HW6Solutions - volume (displaced oil) x rho (oil) = d.m...

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Sheet1 Page 1 Question #1 Err:508 1 pascal = 9.86923267 10-6 atmospheres Question #2 1/2PGWH^2 Question #3 F(new)/F = 4/3 Question #4 (H(2)or(8.8))=644/1000 * H=13.665 H-H(2)=H(1) 13.665-8.8=4.865 Question #5 H(1)=644/1000 * H 5.667=644/1000 * 8.8 H(2)=H-H(1) 3.133=8.8-5.667 Question #6 F1/S1=f2/S2 20/56=X/1.6 0.57 Question #7 877/1296 = .6767 H=12 12*.6767=8.1209 Question #8 8.1209-606/1296*H 8.1209-.468*5.3=5.643 Question #9 (1296-877)/(1296-606)*12=7.287
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Sheet1 Page 2 Question #10 rho (oil) = 906 kg/m3 m (iron) = 1.66 kg rho (iron) = 7860 kg/m3 volume (iron) = volume (displaced oil) = m (iron)/rho(iron) = 1.66/7860 = 2.111E-4 m3
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Unformatted text preview: volume (displaced oil) x rho (oil) = d.m (oil) = 2.111E-4 x 906 = 0.191 kg Buoyancy = d.m (oil) x gravity = 0.191 x 9.8 = 1.8718N Reading on spring = Weight (iron) - buoyancy = 1.66 x 9.8 - 1.8718 = 14.39N Question #11 Total weight in Newtons - Reading of Spring scale Question #12 V1 * D(wide)/D(small) = V2 Question #13 Question #14 V(c)=(2g(H2-H1))^(1/2) Question #15 P(atm)-PGH2 Question #16 b=P(atm)/PG - h2...
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This note was uploaded on 09/08/2008 for the course PHY 2048 taught by Professor Criss during the Spring '08 term at University of South Florida - Tampa.

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HW6Solutions - volume (displaced oil) x rho (oil) = d.m...

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