Bukhovtsev-et-al-Problems-in-Elementary-Physics

262 ANSWERS AND SOLUTIONS
Thus, the water acts on the wall with the hole with a force 2pA smaller
than that acting on the opposite wall, and not with a force smaller by pA
as might be expected. This is due to a reduction in the pressure acting on
the wall with the hole, since the water flows faster at this wall.
The vessel will begin to move if kG < 2pA or
k < 2pghA
G
272. According to Newton's second law, the equality pA o = 2pA should
exist. Therefore, if the liquid flows out through the tube, the cross-sectional
area of the stream should be halved
A _ A o
-2
This compression of the stream can be explained as follows.
The extreme streamlets of the Iiquid approaching the tube from above
cannot, in view of inertia, flow around the edge of the tube directly adhering
to its walls, and move towards the centre of the stream. Under the pressure
of, the particles nearer to the centre of the stream, the lines of flow straighten
out and a contracted stream of the liquid flows along the tube.
273. By neglecting splashing of the water, we thus assume the impact of
the stream against the wall to be absolutely inelastic. According to Newton's
second law. the change in the momentum of the water during the time ~t is
limv=FM, where lim=p n:' vM is the mass of the water flowing during
the time ~t through the cross section of the pipe.
Hence,
plld 2
F=--v 2 === 8 gi
4
274. When the gas flows along the pipe (Fig. 387), its momentum changes
in direction, but not in magnitude.
A mass of pAv passes in a unit of time through cross section I of the
vertical part. This mass brings in the momentum PI=pAUVl where VI is the
J
p
Fig. 387
H-h
A=Po-pgh
Fig. 388
H