413_Physics ProblemsTechnical Physics

413_Physics ProblemsTechnical Physics - Chapter 14 Section...

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Chapter 14 415 Section 14.2 Variation of Pressure with Depth P14.6 (a) PP g h =+ = × + 0 5 1 013 10 1 024 9 80 1 000 ρ .. Pa kg m m s m 32 ej e j bg P 101 10 7 . P a (b) The gauge pressure is the difference in pressure between the water outside and the air inside the submarine, which we suppose is at 1.00 atmosphere. PP P g h gauge Pa =− = = × 0 7 100 10 . The resultant inward force on the porthole is then FP A == × = × gauge Pa m N 1 00 10 0 150 7 09 10 7 2 5 . π af . P14.7 FF el = fluid or kx ghA = and h kx gA = h = × × L N M O Q P = 1000 500 10 10 9 80 1 00 10 162 3 2 Nm m kg m m s m m 2 e j e j e j . . FIG. P14.7 P14.8 Since the pressure is the same on both sides, F A F A 1 1 2 2 = In this case, 15 000 200 3 00 2 = F . or F 2 225 = N P14.9 F g 80 0 9 80 784 kg m s N 2 When the cup barely supports the student, the normal force of the ceiling is zero and the cup is in equilibrium.
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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