course Phy 202 ¥ ·Î–ˆR”ƒñÁôº¶¡¬Šj¶à_µÎµw áÖðassignment #005
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10:44:15 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 --> 'rho, g, and y will remain constant, there is no change in altitude. The change in pressure will be the negative of the change in (1/2) * ('rho) *( v^2) so that the formula will be: 'dP = (1/2) 'rho (v2^2 - v1^2)
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10:44:29 ** 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 --> okay
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10:48:56 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 --> There was a signifcant difference between the KE in the x direction and KE in the y direction. For KE in the x direction my average 355.88, and for KE in the y direction my average was 384.44. I think that the collisions were completely random and if the simulation was run again and the same experiment was repeated, the results could be completely opposite. The balls start out in a relatively orderly pattern of mainly x-directions, but as more and more collisions occur, disorder increases greatly with balls going in every direction, sometimes one direction more than the other, but it is constantly changing.
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10:50:29 ** 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 --> So, the Kinetic Energies cancel eachother out pretty much.
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10:58:28 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 --> The average velocity of the red ball is around 5 or 6 (units ?). The average velocity of the blue ball is around 1 or 2 (units ?). I think the blue particle has a greater mass than the red ball. When the green balls hit the blue ball, the blue ball barely changes direction or velocity, but when the green balls hit the red ball, the red ball changes direction very easily and rapidly. This makes me think that the blue ball is much heavier than the red ball.
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10:59:04 ** 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 --> Okay
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11:00:44 What do you think is the most likely velocity of the 'red' particle?
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RESPONSE --> The velocity of the red ball is constantly changing, because the collisions are so frequent. The red ball went from 1 to 5 then up to 10 then back down to one. It was never really ever constant, but I would say the average velocity was 6.
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11:00:59 ** 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 --> Okay.
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11:02:58 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 --> Would that ever occur?! The chances of that happening have to be slim to none. You would probably have to watch it for the rest of your life, and even if it did happen, you would probably miss it because you had to go to the bathroom or something!
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11:04:29 ** 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 --> Wow...I knew it would be a looooong time, but I never guessed it would be that long! But it is pretty simple to figure out.
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11:07:41 What do you think the graphs at the right of the screen might represent?
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RESPONSE --> I think the graphs represent the time between collisions. Each line will get longer or shorter depending on the length of time between collisions for the red or blue ball.
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11:11:24 ** 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 --> Oh, I was really off on that answer. The two graphs do seem to coincide with eachother though. So I guess the higher the velocity the higher the energy and vice versa. Would that be right?
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11:28:25 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 --> For General Physics we had to do problems 35, 38, 40, 43, and 46. But I will try to do this problem anyway, since it says gen phy. The formula is A1 * v1 = A2 * v2 = V2/t v1 = (V2 / t) / A1 v1 = (207m^3) / [('pi * r^2) * 960s] v1 = (207m^3) / [('pi * .15^2m) * 960s] v1 = 3.1 m/s
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11:28:34 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 --> Okay
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11:32:45 prin phy and gen phy problem 10.40 gauge pressure to maintain firehose stream altitude 15 m ......!!!!!!!!...................................
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RESPONSE --> I had trouble with this question. I started out with finding the velocity of the water by using the equation v = 'sqrt (2gh) v = 'sqrt (2 * 9.8 m/s^2 * 15m) v = 17 m/s Now we have to figure out how much pressure it will take to force the water out at 17 m/s and make the water go up 15 m. But I don't know how to do that without the radius or diameter of the hose.
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11:54:29 ** 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 --> Bernoulli's Equation: P1+(.5*'rho*v1^2)+pgy1 = P2+(.5 *'rho*v2^2)+pgy2 Change in vertical positions is 15 m. Velocities at the two points is pretty much zero. Exit pressure is atmospheric pressure. We need to know the pressure inside the hose. 'd ('rho * g * h) = 1000 kg/m^3 * 9.8 m/s^2 * 15 m = 14700N/m^2 'dP = - 'd('rho * g * h) = -147,000N/m^2 So atmospheric pressure plus 147,000 N/m^2 will give you the pressure inside the hose. 147,000 N/m^2 + 1.013 * 10^5 N/m^2 = 2.5 * 10^5 N/m^2 I understand now.
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11:56:21 Gen phy: Assuming that the water in the hose is moving much more slowly than the exiting water, so that the water in the hose is essentially moving at 0 velocity, what quantity is constant between the inside of the hose and the top of the stream? what term therefore cancels out of Bernoulli's equation?
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RESPONSE --> The velocity term cancels out in Bernoulli's equation because at the top of the stream the velocity of the water is zero before the water starts to fall back down again, the velocity is also zero inside the hose before the nozel narrows the hose.
