Irodov – Problems in General Physics

(Joyce) #1
2.166. R = 2a/pgh = 0.6 mm.
2.167. x = //(1 pod/4a) = 1.4 cm.
2.168. a = [pgh poll(1 — h)] d/4 cos 0.
2.169. h = 4a/pg (d 2 — d 1 ) = 6 cm.
2.170. h = 2a cos 0/pgx&p.

2.171. V 1 = 1/4ld2

20 - 4«.
n4

(n
1

— 1)/pd
0.9 ems/s.
2.172. R^2 R1 %.t'.• 1 / 8 pgh^2 /ct = 0.20 mm.
2.173. m 2nR 2 ccl cos 0 l(n 2 — 1)Igh = 0.7 kg.
2.174. F 2am/ph 2 = 1.0 N.
2.175. F = 2nR 2 alh = 0.6 kN.
2.176. F = 2a 2 1/pgd 2 = 13 N.
2.177. t = 21T1R 4 /ar 4.
2.178. Q = 2na 2 lpg.
2.179. (a) F = nad^2 = 3μJ; (b) F = 2nad^2 = 10 P.
2.180. AF = 2nad 2 (2-h/ 3 — 1) = —1.5 la.
2.181. A' = F pV In (plp,), where F = 8nR^2 a, p =
4aIR, V = 413 nR 3.
2.182. C — Cp = 1 / 2 R/(1 2 /^2 por/a).
2.184. (a) AS = —2 (da/dT) Au; (b) A U = 2 (a — T daldT) X
X Aa.
2.185. A = AmRTIM = 1.2 J.
2.186. m, = (V — inVi)/(V; — VD = 20 g, V, = 1.0 1. Here
17 ; is the specific volume of water.
2.187. m z Mp 1 o (V, — V)IRT = 2.0 g, where Po is the stan-
dard atmospheric pressure.
2.188. (n — 1)I(N — 1); = 1/(N^ 1).
2.189. AS = mq/T = 6.0 kJ/K; AU = m(q — RT/M) = 2.1 MJ,
where T = 373 K.


2.190. h x

(Q —mcAT) = 20 cm, where c is the specific
poS(1+ qM RT)
heat capacity of water, AT = 100 K, q is the specific heat of vapo-
rization of water, T is its boiling temperature.
2.191. A = me (T — To) RTIqM = 25 1, where c is the specific
heat capacity of water, T is the initial vapour temperature equal to
the water boiling temperature, as is seen from the hypothesis, q is
the specific heat of vapour condensation.
2.192. d 4aMITOT = 0.2 am, where p is the density of
water.
2.193. a = 71poYM/25TRT = 0.35 g/(s•cm 2 ), where pc, is the
standard atmospheric pressure.
2.194. p = pr2nRTIM = 0.9 nPa.
2.195. Ap = a/V 2 M = 1.7.10 4 atm.
2.196. pi pq. About 2.10^4 atm.
2.198. a = 27 /64R 2 T2r/Per = 3.6 atm•1 2 /mo1 2 , b= 118 RT„Ip„.
= 0.043 1/mol.
2.199. T 7 ;,. = 3181 :1TcrIMper = 4.7 cm^3 /g.
2.200. (n 3/v 2 ) (3v — 1) = 8T, T = 1.5.

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