Density of Wood Blocks: Archimedes Lab: AP Physics B

Density of Wood Blocks

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Information from wood blocks (taken straight from Nathan Lindquist's post):

Small block:

dimensions: 4.3 cm X 4.3 cm X 7.0 cm
when submerged, .2 cm remained above the water (out of a total height of 7.0 cm)
mass: 129.8 g

Large block:

dimensions: 3.8 cm X 7.5 cm X 7.5 cm
when submerged, .95 cm remained above the water (out of a total height of 3.8 cm)
mass: 163.2 g

The key equation for this problem is

$\frac{V_{disp}}{V_o}=\frac{\rho_o}{\rho_f}$

for the both blocks, this can be easily applied because the density of the fluid is known to be 1. Therefore, the ratio of volumes is equal to the density of the object.

SMALL BLOCK-

Using the definition of density $\rho=\frac{mass}{volume}=\frac{129.8}{129.43}$ , we calculate the density to be 1.0 g/cm^3

However, using Archimedes' definition $\rho=\frac{V_{disp}}{V_o}=\frac{4.3\cdot4.3\cdot\left(7-.2\right)}{4.3\cdot4.3\cdot7}$ , the density of the block is calculated to be .97 g/cm^3

LARGE BLOCK-

Using the definition of density $\rho=\frac{mass}{volume}=\frac{163.2}{3.8\cdot7.5\cdot7.5}$ , we calculate the large block's density to be .76 g/cm^3

However, using Archimedes' definition, $\rho=\frac{V_{disp}}{V_o}=\frac{7.5\cdot7.5\cdot\left(3.8-.95\right)}{7.5\cdot7.5\cdot3.8}$ the density of the block is calculated to be .75 g/cm^3

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[{:section_type=>"rich_text", :content=>"Information from wood blocks (taken straight from Nathan Lindquist's post):\n \nSmall block:\n<ul>\n<li>dimensions: 4.3 cm X 4.3 cm X 7.0 cm</li>\n<li>when submerged, .2 cm remained above the water (out of a total height of 7.0 cm)</li>\n<li>mass: 129.8 g</li>\n</ul>Large block:\n<ul>\n<li>dimensions: 3.8 cm X 7.5 cm X 7.5 cm</li>\n<li>when submerged, .95 cm remained above the water (out of a total height of 3.8 cm)</li>\n<li>mass: 163.2 g</li>\n</ul> \nThe key equation for this problem is \n \n<img class=\"equation_image\" title=\"\\frac{V_{disp}}{V_o}=\\frac{\\rho_o}{\\rho_f}\" src=\"/equation_images/%255Cfrac%257BV_%257Bdisp%257D%257D%257BV_o%257D%253D%255Cfrac%257B%255Crho_o%257D%257B%255Crho_f%257D\" alt=\"\\frac{V_{disp}}{V_o}=\\frac{\\rho_o}{\\rho_f}\">\n \n \nfor the both blocks, this can be easily applied because the density of the fluid is known to be 1. Therefore, the ratio of volumes is equal to the density of the object.\n \nSMALL BLOCK-\n \nUsing the definition of density <img class=\"equation_image\" title=\"\\rho=\\frac{mass}{volume}=\\frac{129.8}{129.43}\" src=\"/equation_images/%255Crho%253D%255Cfrac%257Bmass%257D%257Bvolume%257D%253D%255Cfrac%257B129.8%257D%257B129.43%257D\" alt=\"\\rho=\\frac{mass}{volume}=\\frac{129.8}{129.43}\">, we calculate the density to be 1.0 g/cm^3\n \nHowever, using Archimedes' definition <img class=\"equation_image\" title=\"\\rho=\\frac{V_{disp}}{V_o}=\\frac{4.3\\cdot4.3\\cdot\\left(7-.2\\right)}{4.3\\cdot4.3\\cdot7}\" src=\"/equation_images/%255Crho%253D%255Cfrac%257BV_%257Bdisp%257D%257D%257BV_o%257D%253D%255Cfrac%257B4.3%255Ccdot4.3%255Ccdot%255Cleft%25287-.2%255Cright%2529%257D%257B4.3%255Ccdot4.3%255Ccdot7%257D\" alt=\"\\rho=\\frac{V_{disp}}{V_o}=\\frac{4.3\\cdot4.3\\cdot\\left(7-.2\\right)}{4.3\\cdot4.3\\cdot7}\">, the density of the block is calculated to be .97 g/cm^3\n \nLARGE BLOCK- \n \nUsing the definition of density <img class=\"equation_image\" title=\"\\rho=\\frac{mass}{volume}=\\frac{163.2}{3.8\\cdot7.5\\cdot7.5}\" src=\"/equation_images/%255Crho%253D%255Cfrac%257Bmass%257D%257Bvolume%257D%253D%255Cfrac%257B163.2%257D%257B3.8%255Ccdot7.5%255Ccdot7.5%257D\" alt=\"\\rho=\\frac{mass}{volume}=\\frac{163.2}{3.8\\cdot7.5\\cdot7.5}\">, we calculate the large block's density to be .76 g/cm^3\n \nHowever, using Archimedes' definition, <img class=\"equation_image\" title=\"\\rho=\\frac{V_{disp}}{V_o}=\\frac{7.5\\cdot7.5\\cdot\\left(3.8-.95\\right)}{7.5\\cdot7.5\\cdot3.8}\" src=\"/equation_images/%255Crho%253D%255Cfrac%257BV_%257Bdisp%257D%257D%257BV_o%257D%253D%255Cfrac%257B7.5%255Ccdot7.5%255Ccdot%255Cleft%25283.8-.95%255Cright%2529%257D%257B7.5%255Ccdot7.5%255Ccdot3.8%257D\" alt=\"\\rho=\\frac{V_{disp}}{V_o}=\\frac{7.5\\cdot7.5\\cdot\\left(3.8-.95\\right)}{7.5\\cdot7.5\\cdot3.8}\"> the density of the block is calculated to be .75 g/cm^3\n \n \n \n \n \n "}]

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AP Physics B

Density of Wood Blocks

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