Asked by Teagan
An 80 kg man is put into a tank of water. When completely submerged he displaces
75 L of water. If fatty body tissue has an average density of 0.9 kg/L and non-fatty body tissue
has an average density of 1.1 kg/L, what percentage of the man is composed of fat?
75 L of water. If fatty body tissue has an average density of 0.9 kg/L and non-fatty body tissue
has an average density of 1.1 kg/L, what percentage of the man is composed of fat?
Answers
Answered by
Damon
volume of man = 75 liters
so density of man = 80 kg/75 L = 1.067 kg/L
x(fat density ) + (1-x)(bone density) = 1.067
.9 x + 1.1 (1-x) = 1.067
1.1 - 1.067 = (1.1-.9)x
x = .167 or 17%
so density of man = 80 kg/75 L = 1.067 kg/L
x(fat density ) + (1-x)(bone density) = 1.067
.9 x + 1.1 (1-x) = 1.067
1.1 - 1.067 = (1.1-.9)x
x = .167 or 17%
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.