P15_073 - 73. (a) Using Eq. 15-10, we have pg = ρgh = 1.21...

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Unformatted text preview: 73. (a) Using Eq. 15-10, 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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