A beaker filled to the rim with water weighs 14.70 N. One gently places a cube o

Question

A beaker filled to the rim with water weighs 14.70 N. One gently places a cube of red cedar wood weighing 0.820 N to float on the water surface, letting water flow over the rim slowly so that the water keeps touching the rim. The densities of water and red cedar wood are Pwater= 1.00X10^3 kg/m^3 and Pcedar= 0.380X10^3 kg/m^3

a. How much does the beaker with water and wood weigh now?

b. What is the weight of the displaced water?

c. What fraction of the wood is below the water surface?

d. What is the volume of the wood above the water surface?

Explanation / Answer

Weight of displaced water = weight of wood

a) Weight = 14.70 N (there is no change in the weight because the weight of displaced water = weight of wood)

b) Weight of displaced water = 0.820 N

c) Fraction below the surface of water

volume of displaced water = weight /(density x gravity)

= 0.820 /(1000 x 10) = 8.20 x 10-5 m3

mass of wood under water = density x volume = 380 x 8.20 x 10-5 = 3.116 x 10-2 kg

weight of wood below water surface = mass x gravity = 3.116 x 10-2 x 10 = 0.3116 N

Fraction below the surface = 0.3116 /0.820 = 0.38 or 38%

d) Volume above the surface

Fraction below the surface = 38%

===>> fraction above the surface = 100% - 38% = 62%

38% --------> 8.20 x 10-5 m3

62% --------> Y

38 x Y = 62 x 8.20 x 10-5 m3

Y = (62 x 8.20 x 10-5 m3)/38

Y = 1.338 x 10-4 m3

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