Agraduated cylinder if filled with water to a level of 40.0 ml. when a piece of paper of copper is lowered into the cylinder , the water level rises to 63.4ml. find the volume of the copper sample. if the density of the copper is 8.9g/cm³, what is the mass

Given:

Water level is 40mL

Water level rises to 63.4mL after copper is placed

Density is 8.9g/cm3

Required:

Mass of the object

Solution:

The displaced water level is the volume of the copper

63.4mL – 40mL = 23.4mL

Since 1mL = 1cm3, therefore 23.4mL is 23.4cm3

The density formula is D = M/V where D is density, M is mass of the object and V is the volume of the object. Rearranging we get M = DV.

M = DV

M = (8.9g/cm3)(23.4cm3)

M = 208 g

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pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y