Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
last physics problem need help!!?
A cube with an edge of 2.3 m is completely submerged in freshwater. The density of the cube is 4780 kg/m^3, and the density of freshwater is 1000 kg/m^3.
Weight of the cube-
buoyant force the water exerts on the cube-
calculate the apparent weight of the cube when submerged in freshwater-
- billrussell42Lv 71 month ago
volume is 2.3³ = 12.167 m³
4780 kg/m³ x 12.167 m³ = 58200 kg
weight is 58200 kg x 9.8 N/kg = 570000 N
mass of same volume of water =
1000 kg/m³ x 12.167 m³ = 12167 kg
diff is buoyancy = 45991 kg or 450700 N
apparent weight is 570000 – 450700 = 120000 N
- JohnLv 71 month ago
The buoyancy force is the weight of the displaced fluid (cube volume x water density x gravitational accleration), which is in the upwards direction. The weight of the cube itself is cube volume x cube density x gravitational acceleration, in the downwards direction. The difference is the apparent weight.