acquires during the time dt is equal to dp = m dv dt•u = F dt.
What follows is evident.
1.179. v = —u In (mo/m).
1.180. m = moe-w t /u.
1.181. a = (u/vo) In (mo /m).
1.182. v = —
F
In m0w (^) Rt •
11 me
1.183. v = Ft/m 0 (1 (^) t p,t/m0), w
nzo—
= F/m (^0) (1 (^) Rtimor-
1.184. v (^) 2gh ln (11h).
1.185. N=2b V alb.
1.186. M = i/ 2 mgvot 2 cos a; M = (mv:/2g) sine a cos a =
= 37 kg • m 2 /s.
1.187. (a) Relative to all points of the straight line drawn at
right angles to the wall through the point 0;
(b) I AM I = 2 mv/ cos a.
1.188. Relative to the centre of the circle.
I AM I = 2 V1 — (g/& 2 1) 2 mgl/w.
1.189. I AM I = hmV.
1.190. M = maivgt 2.
1.191. m = 2kr1/v 22.
1.192. vo =1/-2g//cos O.
1.193. F = mcogr:Ir 3.
1.194. M = Rmgt. Z
1.195. M = Rmgt sin a. Will not change.
1.196. M' = M — [rot)]. In the case when p = 0, i.e. in the
frame of the centre of inertia.
1.198. M = 1/ 3 imvo.
1.199. Erna. mil/x102. The problem is easier to solve in the
frame of the centre of inertia.
1.200. T = 23 - iyM/v 3 = 225 days.
1.201. (a) 5.2 times; (b) 13 km/s, 2.2.10- 4 m/s 2.
1.202. T = (r R) 3 /2yM. It is sufficient to consider the
motion along the circle whose radius is equal to the major semi-axis
of the given ellipse, i.e. (r R)12, since in accordance with Kepler's
laws the period of revolution is the same.
1.203. Falling of the body on the Sun can be considered as the
motion along a very elongated (in the limit, degenerated) ellipse
whose major semi-axis is practically equal to the radius R of the
Earth's orbit. Then from Kepler's laws, (2T/T) 2 = [(R12)1R1 3 ,
where i is the falling time (the time needed to complete half a revo-
lution along the elongated ellipse), T is the period of the Earth's
revolution around the Sun. Hence, T = T/4112 = 65 days.
1.204. Will not change.
1.205. 1= f y M (T/231) 2.
1.206. (a) U = — ym 1 m 2 /r; (b) U = —y (mM11) In + 11a);
F yrnM I a (a +1).
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