(1) m  =  V×r  =  108 m3 ×2.7×103kg/m3  =  2.7×1011kg.

(3) V  =  0.6 m ×0.25 m ×0.15 m ×19.3×103 kg/m3  =  430 kg (approx. 960 lb).

(5) V  =  (0.09844 - 0.035)/103 m3  =  6.344×10-5m3
r¢  =  (0.08878-0.035)/6.344×10-5kg/m3  =  848 kg/m3
So the specific gravity is 848 kg/m3/103 kg/m3  =  0.848.

(7) D P  =  r g h  =  1.05×103 kg/m3×9.8 m/s2×1.6 m  =  1.6×104 Pa  =  120 mm of Hg.

(9) F  =  1×105 Pa ×1.6 m×1.9 m  =  3×105 N (down).
(b) The pressure on the bottom side is essentially the same magnitude (up), since the thickness of the table is inconsequential.

(15) h  =  P/(r g)  =  105 Pa/1.29/9.8 m/s2  =  7.9 km.

(21) The specific gravity is given by 22.5/22.9  =  0.983 Thus, since the density of water is 103 kg/m3, the density of the brew is 983 kg/m3.

(23) Vdispl  =  m/rHg, V  =  m/rAl. Thus Vdispl  =  VrAl/rHg, or Vdispl/V  =  rAl/rHg  =  2.7/13.6  =  0.20; i.e., only 20% of the bar is submerged, or 80% above the surface of the mercury.

(35) The volume flow rate is Q  =  9.2×5×4.5 m3/(600 s)  =  0.345 m3/s. This must be equal to A×v, so v  =  0.345/(p×0.17 m2)  =  3.8 m/s.

(61) r g h  =  65×105/760 Pa, and for water r  =  103 kg/m3. So for (a) h  =  0.87 m.
(b) Since water is less dense than Hg by a factor 13.6, h  =  (550/13.6)×105/760/103/9.8  =  0.54 m
(c) h  =  18×105/760/103/9.8 m  =  0.24 m.

File translated from TEX by TTH, version 1.95.
On 21 Nov 1999, 14:18.