course phy122 Hi! assignment #015015. `query 5
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14:49:51 query introset change in pressure from velocity change. Explain how to get the change in fluid pressure given the change in fluid velocity, assuming constant altitude
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RESPONSE --> P = rho gh change in fluid velocity P = F/A If veloctiy changes then pressure in square.
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14:52:08 ** The equation for this situation is Bernoulli's Equation, which as you note is a modified KE+PE equation. Considering ideal conditions with no losses (rho*gy)+(0.5*rho*v^2)+(P) = 0 g= acceleration due to gravity y=altitude rho=density of fluid }v=velocity P= pressure Constant altitude causes the first term to go to 0 and dissapear. (0.5*rho*v^2)+(P) = constant So here is where we are: Since the altitude h is constant, the two quantities .5 rho v^2 and P are the only things that can change. The sum 1/2 `rho v^2 + P must remain constant. Since fluid velocity v changes, it therefore follows that P must change by a quantity equal and opposite to the change in 1/2 `rho v^2. MORE FORMAL SOLUTION: More formally we could write }1/2 `rho v1^2 + P1 = 1/2 `rho v2^2 + P2 and rearrange to see that the change in pressure, P2 - P1, must be equal to the change 1/2 `rho v2^2 - 1/2 `rho v1^2 in .5 rho v^2: P2 - P1 = 1/2 `rho v2^2 - 1/2 `rho v1^2 = 1/2 rho (v2^2 - v1^2). **
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RESPONSE --> P=mv^2/velocity P=mv^2/A*L P directly prop to v^2 Pretty much what I said
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14:53:14 query billiard experiment Do you think that on the average there is a significant difference between the total KE in the x direction and that in the y direction? Support your answer.
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RESPONSE --> No, particles a moving in both x and y direction because of random collisions. There were no biases in bombardment in one direction or another.
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14:53:40 ** In almost every case the average of 30 KE readings in the x and in the y direction differs between the two directions by less than 10% of either KE. This difference is not statistically significant, so we conclude that the total KE is statistically the same in bot directions. **
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RESPONSE --> Great! THat is what I got in the experiment as well!
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14:55:24 What do you think are the average velocities of the 'red' and the 'blue' particles and what do you think it is about the 'blue' particle that makes is so?
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RESPONSE --> Blue is 10X the mass of red. I am not sure what the average velocities were but this would influence them.
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14:56:02 ** Student answer with good analogy: I did not actually measure the velocities. the red were much faster. I would assume that the blue particle has much more mass a high velocity impact from the other particles made very little change in the blue particles velocity. Similar to a bycycle running into a Mack Truck. INSTRUCTOR NOTE: : It turns out that average kinetic energies of red and blue particles are equal, but the greater mass of the blue particle implies that it needs less v to get the same KE (which is .5 mv^2) **
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RESPONSE --> This is good. Blue is heavier, less affected by other particles.
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14:56:18 What do you think is the most likely velocity of the 'red' particle?
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RESPONSE --> I am not sure.
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14:56:27 ** If you watch the velocity display you will see that the red particles seem to average somewhere around 4 or 5 **
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RESPONSE --> OK
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14:57:12 If the simulation had 100 particles, how long do you think you would have to watch the simulation before a screen with all the particles on the left-hand side of the screen would occur?
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RESPONSE --> I don't know if it would ever happen. This would require someting pretty amazing.
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14:57:59 ** STUDENT ANSWER: Considering the random motion at various angles of impact.It would likely be a very rare event. INSTRUCTOR COMMENT This question requires a little fundamental probability but isn't too difficult to understand: If particle position is regarded as random the probability of a particle being on one given side of the screen is 1/2. The probability of 2 particles both being on a given side is 1/2 * 1/2. For 3 particles the probability is 1/2 * 1/2 * 1/2 = 1/8. For 100 particlles the probability is 1 / 2^100, meaning that you would expect to see this phenomenon once in 2^100 screens. If you saw 10 screens per second this would take about 4 * 10^21 years, or just about a trillion times the age of the Earth. In practical terms, then, you just wouldn't expect to see it, ever. **
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RESPONSE --> That is actually funny.
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14:58:46 What do you think the graphs at the right of the screen might represent?
