### Example Problem: Calculating the Force due to Non-Mixing Liquids in a ContainerThree liquids that do not mix are poured into a cylindrical container. The volumes and densities of the liquids are as follows:- Liquid 1: 0.45 L, 2.6 g/cm³- Liquid 2: 0.20 L, 1.0 g/cm³- Liquid 3: 0.40 L, 0.65 g/cm³**Question:**What is the force on the bottom of the container due to these liquids?**Note:**One liter (L) = 1 L = 1000 cm³. Ignore the contribution due to the atmosphere.[**Calculation Box**]_________ N

V1 = 0.45 L = 0.00045 m³ρ1 = 2.6 g/cm³ = 2600 kg/m³V2 = 0.20 L = 0.0002 m³ρ2 = 1 g/cm³ = 1000…

Three liquids that will not mix are poured into a cylindrical container. The volumes and densities of the liquids are 0.45 L, 2.6 g/cm³; 0.20 L, 1.0 g/cm³; and 0.40 L, 0.65 g/cm³. What is the force on the bottom of the container due to these liquids? One liter = 1 L - 1000 cm³. (Ignore the contribution due to the atmosphere.)

Three liquids that will not mix are poured into a cylindrical container. The volumes and densities of the liquids are 0.45 L, 2.6 g/cm³; 0.20 L, 1.0 g/cm³; and 0.40 L, 0.65 g/cm³. What is the force on the bottom of the container due to these liquids? One liter = 1 L - 1000 cm³. (Ignore the contribution due to the atmosphere.)

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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