12 Dec 07
Originally posted by Ramned17.0 cm, by symmetry. 🙂
A water tank open to the atmosphere at the top has two small holes punched in its side, one above the other. The holes are 5.00 cm and 12.0 cm above the floor. How high does the water stand in the tank if the two streams of water hit the floor at the same place?
I did do the math using Bernoulli's equation to confirm (which I can post if you want), but I like the simplicity of the above solution.
12 Dec 07
Originally posted by PBE6ha ha, you find alot of loopholes dont you 😉
17.0 cm, by symmetry. 🙂
I did do the math using Bernoulli's equation to confirm (which I can post if you want), but I like the simplicity of the above solution.
I dont think there's anything to look for in this one except Fluid physics
A 2.0 cm thick bar of soap is floating on a water surface so that 1.5 cm of it under water. Bath oil of specific gravity .60 is pured into the water and floats on top of it. What is the depth of the oil layer when the top of the soap is just level with the upper surface of the oil?
12 Dec 07
Originally posted by RamnedFrom the first point, specific gravity of the soap is 0.75.
I dont think there's anything to look for in this one except Fluid physics
A 2.0 cm thick bar of soap is floating on a water surface so that 1.5 cm of it under water. Bath oil of specific gravity .60 is pured into the water and floats on top of it. What is the depth of the oil layer when the top of the soap is just level with the upper surface of the oil?
Since the soap is denser than the oil, we know the thickness of the oil is less than the width of the soap. Let's call it d.
Then:
2 x density of soap = d x density of oil + (2 - d) x density of water
1.5 = 0.6d + (2 - d)
d = 1.2 cm.
12 Dec 07
good. i cant make a tough enough problem!
A hollow object with an average density of 930 kg/m^3 floats in a pan containing 500 cm^3 of water. Ethanol is aded to the water and mixed until the object is just on the verge of sinking. What volume of ethanol has been added? Disregard the loss of volume caused by the mixing.
13 Dec 07
Originally posted by RamnedI could do that. I am too lazy to look up the specific gravity of ethanol though. It's like 0.7 isn't it?
good. i cant make a tough enough problem!
A hollow object with an average density of 930 kg/m^3 floats in a pan containing 500 cm^3 of water. Ethanol is aded to the water and mixed until the object is just on the verge of sinking. What volume of ethanol has been added? Disregard the loss of volume caused by the mixing.
13 Dec 07
Originally posted by RamnedI'm not sure what the significance of the hollowness of the object is, but here's my solution:
good. i cant make a tough enough problem!
A hollow object with an average density of 930 kg/m^3 floats in a pan containing 500 cm^3 of water. Ethanol is aded to the water and mixed until the object is just on the verge of sinking. What volume of ethanol has been added? Disregard the loss of volume caused by the mixing.
The density of water is 1.00 g/cm3 and the density of ethanol is 0.789 g/cm3 at STP.
vW = 500 cm3
vE = ?
v = vW + vE
Also:
mW = vW * rhoW
mE = vE * rhoE
m = mW + mE = (vW * rhoW) + (vE * rhoE)
The overall density, neglecting the change in volume from mixing, is:
rho = m / v = ((vW * rhoW) + (vE * rhoE)) / (vW + vE)
Solving for vE, we get:
vE = vW * (rhoW - rho) / (rho - rhoE)
Subbing in the values, we get:
vE = 500 * (1.00 - 0.930) / (0.930 - 0.789) = 248.227 = 248 cm3
13 Dec 07
1 edit
Originally posted by RamnedIn the above solution, "m" is mass (g), "v" is volume (cm3) and "rho" is density (g/cm3). The subscripts "W" and "E" refer to water and ethanol, respectively. Those figures without a subscript are overall values.
No - I can't tell what some of your terms are, but it looks like you are not looking at an assumption that needs made. I'll give more help if no-one gets it. looks like you are headed in the right direction though, using the densities = mass / VOLUME.
I tried plugging the final answer back into Excel and it seems to work just fine:
v = vW + vE = 500 + 248 = 748 cm3 (assuming the volumes are additive, as stated in the question)
m = mW + mE = (vW * rhoW) + (vE * rhoE) = (500 * 1) + (248 * 0.789) = 695.7 g
rho = m / v = 695.7 / 748 = 0.930 g/cm3
It's my understanding that when an object has the same density as the surrounding fluid, it's on the verge of floating/sinking as required by the question. What am I missing here?
13 Dec 07
Originally posted by RamnedI must admit I'd agree with the solution given. Unless I'm misunderstanding what it meant by "on the verge of sinking", Archimedes' principle states that the mass of displaced fluid equals the mass of the floating object.
I'll give more of a hint later if nobody gets it. I don't think you are looking at enough components besides the densities and masses, volumes.
Which seems to be what PBE6 does - with the assumption that the fluids are well mixed and no volume is lost.
13 Dec 07
You are also assuming that your hollow object is small enough to be submerged in the volume of ethanol/water mixture, which is not given, but this seems like a reasonable assumption. The hollow object is also completely sealed, right? The fluid cannot get in?
I would have agreed that the above solution was reasonable given the stated problem.
14 Dec 07
5 edits
Originally posted by mtthwArchimedes' principle is saying the Buoyant force, B, is equal to the amount of water displaced. You can find the amount of water displaced, through measuring the volume of the object (= volume of fluid displaced), the density of the fluid (1) and then the gravity.
I must admit I'd agree with the solution given. Unless I'm misunderstanding what it meant by "on the verge of sinking", Archimedes' principle states that the mass of displaced fluid equals the mass of the floating object.
Which seems to be what PBE6 does - with the assumption that the fluids are well mixed and no volume is lost.
Thus, B = Dfluid * Vobject * g. This is submerged. The volume of the object = volume of the fluid.
But the object is floating. So the upward buoyant force is balanced by the downward weight of the object.
Dobj / Dfluid = Vfluid / Vobj (neglects Patm.)
Let me ask you this: The density of lead is > iron > water. Is the buoyant force on a solid lead object greater than or less than the buoyant force on a solid iron object of the same dimensions?
14 Dec 07
Originally posted by RamnedSince the object has the same volume as the water it displaces, your equation needs one more step:
Archimedes' principle is saying the Buoyant force, B, is equal to the amount of water displaced. You can find the amount of water displaced, through measuring the volume of the object (= volume of fluid displaced), the density of the fluid (1) and then the gravity.
Thus, B = Dfluid * Vobject * g. This is submerged. The volume of the object = ...[text shortened]... ject greater than or less than the buoyant force on a solid iron object of the same dimensions?
Dobj / Dfluid = Vfluid / Vobj = 1
And this only holds when the object has the same density as the fluid, which is what I calculated above.
In answer to your other question, since the lead object and the iron object have the same dimensions, and displace the same amount of water, the buoyant force is the same. However, a greater buoyant force is required to keep the lead object afloat because it's heavier.
14 Dec 07
1 edit
The answer is 532 cm^3. turns out I missed it.
I do not know where to head now. I've got the physics part of it down, it's just the math, or so I think! How much ethanol must be added to get the new total density = to the object's density of 900 kg/m^3 (needs converted.)?
So initially, the object's volume can be found looking at the volume it displaces?
Vwater * Dwater + Vobj (AKA Vwater?) * Dobject = Mass of ...?
We need to figure it out. The answer again is 532 cm^3. Today I got 499 cm^3 - close? I used density of ethanol as .806e3 kg/m^3. Perhaps I am even making a rounding error.
If someone gets it, PM me. But I have tried it, you 3 have tried it, and also my Physics professor tried it. We all got differing results! I'm stumped.