A person who weights 740 N in air is lowered into a tank of water up to chin level. He sits in a harness of neligible mass suspended from a scale that reads his apparent weight. He now exhales as much air as possible and dunks his head under the water, submerging his entire body. His apparent weight while submerged is 35.5 N.
1)What is total net force on the man when he is suspended and submerged?
Fnet =
N
2)What is the total volume of the person?
Vp =
m3
3)What is the average density of the person?
ρp =
kg/m3
Apparent Weight in Water?
2020-07-14 2:02 am
回答 (2)
2020-07-14 2:27 am
✔ 最佳答案
1.When the person is suspended and submerged, he is at rest.
Hence, total net force, Fₙₑₜ = 0
====
2.
Take g = 9.8 m/s²
Weight of water displaced = 740 - 35.5 = 704.5 N
Mass of water displaced = 704.5 / 9.8 = 71.9 kg
Density of water = 1000 kg/m³
Total volume of person, Vₚ = 71.9/1000 = 0.0719 m³
====
3.
Mass of the person = 740 / 9.8 = 75.5 kg
Density of the person, ρ = 75.5 / 0.0719 = 1050 kg/m³
2020-07-14 10:55 am
1) F(net) = 0
2) Let the volume of the person is Vp, By Archimedes' principle:Apparent immersed weight = weight - weight of displaced fluid=>35.5 = 740 - Vp x 997=>Vp = 0.71 m^33)By Archimedes' principle:density/density of fluid = weight/weight of displaced fluid=>dp/997 = 740/(740-35.5)=>dp = 1047.24 kg/m^3
2) Let the volume of the person is Vp, By Archimedes' principle:Apparent immersed weight = weight - weight of displaced fluid=>35.5 = 740 - Vp x 997=>Vp = 0.71 m^33)By Archimedes' principle:density/density of fluid = weight/weight of displaced fluid=>dp/997 = 740/(740-35.5)=>dp = 1047.24 kg/m^3
收錄日期: 2021-04-18 18:36:10
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