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Select Buoyy 1 conducting density the Jadum ] cylinder 3 eqdaus 8 23 82 Jaino 1 radlus 2 43cvlinder 1 total 1 70l...

Question

Select Buoyy 1 conducting density the Jadum ] cylinder 3 eqdaus 8 23 82 Jaino 1 radlus 2 43cvlinder 1 total 1 70l

Select Buoyy 1 conducting density the Jadum ] cylinder 3 eqdaus 8 23 82 Jaino 1 radlus 2 43 cvlinder 1 total 1 70l



Answers

You measure the masses and volumes of two cylinders. The mass of cylinder 1 is 1.35 times the mass of cylinder 2. The volume of cylinder 1 is 0.792 times the volume of cylinder 2. If the density of cylinder 1 is 3.85 $\mathrm{g} / \mathrm{cm}^{3}$, what is the density of cylinder $2 ?$

Static, the cylinders and static equilibrium the some of the forces in the Y direction would be equaling the first buoyancy force close Epsom beat too, minus the gravitational force. Essentially, these the the buoyancy force. Someone would be essentially one that corresponds to d someone the length of the cylinder in the less desk and the less dense liquid. And then we could say the buoyancy force up to would correspond to descend to the length of the cylinder in the more dense liquid, uh, rose up to. And so we can say that then, Rose, this is gonna equal zero Newtons, of course. Static equilibrium. Rows of one multiplied by the cross sectional area. Dese of one again, this is the less dense liquid G plus arose of to a decent to G Equalling, then row. Hey, this would be decent one, plus decent to the entire volume of the cylinder multiplied by G. This is, of course, the gravitational force of the cylinder. So now we can say roast of one divided by Rosa two multiplied by dese of one plus dieser two would be equaling two row divided by Rosa too. I nd someone plus decent too now because the length Al is equaling two dese of one plus decent to weaken, simplify and we can solve For geese of one, this would be equaling toe l minus decent, too, and so diese to is then equaling two l multiplied by ro minus roast of one. This would be divided by rows of two minus roasts of one, essentially the density of the cylinder, minus the density of the less dense liquid, divided by the density of the denser liquid, minus the density of the less dense liquid. And so we can say that then the fraction of the cylinder in the more dense liquid, so fraction of the cylinder in two, the more the more dense liquid this would be equaling. Then ro, minus rows of one, divided by Ro said to minus rows of one. That is the end of the solution. Thank you for watching

Density is even by mass over volume equals to 87 grand, divided by 10 centimeter cube equals 2.7 gram per centimeter cube. I am defying it from the table. It comes out to be element.

Alright problems 73 is there following integral and to go from 0 to 2 pi and you go from 0 to 2 and from 0 to 8 There are times one minus 0.5 e five negative 0.1 r squared Deasy, drd, Theodora And evaluating this we've got approximately 95.60 362

Why? So this question is asking you to find the density of silver. They give you the mass of a piece of solder, and then they put it in water, and then they show you the change in the volume of water, and they want you to find the density based on that. So an important equation we're gonna need here is what density is. Density is mass over volume. Okay, so what are the two things were going? Fine. We're going to find the mass, right to give us the mass here, and then we're gonna have to find the volume. How do we know the volume of the silver? We can find that by how much water it displaced. So basically, if you've changed the volume of the water by a certain volume, then you have changed that then by adding the silver, then that is the volume of the silver. The difference in the volume of water after you add the silver is the volume of the silver. Because you've taken up that much space which has pushed the water to a higher level. All right, so we can write out the equation like this. Um, volume of silver equals volume of I guess. Water final, minus volume of water initial. Okay. Now, Yes, please. And here to give us the values she just plug in the final volume in the graduated cylinder, and then you subtract the big initial from it. So to 60 0.5 no years minus to sort of minus 2 40 to 0.0 their leaders. Okay. And then you get the volume of silver. Now, notice. I'm actually gonna change the unit from new leaders. Two centimeters cube, because milliliters and sent rescuers are exactly the same volume. But you use seven years cute for solids and use milliliters for liquids. And since silver is a solid, we're gonna have to use centimeters cute instead of, so you get 18.5 centimetres cute as your volume of silver and you can just plug us into our density equation. Here. What is the master? Give us 1 94.3 Right. You put this all over. Why? 18.57 years. Cute. Get grounds per centimeters cube. Right. Which is the correct number for never unit. And you should get 10 0.5 grounds per centimeter. Cute notice. I have three sig figs because my volume number was 366 here. Right. And the reason this was three sig figs instead of four. Because when you do additions attraction, you you count decimal places, right? So one decimal place, and then you change the six based on that. Okay, so that's pretty much this whole question. And, um, yeah, hope is a helpful and thank you for listening.


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