Physics questions.
Pages
- 1
- 2
| Author | Topic: Physics questions. |
|---|---|
|
Shaper
Member # 247
|
written Monday, December 5 2005 12:49
Profile
Homepage
Good explanation ^. Any way I just wrote my Physics final. I killed the written portion of it. But the multiple choice was slightly harder as it always seems to be. I thought I'd see if any of you could answer the bonus question. I think I did it right. Aight: You have a tank which is 65 cm tall which rests on a stool 75 cm high. The tank is open to the atmosphere, not sealed. The tank has sprung a leak at its bottom corner. How far away from the tank will the water hit the ground initially? -------------------- The Knight Between Posts. Posts: 2395 | Registered: Friday, November 2 2001 08:00 |
|
...b10010b...
Member # 869
|
written Monday, December 5 2005 13:53
Profile
Homepage
quote:To see why the velocity must be higher and the pressure lower in part of an artery that's narrowed, consider an artery where a small region in the middle is narrower than the rest. The rate of blood flow must be constant along the entire artery, because any fluid that goes into one end of the artery has to come out the other end. Flow rate through a tube is proportional to both cross-sectional area and velocity. Since the cross-sectional area of the narrowed region is less than that of the surrounding areas, the velocity in that region must be higher in order for the same flow rate to be maintained -- and higher velocity means lower pressure. The reason this translates to higher pressure in the cardiovascular system as a whole is that the narrowed region increases the overall resistance of the artery. So even though the pressure in the narrowed region is lower than the surrounding areas, the average pressure throughout the artery as a whole is higher than it would be if the narrowed region weren't there. [ Monday, December 05, 2005 13:57: Message edited by: Thuryl ] -------------------- The Empire Always Loses: This Time For Sure! Posts: 9973 | Registered: Saturday, March 30 2002 08:00 |
|
Law Bringer
Member # 335
|
written Monday, December 5 2005 14:55
Profile
Homepage
In a simple model, the water can be considered as a continuous stream of projectile particles and the usual kinematic equations hold. Let's suppose we're ignoring friction, the attraction between polar water molecules, and other more complicated forces, leaving nothing but collisions and gravity. Then the path of the water depends entirely on the acceleration of gravity and the velocity of a particle of water as it is forced from the tank by water pressure and/or gravity. I don't know how to calculate the forces from pressure, and this approximation makes a lot of assumptions, and it requires more information about the hole. I think I may be missing something. After rereading your first post, I think I'm overthinking this. You can find the time from ?y = (v_0)(t) + 1/2(g)(t^2) and the horizontal displacement from displacement from ?x = (v_0)(t). The only thing I'm missing is the velocity of the water coming out of the hole, which I still can't calculate from the information given. —Alorael, who suspects he's both forgetting physics and missing easier methods. Posts: 14579 | Registered: Saturday, December 1 2001 08:00 |
|
Warrior
Member # 5886
|
written Monday, December 5 2005 16:14
Profile
Hmmmm, mustagacheed lanyo chejetna, gourdesh nidi! Hot Hot Hot lamb curry over basmati rrrrice! Ok now that that is out of the way.... For the answer to the water spout from a tank problem... Once you have formulated the projectile equation, it is a trivial matter to find the distance in question. So what paramter is missing from the equation? I am assuming that the leak is small enough that it does not drastically change the level of the water in the vessel, and large enough that wierd fluid viscosity effects (turbulence) and air resistance don't effect the stream too much. Also I am assuming that the leak is perfectly horizontal. Treat the water like a projectile...need to know initial velocity horizontally. Is there any horizontal acceleration? No, not once the water has left the vessel. So what should be the value of the initial velocity....Hmmm...The pressure of the water column above the leak is..Well we don't need to take into account atmospheric pressure, as the leak and the top of the water level are both subject to it. So it is just the pressure of the water column. Let rho be the density of water, V be the total volume. The rho*V = the mass of the water. Weight of the water = rho*V*g (g = gravitational acceleration approximate value). The pressure that the water exerts is rho*V*g/A. Now what is the relationship between V and A? If A is the cross