$
#
Ft1
103 ⋅ 5
F =
=
= 330
t2 + t1
5
3. (
)
300
10
,
?
"
;
;
10
:
v3 = 2 gR3 − 2
mg3 = γ
4. '
;
mM
−
R32
"
g =γ
M
−
R2
v
=
v3
2g R
=
2 g3 R3
;
"
)
;
M R R32
300 M ⋅ R3
=
= 30 ≈ 5, 48
R 2 R3 M
10 R3 ⋅ M
,
5
10 % . !
7 º*
0,51 "/ 3.
32 ",
$
.
:
ν1 RT
ν RT
;
P2 = 2
V
V
RT
P =
(ν1 + ν 2 ); (1)
V
m
m
ρ1 = 1 ρ2 = 2 (2)
V
V
ρ = ρ1 + ρ2
(3)
P1 =
1,2,3
m2, . .
.
: m2 = 4⋅10-3 "
5. &
+
.%
"
,
"
3
10
"
3
.&
"
3000
,
"
"
.-
"
.
10
:
& ,
"
PV
300 ⋅105 ⋅103
1 1
V2 =
=
= 300 ⋅103
5
10
P2
PV
1 1 = PV
2 2
: 300000
6. !
.
#
.
U
r
! "
#
.
"
16
#
,
"
,
"
,
,
"
,
25
.
,
R/2?
:
I=
U
;
r
U2
U2
Q = I rt ;
rt =
t1
r
r2
U2
Q2 = Q1 =
t2
r+R
U2
2U 2
U2
Q3 = Q1 =
t;
t3 =
t1
R 3
+
2
r
R
r
r+
2
2r (t3 − t1 )
2rt3 = 2rt1 + Rt1 ;
R=
t1
2
U 2t2
U2
t1 =
; 2t3 − t1 = t2
r
t1 r + 2 r (t3 − t1 )
t3 =
t2 + t1 25 + 16
=
= 20,5
2
2
3