Question

PH105 Physics for Life Sciences Week 10. Labs 49: Archimedes Principle Fall 2019 Pre-laboratory Assignment Team Total credit: 20 points Student names: 1. (10 pts) The dam below is 400 m wide, and the water is 60.0 m deep. The density of water p = 1.00 x 10' Atmospheric pressure is Parm = 1.01 X 10' Pa. a) What is the total pressure on the dam at the deepest point? b) What is the average pressure on the dam due to the water? c) Calculate the force exerted against the dam. 2. (10 pts) The mass of an ancient Greek coin is determined in air to be 8.630 g. When the coin is submerged in water as shown, its apparent mass is 7.800 g. Calculate its density, given that water has a density of 1.000 g/cm3 and that effects caused by the wire suspending the coin are negligible.

Answer 1

" Patm - 1.01 X 105 Pa Sweater =103 kg/m3 W = 40om Н= $om dy a DA (6) "p=F/A = F = PA & df = (w .gy). da = ({w.gy) Wody Y=H I tweg w Een twy w gdy - Mw3W) [+]" Gdy i Farg = Sw d w H² - .. Farg = + (49H). HW Area (A) = HW Parg x Area - 0 - Farg = .. Parg - SwidH 163 x 9o8x60 - 2 in Parg = 2.94 x 105 pa

" Real weight Wreal = 8.638 16-338 • Wapparent) in water = 7.8 x 103 x 9.8 NB 8 (Buayant force) ' = WReal Wapparent fua xgxv = (8.63-78) 9 x 1003 v = volume of con Scoin - filed xm = 0.83 X 16-3 Scoin Afluid = 1 gm1cm3 = 10m3 = 163 kg/m² (26-2)3 = my real mass = 8. 63 X 16-3 kg = = 0.83 X 16-3 103 x 80 63 X 10 3 Scoln Scoin I - 8.63 x 16 0083 i Scoin = 1003975 x 103 kg/m²