Question

a) You come across an open container that is filled with two liquids. Since the two liquids have different densities, there is a distinct separation between them. Water, which has a density of ρw=1.00×103 kg/m3, fills the lower portion of the container to a depth of 0.215 m. The fluid that is floating on top of the water is 0.349 m deep. If the absolute pressure on the bottom of the container is 1.049×105 Pa, what is the density, ρl, of the unknown fluid? The acceleration due to gravity is g=9.81 m/s2 and atmospheric pressure is P0=1.013×105 Pa. b) A child playing in a swimming pool realizes that it is easy to push a small inflated ball under the surface of the water whereas a large ball requires a lot of force. The child happens to have a styrofoam ball (the shape of the ball will not distort when it is forced under the surface), which he forces under the surface of the water. If the child needs to supply 554 N to totally submerge the ball, calculate the diameter d of the ball. The density of water is ρw=1.000×103 kg/m3, the density of styrofoam is ρfoam=95.0 kg/m3, and the acceleration due to gravity is g=9.81 m/s2.

          a) You come across an open container that is filled with two liquids. Since the two liquids have different densities, there is a distinct separation between them. Water, which has a density of ρw=1.00×103 kg/m3, fills the lower portion of the container to a depth of 0.215 m. The fluid that is floating on top of the water is 0.349 m deep. If the absolute pressure on the bottom of the container is 1.049×105 Pa, what is the density, ρl, of the unknown fluid? The acceleration due to gravity is g=9.81 m/s2 and atmospheric pressure is P0=1.013×105 Pa.
b) A child playing in a swimming pool realizes that it is easy to push a small inflated ball under the surface of the water whereas a large ball requires a lot of force. The child happens to have a styrofoam ball (the shape of the ball will not distort when it is forced under the surface), which he forces under the surface of the water. If the child needs to supply 554 N to totally submerge the ball, calculate the diameter d of the ball. The density of water is ρw=1.000×103 kg/m3, the density of styrofoam is ρfoam=95.0 kg/m3, and the acceleration due to gravity is g=9.81 m/s2.
        
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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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a) You come across an open container that is filled with two liquids. Since the two liquids have different densities, there is a distinct separation between them. Water, which has a density of ρw=1.00×103 kg/m3, fills the lower portion of the container to a depth of 0.215 m. The fluid that is floating on top of the water is 0.349 m deep. If the absolute pressure on the bottom of the container is 1.049×105 Pa, what is the density, ρl, of the unknown fluid? The acceleration due to gravity is g=9.81 m/s2 and atmospheric pressure is P0=1.013×105 Pa. b) A child playing in a swimming pool realizes that it is easy to push a small inflated ball under the surface of the water whereas a large ball requires a lot of force. The child happens to have a styrofoam ball (the shape of the ball will not distort when it is forced under the surface), which he forces under the surface of the water. If the child needs to supply 554 N to totally submerge the ball, calculate the diameter d of the ball. The density of water is ρw=1.000×103 kg/m3, the density of styrofoam is ρfoam=95.0 kg/m3, and the acceleration due to gravity is g=9.81 m/s2.
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A child playing in a swimming pool realizes that it is easy to push a small inflated ball under the surface of the water while a large ball requires a lot of force. The child happens to have a styrofoam ball (this way the shape of the ball will not distort when it is forced under the surface) which is being forced under the surface of the water. If the child needs to supply 5.00 × 10^2 N to totally submerge the ball, calculate the diameter of the ball. The density of water is ρwater = 1.000 × 10^3 kg/m^3, the density of styrofoam is ρfoam = 95.0 kg/m^3, and the acceleration due to gravity is g = 9.81 m/s^2.

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Transcript

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00:01 Hello student, to solve the given question, let us write the formula for absolute pressure p equals p .0, that is atmospheric pressure plus ro gh.
00:10 So using this relation, we can solve the given question.
00:14 Firstly, for part a, at the bottom of container, the pressure p2 will be equal to p1 plus row w, that is density of water, gh2.
00:26 Now here we can use the given relation in the formula and plug in here we will get p .0 plus row gh1 plus row w gh2 equals the absolute pressure.
00:42 Now p2 here is the absolute pressure.
00:45 So according to the values given in the statement, we can find the density of the liquid or fluid as plug in the values 1 .049 into 10 .5 .5.
00:59 That is absolute pressure equals atmospheric pressure is 1 .013 into 10 to power 5 plus density multiply by 9 .8 into h1 that is 0 .349 meters plus the last term will be 1 into 1010 to par 3 density of water into 9 .8 multiply by h2 0 .215 meters.
01:27 So after this calculation we get the final answer for the density for part a equals 436 .5 kilogram per meter cube.
01:42 Next, for part b to calculate the diameter, we can use the relation for mass that is equal to density multiply by volume.
01:53 So using this relation we can calculate part b according to which let us first write the force equation.
02:02 M1g plus f equals m2g...
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