Гражданская Оборона. Расчет параметров ядерного взрыва
( )
()
ר
:
,
( 1.4)
29
55 (.
-1-95)
..
..
1997
,
-
,
.
,
, .
,
,
.
,
.
I
:
105 410
1370
(P = q1/3.. + q2/3.. + q .
R R2
R3
q . = 0,5 q = 0,5 * 20 = 10
105 410
1370
(P = *(10)1/3 + *(10)2/3. + *10
=
2*103 (2*103)2
(2*103)3
= 17,781
(c , P0=101,3 ):
2,5 (P2
2,5*(17,781)2
(Pck = = = 1,087
(P+7P0
17,781+7*101,3
: ,
,
, ,
,
.
II
(c
, r = 0,068 q0,4 = 0,225 ):
111 q
111*20
U = exp [-K(R- r)] = exp[-0.4*(2-0,225)]
=
R2
(2)2
= 272,904 /2
:
, -
.
III
:
- :
7,5*1022
= q exp (-R/190 ) =
R2
7,5*1022
= *20 * exp(-2*103/190) = 1013 /2
(2*10 3)2
- -:
1013
P( = q exp (-R/200 ) =
R2
1013
= *20 *exp(-2*103/200) = 2269,996 /
(2*103)2
- -:
D( = D + D
5*108
D = q exp (-R/410 ) =
R2
5*108
= *20 *exp(-2*103/410) = 19,03
(2*103)2
1,4*109 q(1+0,2q0,65)
D = exp (-R/300 ) =
R2
1,4*109 * 20*(1+0,2*(20)0,65)
= exp (- 2*103/300 ) = 21,4
(2*103)2
D( = 19,03 + 21,4 = 40,43
:
, .
.
IV
():
5*109 *(1+2R)
E = lg 14,5 q
R2
:
5*109 *(1+2*2)
E = lg 14,5 *20 = 7694,994 /
(2)2
:
5*109 *(1+2*25)
E = lg 14,5 *20 = 40,186 /
(25)2
:
. ,
23 200 .
V
:
10 q
P0 = tg4 (45-2()
(R/22)1,5*exp( R/V
:
10 * 20
P0 = tg4 (45-2*5) = 1436,042
/
(2/22)1,5*exp( 2/(50/)
:
10 * 20
P0 = tg4 (45-2*5) = 19,569 /
(25/22)1,5*exp( 25/(50/)
:
( ).
, .
( ).
,
.
1.4, 29
-
-
P0, / E, / (P, U, /2 .
P(, / .
D(,
,/2
.
1436,042
19,569
40,186 7694,994 17,781
272,904 2269,996 40,426 1013