419_Physics ProblemsTechnical Physics

419_Physics ProblemsTechnical Physics - Chapter 14 P14.27...

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Chapter 14 421 P14.27 (a) PP g h =+ 0 ρ Taking P 0 5 1013 10 . N m 2 and h = 500 c m we find P top 2 Nm 10179 10 5 . For h = 17 0 . cm, we get P bot 2 10297 10 5 . Since the areas of the top and bottom are A == 0100 10 2 2 m m 2 af we find FP A top top N × 3 . and F bot N 3 . (b) TBM g +− = 0 where BV g w × = 10 1 20 10 9 80 11 8 33 kg m m m s N 2 ej e j e j .. . FIG. P14.27 and Mg 10 0 9 80 98 0 . N Therefore, TM gB =− = −= 98 0 11 8 86 2 . N (c) FF bot top N N × = 1 029 7 1 017 9 10 11 8 3 . bg which is equal to B found in part (b). P14.28 Consider spherical balloons of radius 12.5 cm containing helium at STP and immersed in air at 0°C and 1 atm. If the rubber envelope has mass 5.00 g, the upward force on each is BF F V g V gm g Fr g m g F gg e n v e n v up env up −− = F H G I K J L N M O Q P −× = ,, . . . . He air He air He 322 kg m m m s kg 9.80 m s N ρρ π a f 4 3 129 0179 4 3 0 125 9 80 5 00 10 0 040 1 3 3 3 If your weight (including harness, strings, and submarine sandwich) is 70 0 9 80 686 kg m s N 2 = you need this many balloons:
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