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11:57:46 ** Velocity is 0 at top and bottom; pressure at top is atmospheric, and if pressure in the hose was the same the water wouldn't experience any net force and would therefore remain in the hose **
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RESPONSE --> Pressure is not constant because the pressure has to be higher than atmospheric to get to the 15 m height. So, only velocity is constant.
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12:01:47 query gen phy problem 10.43 net force on 240m^2 roof from 35 m/s wind. What is the net force on the roof?
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RESPONSE --> I was not sure how to go about this problem either. I know we are given the area and the speed of the wind. We also know the coefficient of viscocity of air from Table 10-3, which is 0.018 * 10^-3. The formula used is F = 'eta * A * (v/l) I know everything except separation (l), and I am nost sure how to find that. (v/l) is the velocity gradient.
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12:15:58 ** air with density around 1.29 kg/m^3 moves with one velocity above the roof and essentially of 0 velocity below the roof. Thus there is a difference between the two sides of Bernoulli's equation in the quantity 1/2 `rho v^2. At the density of air `rho g h isn't going to amount to anything significant between the inside and outside of the roof. So the difference in pressure is equal and opposite to the change in 1/2 `rho v^2. On one side v = 0, on the other v = 35 m/s, so the difference in .5 rho v^2 from inside to out is `d(.5 rho v^2) = 0.5(1.29kg/m^3)*(35m/s)^2 - 0 = 790 N/m^2. The difference in the altitude term is, as mentioned above, negligible so the difference in pressure from inside to out is `dP = - `d(.5 rho v^2) = -790 N/m^2. } The associated force is 790 N/m^2 * 240 m^2 = 190,000 N, approx. **
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RESPONSE --> Oh, so I was supposed to use Bernoulli's equation for this problem and not the viscosity formula. So, velocity over the roof is 35 m/s and velocity under the roof is 0 m/s. So, the difference in the velocity component of Bernoulli's equation is: 'd [(1/2) * 'rho * v^2] = (1/2) * (1.29 kg/m^3) * (35 m/s)^2 = 790 N/m^2 So, the difference in the pressure from the inside to the outside is: 'dP = -'d [(1/2) * 'rho * v^2] = - 790 N/m^2 Force = 790 N/m^2 * 240 m^2 = 189600N
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12:18:44 gen phy which term cancels out of Bernoulli's equation and why?
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RESPONSE --> The altitude portion of Bernoulli's equation cancels out because it is unknown and it does not need to be known because it is not useful to know with this particular problem.
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12:18:59 06-14-2006 12:18:59 ** because of the small density of air and the small change in y, `rho g y exhibits practically no change. **
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NOTES -------> Okay
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12:19:17 ** because of the small density of air and the small change in y, `rho g y exhibits practically no change. **
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RESPONSE --> Okay
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12:19:32 univ phy problem 14.67: prove that if weight in water if f w then density of gold is 1 / (1-f). Meaning as f -> 0, 1, infinity. Weight of gold in water if 12.9 N in air. What if nearly all lead and 12.9 N in air?
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RESPONSE --> Gen Phy student.
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12:19:35 ** The tension in the rope supporting the crown in water is T = f w. Tension and buoyant force are equal and opposite to the force of gravity so T + dw * vol = w or f * dg * vol + dw * vol = dg * vol. Dividing through by vol we have f * dg + dw = dg, which we solve for dg to obtain dg = dw / (1 - f). Relative density is density as a proportion of density of water, so relative density is 1 / (1-f). For gold relative density is 19.3 so we have 1 / (1-f) = 19.3, which we solve for f to obtain f = 18.3 / 19.3. The weight of the 12.9 N gold crown in water will thus be T = f w = 18.3 / 19.3 * 12.9 N = 12.2 N. STUDENT SOLUTION: After drawing a free body diagram we can see that these equations are true: Sum of Fy =m*ay , T+B-w=0, T=fw, B=(density of water)(Volume of crown)(gravity). Then fw+(density of water)(Volume of crown)(gravity)-w=0. (1-f)w=(density of water)(Volume of crown)(gravity). Use w==(density of crown)(Volume of crown)(gravity). (1-f)(density of crown)(Volume of crown)(gravity) =(density of water)(Volume of crown)(gravity). Thus, (density of crown)/(density of water)=1/(1-f). **
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12:19:44 univ phy What are the meanings of the limits as f approaches 0 and 1?
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RESPONSE --> Gen Phy student
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12:19:46 ** GOOD STUDENT ANSWER: f-> 0 gives (density of crown)/(density of water) = 1 and T=0. If the density of the crown equals the density of the water, the crown just floats, fully submerged, and the tension should be zero. When f-> 1, density of crown >> density of water and T=w. If density of crown >> density of water then B is negligible relative to the weight w of the crown and T should equal w. **
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