What is the volume of a body that stretches a dynamometer in air, on which it is suspended by a force of 2.5 N, and if it is immersed in alcohol with a density of 800 kg/m3, does it tension the dynamometer with a force of 1.3 N?

Correct answer:

V =  0.15 l

Step-by-step explanation:

F1=2.5 N F2=1.3 N ρ=800 kg/m3  g=10 m/s2  F=mg m=Vρ F=Vρg  V1=(F1F2)/(ρ g)=(2.51.3)/(800 10)=2000030.0002 m3  V=V1 l=V1 1000  l=0.0002 1000  l=0.15 l=0.15 l

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