sectional area of an arbitrary column then A*h = V, where h is the hieght of the column. So the pressure now becomes rho*g*h. So rho = 1 g/cm^3 for water at stp, g = 9.81m/s^2, and h = 65 cm. rho*g*h = 65 g/cm^2 * 981 cm/s^2 = 63765 g/(cm*s^2). Ok great, now how to translate this into initial velocity is a feat beyond me at the moment. Ok, I cheated and looked up Bernoulli's equation. It works out that the velocity of the water leaving the tank is (2*g*h)^0.5. Neat. So now I want to know how long it takes for the water to hit the ground from 75 cm. Zero initial vertical velocity so: 75 = 0.5*g*t^2. t = (150/g)^0.5 = 0.391 seconds. The water travels (2*g*h)^0.5 * 0.391 cm horizontally before it touches the ground, or about 140 cm..hmmm that seems wrong. Oh well. [ Monday, December 05, 2005 16:51: Message edited by: Aranea Hirsuta ] -------------------- We don't make the white chalky excrement that splats down and ruins your car's paint job. We make it stronger. Posts: 52 | Registered: Friday, June 3 2005 07:00 |
|
Law Bringer
Member # 335
|
written Monday, December 5 2005 17:07
Profile
Homepage
Don't work with non-SI units. Convert to meters and kilograms. I still have trouble with a hole in the corner of the tank. Is that a horizontal hole, a vertical hole, or both? It's a point-like hole, maybe, but it still needs a direction, because water can't leave a tank in a direction blocked by the tank. —Alorael, who might as well set up an ideal tank of water, set it up on an appropriate stool, make a hole exactly large enough for one water molecule, and see how far the rather small stream goes. He should do it in a vacuum, too, so that the stream of molecules doesn't encounter intereference from air. Hm. High pressure vacuum to keep the water in liquid phase. Posts: 14579 | Registered: Saturday, December 1 2001 08:00 |
|
Warrior
Member # 5886
|
written Monday, December 5 2005 17:14
Profile
quote::P Or you could just lower the temperature sufficiently. Oh, but wait, you run into that nasty little phase change type called sublimation at some point. :mad: [ Monday, December 05, 2005 17:26: Message edited by: Aranea Hirsuta ] -------------------- We don't make the white chalky excrement that splats down and ruins your car's paint job. We make it stronger. Posts: 52 | Registered: Friday, June 3 2005 07:00 |
|
Master
Member # 4614
|
written Monday, December 5 2005 18:43
Profile
Homepage
Seems like it would depend on how full the tank is. As the water continued to flow out, the distance with shorten with suit. Anyway, just find the horizontal velocity of the water as it comes out, find the time in the air from the hole to the ground, and multiply the two together. -------------------- -ben4808 Posts: 3360 | Registered: Friday, June 25 2004 07:00 |
|
Law Bringer
Member # 335
|
written Monday, December 5 2005 20:53
Profile
Homepage
Yes, the pressure depends on how much water is pressing on the hole. Everything vaporizes in a vacuum at any temperature, I think. Well, anything not experiencing enough gravity or or other forces to hold it together. I suppose we'll have to posit water molecules that are exactly like microscopic balls and only able to interact through perfectly elastic collisions. No, I really think a high pressure vacuum is the best solution. —Alorael, who can only conclude that it is in the best interests of everyone never to put water in a tank again. Posts: 14579 | Registered: Saturday, December 1 2001 08:00 |
|
La Canaliste
Member # 5563
|
written Tuesday, December 6 2005 02:45
Profile
You have to assume the hole is like a tiny pipe sticking horizontally out of the side of the lower edge of the tank, otherwise the answer is either indeterminate, or directly below. Anyway, it is an energy balance question with a bit of projectiles in it, as Hairy Spider points out. Also, you do need to use a consistent set of units, either cgs (centimeters, grams, seconds) which has silly derived units like ergs (the amount of work done bt a bored student in one hour of lectures) or else mks (meters, kilograms, seconds). Actually, I really can't be bothered with an exposition of my encyclopaedic research into the theory of leaky tanks. You will just have to imagine it, and let it remain unsullied by translation from the Platonic ideal into earthly real form. -------------------- I am a mater of time and how . Deep down, you know you should have voted for Alcritas! Posts: 387 | Registered: Tuesday, March 1 2005 08:00 |
|
Infiltrator
Member # 4256
|
written Tuesday, December 6 2005 06:22
Profile
