## Problem StatementYour friend of mass 100 kg can just barely float in fresh water (the buoyant force is equal to her weight). What is her volume in units of L (Liters)?## ExplanationSince the friend barely floats, it implies that the weight of the displaced water is equal to the weight of the person. The density of fresh water is approximately 1 kg/L. Therefore, the volume of water displaced (and hence the volume of the friend) is equal to her mass.Using the formula:\[ \text{Volume (L)} = \frac{\text{Mass (kg)}}{\text{Density (kg/L)}} \]For a mass of 100 kg:\[ \text{Volume} = \frac{100 \text{ kg}}{1 \text{ kg/L}} = 100 \text{ L} \]Thus, the volume of your friend is 100 liters.

Given Mass of the friend is m= 100 kg Acceleration due to gravity is g=9.80 m/s2.The weight of the personal is balanced with the Buoyant force. So, W=FBmg=ρVg The volume of the personal is V=mρ =100 kg1000 kg/m3 =0.10 m3 =100 L

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Your friend of mass 100 kg can just barely float in fresh water (the buoyant force is equal to her weight). What is her volume in units of L (Liters)?

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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