\fdO'^ - Old Forge Coal Mines

208 PNE-aMATlCS.
(548) CO' — 50' = 10'. Since the volumes are proper
tional to the lengths of the spaces between the piston and
the end of the stroke, we may apply formula 62,
14.7 iV^AZi, X GO ^ /. X 10
' T r,
^^ 460 + 60 460 + 130*
^
14.7 X COX 590 ,^^^^1, . .
-
_,,
Therefore, p, = -— r = r^r 100. 07 lb. per sq. m. Ans.
''^' 520
r M
X 10
(549) T = 127° + 460° = 587°. Using formula 60,
p V= .37052 r, or p^^ -37052 X 587 ^ ^^^^ ^^ ^^ ^^^^
A/ I
(550) T - 100° + 4C0° = 5C0°.
Substituting in formula ei,/>V= .37052 W T, or
rr .37052 W T .37052 X .5 X 560
p 4,000
144
= 3.735 cu. ft. Ans.
(551) Use formula 64. PV=U^+^^T.
r= 110° + 460° = 570°; T^ = 100° + 460° = 560°; T, =
130° + 460° = 590°.
/ 90 X 40 80 X 57 \^yQ
Therefore, V= \-^ ^^^ / = 67.248 cu. ft.
^^^
Ans.
(552) The pressure exerted by squeezing the bulb may
be found from formula 53, in which / is 14.7, v, the orig-
inal volume = 20 cu. in., and v^, the new volume, = 5 cu. in.
pV 14.7X20 r,o o ^u n.u a ^ *u
p,-=-— = = = 58.8 lb. The pressure due to the
atmosphere must be deducted, since there is an equal pressure
on the outside which balances it. 58.8 — 14.7 =44.1
lb. per sq. in. = pressure due to squeezing the bulb.
3'X .7854= 7.0686 sq. in. = area of bottom. 7-0686 X 44.1 =
311.725 lb. 7.0686 X .434= 3.068 lb. = pressure due to
weight of water. 311.725 + 3.068 = 314.793 lb. Ans.
(553)
Use formula 58.
Z//4C0
'4G0_+ + /A . .
,/460+115\ ,^
,4C0 +