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RESPONSE --> KE along the x axis versus spped and KE on the Y axis versus speed
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14:59:31 ** One graph is a histogram showing the relative occurrences of different velocities. Highest and lowest velocities are least likely, midrange tending toward the low end most likely. Another shows the same thing but for energies rather than velocities. **
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RESPONSE --> One is velocity and one is energy. Got it.
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16:52:03 prin phy and gen phy problem 10.36 15 cm radius duct replentishes air in 9.2 m x 5.0 m x 4.5 m room every 16 minutes; how fast is air flowing in the duct?
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RESPONSE --> Room Volume = r = 15cm v = 9.2 *5*4.5 t = 16 minutes P = mv^2/v ???
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16:54:25 The volume of the room is 9.2 m * 5.0 m * 4.5 m = 210 m^3. This air is replentished every 16 minutes, or at a rate of 210 m^3 / (16 min * 60 sec/min) = 210 m^3 / (960 sec) = .22 m^3 / second. The cross-sectional area of the duct is pi r^2 = pi * (.15 m)^2 = .071 m^2. The speed of the air flow and the velocity of the air flow are related by rate of volume flow = cross-sectional area * speed of flow, so speed of flow = rate of volume flow / cross-sectional area = .22 m^3 / s / (.071 m^2) = 3.1 m/s, approx.
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RESPONSE --> cross section is pi r ^2 = 0.71m^2 v= 210m^3 rate = 210/60 * 16 =0.22m^3 Got it
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17:04:50 prin phy and gen phy problem 10.40 gauge pressure to maintain firehose stream altitude 15 m ......!!!!!!!!...................................
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RESPONSE --> Previous flow problem problem add: Rate of volume flow = area of cross section * speed of flow. Find speed of flow and divide area by each side. So speed of flow = rate of volume divided by area of cross section of duct This prroblem: Does not seem to be enough data. guage pressure? fire hose stream 15m atmospheric pressure plus extra pressure rho g h P = 1.2 kg/m^3 * 9.8 m/s^2 * 15 P = 176.4 newton per meter^2 or pa
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17:06:11 ** We use Bernoulli's equation. Between the water in the hose before it narrows to the nozzle and the 15m altitude there is a vertical change in position of 15 m. Between the water in the hose before it narrows to the nozzle and the 15 m altitude there is a vertical change in position of 15 m. Assuming the water doesn't move all that fast before the nozzle narrows the flow, and noting that the water at the top of the stream has finally stopped moving for an instant before falling back down, we see that we know the two vertical positions and the velocities (both zero, or very nearly so) at the two points. All that is left is to calculate the pressure difference. The pressure of the water after its exit is simply atmospheric pressure, so it is fairly straightforward to calculate the pressure inside the hose using Bernoulli's equation. Assuming negligible velocity inside the hose we have change in rho g h from inside the hose to 15 m height: `d(rho g h) = 1000 kg/m^3 * 9.8 m/s^2 * 15 m = 147,000 N / m^2, approx. Noting that the velocity term .5 `rho v^2 is zero at both points, the change in pressure is `dP = - `d(rho g h) = -147,000 N/m^2. Since the pressure at the 15 m height is atmospheric, the pressure inside the hose must be 147,000 N/m^2 higher than atmospheric. **
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RESPONSE --> i used air not water! Whoops! Use 1000 kg/m^3 and I would be ok.
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Ñ{£„ê~íà›T¡§â±à…ÓŒyý™£ÔÍx×} assignment #016 016. `query 6 Physics II 09-25-2007
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17:10:05 query introset change in pressure from diameter change given original vel and diameter
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RESPONSE -->
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17:10:43 ** The ratio of velocities is the inverse ratio of cross-sectional areas. Cross-sectional area is proportional to square of diameter. So velocity is inversely proportional to cross-sectional area: v2 / v1 = (A1 / A2) = (d1 / d2)^2 so v2 = (d1/d2)^2 * v1. Since h presumably remains constant we have P1 + .5 rho v1^2 = P2 + .5 rho v2^2 so (P2 - P1) = 0.5 *rho (v1^2 - v2^2) . **
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RESPONSE -->
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17:10:54 query video experiment 4 terminal velocity of sphere in fluid. What is the evidence from this experiment that the drag force increases with velocity?
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RESPONSE -->
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