A cube of plastic 0.3 times 0.3 times 0.3 m. is placed in a large vessel contain
ID: 1623776 • Letter: A
Question
A cube of plastic 0.3 times 0.3 times 0.3 m. is placed in a large vessel containing alcohol whose specific gravity is 0.785. The plastic floats such that 0.1 m is above the liquid surface. a) What is the density of the plastic? b) What force must I apply so that 0.25 m of the plastic is submerged, leaving 0.05 the surface. c) Given that the density of Hg is 13, 500 kg/m^3 Calculate the atmospheric pressure when the barometic pressure is 76 cm. Hg. d) Suppose superman tries to drink this alcohol through a straw. What is the length of the longest straw he could possibly use on a day when the barometric pressure reads 75 cm Hg? e) How long would the straw have to be on Mars, where the barometric pressure reads 20 cm. Hg?Explanation / Answer
Part A dimensions of cube= 0.3*0.3*0.3
specific gravity of alcohol=0.785
as weight of alcohol displaced = weight of cube ….(1)
weight of alcohol displaced= volume of cube immersed in alcohol* density of alcohol*g
= (0.3*0.3*0.2) * (0.785*1000) * 9.8=138.474 N
Weight of cube= volume of cube* density of plastic cube*g
From ….(1)
0.3*0.3*0.3*density of plastic cube * 9.8 = 138.474
density of plastic cube=523.333 Kg/m^3
part b
let we apply a force F on cube . and a buoyancy force B act on cube …then
at equilibrium
F + weight of box = B
F + 138.474 = volume of cube immersed in alcohol* density of alcohol*g
F + 138.474 = 0.3*0.3*0.25*785*9.8
F=173.0925 – 138.474
F=34.6185 N
Part c
Atmospheric pressure= height* density of mercury * 9.8
= 100548 pascal
Part d
Barometric pressure = height of straw * density of alcohol * 9.8
Height of straw= 12.898 m
Part e
On moon Barometric pressure = height of straw * density of alcohol * 9.8
Height of straw= 3.43949 m
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