WEBVTT

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This is Giancoli Answers
with Mr. Dychko.

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When this athlete is submerged,

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there's going to be two forces
upwards on him:

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there's gonna be the buoyant force
due to the water

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and it's gonna be this apparent
weight due to

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the spring scale that's also pulling him up
and those two forces upwards

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are gonna balance the force of
gravity that's downwards.

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And we can substitute for
the apparent weight

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as being the apparent mass times <i>g</i>

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and then the buoyant force is gonna be

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the weight of the water displaced
by the athlete

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which will be the mass of water which is
gonna be the density of the water

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multiplied by the volume of water

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which is the same as the volume of
the athlete since

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the athlete is totally submerged

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and times that mass of water by <i>g</i> to
get the water's weight

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and that equals <i>mg</i>—the weight
of the athlete—

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and this <i>m</i> is the true mass

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measured in a vacuum although measured
in air is close enough.

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So we can rearrange this after
canceling the <i>g</i>'s

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and subtracting <i>m prime</i> from both sides

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and then divide by the density of water
on both sides

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and we get the volume is the true mass
minus the apparent mass

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divided by the density of water.

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That's 70.2 kilograms minus 3.4 kilograms

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divided by 1.00 times 10 to the 3
kilograms per cubic meter

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which gives 6.68 times 10 to the minus 2
cubic meters of volume.

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The specific gravity of this athlete

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is density of their body divided by
density of water

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and density of the body is their true mass
divided by the volume of their body

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and the volume of the body we get by

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taking this number we calculated

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and then subtracting away the volume of air
that's still in their lungs

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and this gives us just the volume
of only their body.

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So 70.2 divided by this difference
divided by the density of water

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gives 1.07 for the specific gravity.

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And percent body fat, we take this
formula 495 divided by

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specific gravity 1.0718

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and then minus 450 and we get about
12 percent of their body is fat.