Here's one you can try on for size if your feeling really perky. Twas on an Estats test last semester. There are two cocentric spheres (centered at the orgin), with radii, a and b respectively, with surface charge distributions (sigma)_1, and (sigma)_2 respectively. The inner sphere is rotating with an angular velocity of (omega)_1, with its axis of rotation along the z-axis. The outer sphere has an axis of rotation at an angle (phi)from the z-axis, and an angular velocity of (omega)_2. (In the original problem the cavity was filled with fluid of a certain viscocity, and a certain permitivity and yady yady ya, which is what reminded me of the problem) Find the Capacitance, the E,B fields, (D, and H as well with the fluid included) and any other relevant information.(I think these included the bound charge/free charge, and sich, but I don't believe that I remember the info that was given to allow you to find it.) There were extra points for anyone that included the relativistic effects of rotational motion, but I don't think a single person in the class got them. Stupid Tensors. You have to love take home finals, because it allows the teachers to go to town with the problems. [ Tuesday, December 06, 2005 06:28: Message edited by: AxB ] -------------------- "Let's just say that if complete and utter chaos was lightning, he'd be the sort to stand on a hilltop in a thunderstorm wearing wet copper armour and shouting 'All gods are false'." Posts: 564 | Registered: Wednesday, April 14 2004 07:00 |
|
Shaper
Member # 247
|
written Tuesday, December 6 2005 22:41
Profile
Homepage
P1+ DgV1^2/2 + DgY1= P2+ DgV2^2/2 + DgY2 V1 is 0 at the tank's top. P1 and P2 = Patm So Dgy1 = DgV2+ Dgy2 V2=Dgy1-Dgy2/Dg YF= Yi+Viy (x)T - g(t^2)/2 T=? XF= Vix (x) T -------------------- The Knight Between Posts. Posts: 2395 | Registered: Friday, November 2 2001 08:00 |
|
The Establishment
Member # 6
|
written Wednesday, December 7 2005 00:33
Profile
The simplest analysis would be to apply the conservation of energy and assume adiabatic reversible conditions (i.e. Bernoulli's equation) for the tank problem. Of course, this is incorrect as the flow would be turbulent and Bernoulli's is sort of only valid for laminar flows. Ah, the worderful word of form loss coefficients and empirical formulas... :) -------------------- Your flower power is no match for my glower power! Posts: 3726 | Registered: Tuesday, September 18 2001 07:00 |
|
Electric Sheep One
Member # 3431
|
written Wednesday, December 7 2005 00:50
Profile
Actually the streams I have seen pouring out of punctured water containers have always looked pretty laminar. A Reynolds number thing, evidently. -------------------- We're not doing cool. We're doing pretty. Posts: 3335 | Registered: Thursday, September 4 2003 07:00 |
|
Shaper
Member # 247
|
written Wednesday, December 7 2005 01:12
Profile
Homepage
It is a laminar flow. -------------------- The Knight Between Posts. Posts: 2395 | Registered: Friday, November 2 2001 08:00 |
|
Apprentice
Member # 6009
|
written Wednesday, December 7 2005 02:06
Profile
Only laminar if the hole is very small... very very small. Posts: 18 | Registered: Friday, June 24 2005 07:00 |
|
Electric Sheep One
Member # 3431
|
written Wednesday, December 7 2005 02:18
Profile
Relative to what? I repeat, it's a Reynolds number thing. -------------------- We're not doing cool. We're doing pretty. Posts: 3335 | Registered: Thursday, September 4 2003 07:00 |
|
The Establishment
Member # 6
|
written Wednesday, December 7 2005 19:15
Profile
Re = density*flow velocity*flow diameter/viscosity If Re > 2300ish it is turbulent, below that the flow becomes more and more laminar. Although the transistion region is still a very important theoretical problem (as is the turbulent region, but we know more about that). Given some values of water at 20 C and a modest hole size of 1 cm and a flow velocity of 1 m/s: density = 1 000 kg/m^3 diamter = 0.01 m velocity = 1 m/s viscosity = 0.001 N-s/m^2 Re = 10 000 or well into the turbulent region. It is possible for flow to be laminar, but you would need a pretty small hole and slow velocities. -------------------- Your flower power is no match for my glower power! Posts: 3726 | Registered: Tuesday, September 18 2001 07:00 |
|
Shaper
Member # 247
|
written Thursday, December 8 2005 18:48
Profile
Homepage
Just got my mark for the Physics final. Oh yea 86% highest, in the class. Oh yea I had 70% combined before the Exam. Oh yea, my final mark is higher than all areas combined so I get that mark. Bo! :) -------------------- The Knight Between Posts. Posts: 2395 | Registered: Friday, November 2 2001 08:00 |
|
The Establishment
Member # 6
|
written Thursday, December 8 2005 21:30
Profile
Goodie goodie! -------------------- Your flower power is no match for my glower power! Posts: 3726 | Registered: Tuesday, September 18 2001 07:00 |
Pages
- 1
- 2