Unformatted text preview: 73. (a) Using Eq. 1510, we have pg = ρgh = 1.21 × 107 Pa.
(b) By deﬁnition, p = pg + patm = 1.22 × 107 Pa.
(c) We interpret the question as asking for the total force compressing the sphere’s surface, and we
multiply the pressure by total area:
p 4πr2 = 3.82 × 105 N .
(d) The (upward) buoyant force exerted on the sphere by the seawater is
Fb = ρw gV where V = 43
πr .
3 Therefore, Fb = 5.26 N.
(e) Newton’s second law applied to the sphere (of mass m = 7.0 kg) yields
Fb − mg = ma
2 which results in a = −9.04, which means the acceleration vector has a magnitude of 9.04 m/s and
is directed downward. ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Force

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