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by: Malvina Orn


Malvina Orn
GPA 3.77

Jack Ritchie

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About this Document

Jack Ritchie
Class Notes
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This 5 page Class Notes was uploaded by Malvina Orn on Monday September 7, 2015. The Class Notes belongs to PHY 317K at University of Texas at Austin taught by Jack Ritchie in Fall. Since its upload, it has received 37 views. For similar materials see /class/181822/phy-317k-university-of-texas-at-austin in Physics 2 at University of Texas at Austin.




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Date Created: 09/07/15
Example 1 The density of iron is 787 gcm3 and the density of lead is 1135 gcrng What is the volume of a 1000 lb block of iron What is the weight of a cubic meter of lead in pounds Recall 44 N 1 lb Solution W mg pVg so V Wpg Then for the iron block 10001b 10001b M V g In g 1kg 110th 3 m 787 QXQBS Q 7871000g 9857 For the lead 1k 100 3 W pVg 1135 3 ix cm 11113 989 111x105N crn3 1000g lrn s2 ln pounds 11b W111gtlt105Nlt gt25x1041b 44N Example 2 What is the pressure at the bottom of a column of water that is 30 ft in height assuming the water is open to the atmosphere at the top of the column Solution Use the fact p pext pgh to nd the pressure at any depth in the water Here pext latm 101 x 105 Pa the pressure of the atmosphere assuming usual conditions h30ftlt121ngt lt264pmgtlt 1m gt91m 1ft lin lOOom p pm pgh 101 x 105 Pa 1000 kgm398ms29lm Then p 101 x 105 Pa 089 x 105 Pa 190 Pa 188 atm Example 3 Approximately What is the air pressure at the top of Mount Everest Solution Pressure p at an altitude h is given by 7 p h p 2906 0 7 Where p0 and p0 are the density and pressure of air at h 0 ie sea level Using p0 101 x 105 Pa and p0 121 kgrng p0 101 x 105 Pa 3 7 85 10 85k gpo 98ms2121kgn13 x m m h p p0 85km The altitude of Mount Everest is about 88 km 29028 ft giving 19 pOe 036190 036 atrn 36 X 104 Example 4 A U tube is partly lled with mercury Water which does not mix with mercury is poured into one side If the water column is 30 cm in height how much does the mercury rise above its original level on the other side of the U tube Solution Draw a gure If the weight of the water pushes the mercury down by a distance d on one side then it also pushes it up by a distance d on the other At the bottom of the water the pressure in the mercury is the same as on the other side of the tube at the same level Thus the weight of the liquids above that level must be the same pwaterg30 cm pmercuryg2d Solve for d 3 d ipwater 30 cm W 15cm 11cm pmercury 2 Example 5 A block of wood density 750 kgmg is placed in a tank of water The block oats What fraction of the volume of the block is submerged below the surface of the water Solution Draw a gure Take the density of water to be 1000 kgmg lf Vdis is the volume of water displaced by the wood as it oats then the bouyant force Fb is given by Fb pwaterldisg according to Archimedes7 Principle lf Vtot is the total volume of the wooden block the weight of the block is W pblockltotg When an object oats the magnitudes of these opposing forces are equal Fb W Then pwatervdisg pblockltotg7 so the fraction of the blocks volume under water is Vdis i pblock i 750 kg 1113 i 075 ltot pwater


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