A woman floats in fresh water of
density 1.00 times 10 to the third kilograms per cubic meter. 4.00 percent of her volume is above
the water’s surface. What is the density of the
woman? What percent of the woman’s volume
would be above water if she floated in sea water of density 1.03 times 10 to the
third kilograms per meter cubed?
In part one, we want to solve for
the density of the woman. We can call that 𝜌 sub 𝑤. And in part two, we want to solve
for the percent of the woman’s volume that would be above water if she was in sea
water. We’ll call that percent capital
𝑃. We’re told the density of fresh
water as well as salt water. And we can name those densities 𝜌
sub 𝑓 and 𝜌 sub 𝑠, respectively.
In the initial example, where the
woman is in fresh water, we’re told that 4.00 percent of her volume is above that
water when she floats. This means that, compared to the
density of fresh water, the density of the woman, 𝜌 sub 𝑤, is four percent
less. Written as an equation, we can say
that 𝜌 sub 𝑤 equals 𝜌 sub 𝑓 minus four percent of 𝜌 sub 𝑓, 0.04 times that
density. This is equal to 0.96 times 𝜌 sub
𝑓, or 0.96 times 1.00 times 10 to the third kilograms per cubic meter. This product equals 960 kilograms
per cubic meter. That’s the density of the
In part two, we imagine that the
woman moves from fresh water to salt water. And now we want to know what
percentage of her volume is above that level of the water. We’ll figure this out using the
density of salt water, 𝜌 sub 𝑠, and the density of the woman from part one, 𝜌 sub
𝑤. Here’s one way to think of
this. Imagine that 𝜌 sub 𝑤 and 𝜌 sub
𝑠 were the same, so that this fraction is equal to one. In that case, 100 percent of the
woman would be submerged and zero percent would be above water.
So, to solve for 𝑃, the percent
above water, we’ll want to subtract this ratio from 100 percent. And then, more than that, we’ll
want to turn this ratio from a decimal into a percentage value. And we’ll do that by multiplying it
by 100. Now, we have an expression, which
when we calculate it, we’ll solve for the percentage of the woman’s volume that is
above the level of the saltwater.
When we plug in the densities for
the woman and for salt water and calculate this expression, we find it’s equal, to
three significant figures, to 6.80 percent. That’s the percentage of the
woman’s volume that would be above water level if she was floating in sea water.