5.217. / = / 0 (1 p) (^) e
(x2
5.218. AX
5.219. I= -4:+2 e-x(b-a).
5.220. Will decrease exp (Rd) = 0.6.10 2 times.,
5.221. d = 0.3 mm.
5.222. d = (In 2)/p, = 8 mm.
5.223. N = (In 71)/In 2 = 5.6.
5.224. c = 2/z (n 2 — n 1 ) = 3.0.10^8 m/s.
5.225. First of all note that when v < c, the time rate is practic-
ally identical in the reference frames fixed to the source and to the
receiver. Suppose that the source emits short pulses with the inter-
vals To. Then in the reference frame fixed to the receiver the distance
between two successive pulses is equal to X = — v,-To, when
measured along the observation line. Here vr is the projection of the
source velocity on the observation line (v,. = v cos 0). The frequency
of received pulses v = clk = v^0 1(1 — vile), where vo = 1/To. Hence
(v — vo)/vo = (v/c) cos 0.
5.226. AX —X V2T/mc 2 cos 0 = —26 nm.
5.227. T = 4nRX c6X. = 25 days, where R is the radius of the
Sun.
5.228. d = (AX.IX),,ctIn = 3.10 7 km, m = (AX/X4c1/4/27cy
= 2.9.10 29 kg, where -v is the gravitational constant.
5.229. co =- coo (1 ± ()1(1 — (3), where^ 13 = V/c;
-7-- coo (1 + 2V /c).
5.230. v = 1 / 2 kAv 900 km per hour.
5.231. Substituting the expressions for t' and x' (from the Lorentz
transformation) into the equation cot — kx = co' t' — k' x' , we
obtain
= (1 -1-13)11/ 1 —13^2 , k = k' (1 + ()/V 1 —13^2 ,
where 13 = V/c. Here it is taken into account that co' = ck'.^
5.232. From the formula co' = w V(1 —13)/(1 +13) we get
= v/c = 0.26.
5.233. v= c "12-1 = 7' 1.10 4 km/s.
OAT-HI
5.234. co = coo^
5.235. AX = kT/moc 2 = 0.70 nm, where mo is the mass of the
atom.
5.236. (a) co coo/111-13 2 =5.0 .10^10 s-1 ; (b) Co = cooli1 —
=1.8.10i0 s-1. Here 6 = v/c.
5.237. The charge of an electron and the positive charge induced
in the metal form a dipole. In the reference frame fixed to the elec-
tron the electric dipole moment varies with a period T' = d'/v,
where d' = d yi - (v/c)^2. The corresponding "natural" frequency
347
joyce
(Joyce)
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