Irodov – Problems in General Physics

(Joyce) #1

1.328. N = plQ 2 /nr 2 = 0.7 N•m.
1.329. F = pgh (S — s) 2 / = 6 N
1.330. (a) The paraboloid of revolution: z = (6) 2 /2g) r 2 , where z
is the height measured from the surface of the liquid along the axis
of the vessel, r is the distance from the rotation axis; (b) p = Po +
ii2P( 02 / 4 •
1.331. P = nrico 2 R 4 /h = 9 W.
1.332. v— v0 ilancrdo
1.333. (a) 0)=032R/rill Rti^


1.334. (a) Q (^1) / 2 stvoR 2 ; (b) T 116 saR (^2) p4; (c) Ffr
= 4nil/v 0 ; (d) Ap =- 411/v 0 /R 2.
1.335. The additional head Ah = 5 cm at the left-hand end of
the tube imparts kinetic energy to the liquid flowing into the tube.
From the condition pv 2 /2 = pgAh we get v = If 2g Ah = 1.0 m/s.
1.336. eCC Ax = 5.
1.337. r^1 p^1112 μm/s.
1.338. d V 18 Re Ti (^2 5) mm,
Po) Pog
sities of glycerin and lead.
1.339. t = —P 181 d2 In — n 0.20 — S.
where p 0 and p are the den-
1.340. v = c yin (2 — = 0.1c, where c is the velocity of
light.
1.341. (a) P=a (1 +y4-313 2 ); (b) P=a 0/-1— [3 2 +114— [3 2 ).
Here i = V/c.
1.342. (^10) = 1 ji(1 —13 2 sin 2 0)1(1-13 2 )=1.08 m, where [3= v/c.
1.343. (a) tan 0' = Hence 0' = 59 0 ; (b) S =
—132cos20= 3.3 m 2. Here [3= v/c.
1.344. v=c17(2—-1-f-)4. t-. =0.6.10 8 m/s.
1.345. lo = cAt' 1/ 1 — (At/At')2 = 4.5 m.
1.346. s= cAt 1/ 1 — (Ato/At)^2 = 5 m.
1.347. (a) At 0 =(//v) 171—(v/c) 2 .1.4 its;
(b) /' = / —(v/c) 2 = 0.42 km.
1.348. / (^0) = — (v/c) 2 =17 m.
1.349. / 0 = liAxiAx 2 =6.0 m, v = c y1— Axil Ax 2 =2.2.10 8 m/s.
1.350. v
2t°/At
1+(t 0 /cet) 8 •
1.351. The forward particle decayed At = /13/c (1 — 32) = 20 Its
later, where (3 = v/c.
1.352. (a) 10 — SA-2g—Z,^ tB) ;
'V 1— (V/C)2
; (b) N = 43rivo2 RR2fRit?^

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