(II) On dry land, an athlete weighs 70.2 kg. The same athlete,...

Question

(II) On dry land, an athlete weighs 70.2 kg. The same athlete, when submerged in a swimming pool and hanging from a scale, has an “apparent weight” of 3.4 kg. Using Example 13–10 as a guide,

(c) What is the athlete’s percent body fat assuming it is given by the formula (495 / SG) - 450?

Accepted Answer

Welcome back. Everyone in this problem. A biophysicist is trying to calculate the percentage of bones in the body of a tuna fish on land. The tuna weighs 790 kg. When submerged in water, its apparent weight is 15 kg. Assuming the volume of the swim bladder of the tuna while submerged in water is 3.9 multiplied by 10 to the negative two cubic meters. What would be the calculated value of the bone percentage by the biophysicist? If he uses the formula 680 divided by SG minus 620 for his calculation where SG is the specific gravity of the tuna fish A says it's 10% b 11% c 14% and D 47%. Now, our problem tells us that to find a bone percentage. Ok. Then the biophysicist plan to plans to use the formula 680 divided by the tuna fishes specific gravity minus 620. That's as a percentage. Now, if we are going to figure out the bone percentage first, we need to find the tuna fishes specific gravity, what do we know about specific gravity? Well, recall? Ok. That the specific gravity. In this case, for the tuna fish is going to be equal to the density of tuna divided by the density of water. Because it's just a comparison of an object to the density of water. OK. In our problem, we're not told what the density of the tuna fish is, right. But we know that density is going to be equal to mass divided by volume. So for our tuna fish, the density is going to be equal to the tuna fish's mass divided by the volume of for our bones. OK. And when we think about it, that volume is going to be equal to the volume of the water. We can call V minus the swim bladder volume. That is basically the volume, that's not a part of the fish. The fish is bones. That's how I try to think about it. So we can call that V minus V prime. So if we're going to figure out the density of our tuna fish, we need to find the volume of all of the water displaced by the fish. OK. And then subtract the volume of the swim bladder from that value. So how can we find the volume of water displaced by the fish? Well, we can use what we know about the buoyant forces to help us. So now let's say we're solving for Z OK. Now, if we think about our fish, so let's just draw a quick diagram of our fish here. So here we have our fish, OK? And four oil fish, then we know that the weight of the fish is acting downwards. And now we're imagining that we have our fish in water, the weight of the fish is acting downwards while the buoyant force and the fish is acting upward, OK? And now based on what we know, we know that our buoyant force, OK, is going to be equal to the force or the weight, sorry, the weight of the fish. And in this case, that weight is going to be equal to the fish's weight minus its apparent weight. So that's going to be MG minus M prime G. OK. Now, we also know that the buoyant force equals the density of the water multiplied by the volume of the displaced water multiplied by the acceleration due to gravity. In other words, density multiplied by G sorry density multiplied by V. That's the volume we're looking for multiplied by G equals MG minus M prime G. And now from this idea, we can divide G throughout our equation, OK. They get the density of water multiplied by the volume being equal to M minus M prime and thus, or volume. Yeah, our volume is going to be equal to M minus M prime, the apparent mass divided by the density of water. Now, we can go ahead and solve for V because we know the fish's mass is 790 kg. Its apparent mass is 15 kg and the density of water is 1000 kg per cubic meter. When we go ahead and divide, then we should get our volume to be 0.775 cubic meters. Know that we have the volume of the water that is displaced by the fish. Then we can substitute it to solve for our specific gravity of our tuna fish. Because no, this means then that the specific gravity which was equal to the density of tuna that's M divided by V minus V prime, divided by the density of water which you can write as being multiplied by its reciprocal is not going to be equal to 790 kg. OK? Divided by or volume of the water displaced by the fish is 0.775 cubic meters minus the volume of the swim bladder. That's 0.039 cubic meters. OK? Multiplied by the reciprocal of the density of water. So that's one divided by 1000 kg per cubic meters. I know when we work that out, then we should get our specific gravity to be 1.0734. Know that we have the specific gravity, we can go ahead and substitute into our formula for the bone percentage because no, this means our bone percentage then is going to be equal to 680. OK. Divided by our specific gravity of 1.0734 minus 620 which after we solve, we're going to convert to a percentage. And now when we do that, ok, then we should get 13.501% or to two significant figures, approximately 14%. Therefore, the bone percentage, the bone percentage of our tuna fish is approximately 14%. That means c is the correct answer. Thanks for watching everyone. I hope this video helped.

(II) On dry land, an athlete weighs 70.2 kg. The same athlete, when submerged in a swimming pool and hanging from a scale, has an “apparent weight” of 3.4 kg. Using Example 13–10 as a guide,

(c) What is the athlete’s percent body fat assuming it is given by the formula (495 / SG) - 450?

Key Concepts

Buoyancy

Specific Gravity (SG)

Body Fat Percentage Calculation

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