course Phy 122 È[Ë»æ|§‘Æܵ»ŒèúÎà䂨ýassignment #015
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14:10:13 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 --> I used Bernoulli's Equation to solve this problem. The formula reads: (density)(g)(altitude) + 1/2(density)(velocity^2) + Pressure = constant (rho)(g)(h) + 1/2(rho)(v^2) + P = constant In this problem the altitude remained constant, therefore, the part of the equation 1/2(rho)(v^2) can be cancelled. We are then left with: 1/2(rho)(v^2) + P = constant P = -1/2(rho)(v^2) We arrange the equation to solve for pressure change. Since the velocity changes, we can write the change in velocity into our equation. `dP = P2-P1 = -1/2(rho)(v2^2 - v1^2) In the introductory problem set we calculate: `dP = -1/2(1000kg/m^3)[(2.7m/s)^2 - (8.2m/s)^2] = 29975 N/m^2
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14:10:55 ** 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 --> I answered this problem correctly and provided sufficient explanation, but I could have been a little more detailed.
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14:18:33 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 --> I have only completed the preliminary observations for this experiment and emailed them to the professor. However, I would like to attempt to answer the following questions on this experiment. The KEx and KEy values on the simulation changed very rapidly, but it seems that the values were usually very close to the same. So I say that there is no significant difference between the total KE in x direction and y direction.
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14:20:04 ** 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 --> I answered this problem correctly.
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14:21:58 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 --> If I was looking at the correct values for each particle, the red particle seemed to have an average velocity of 4 and the blue averaged about 1 or 2. The blue particle has a slower velocity because it has a mass that is twice that of the red particle. However, if the red particle is twice as fast (which it seems to be on average) the two particles would average around the same KE because KE is a function of both mass and velocity: KE = 1/2mv^2
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14:22:32 ** 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 --> I answered this problem correctly, but I am not certain about the velocities of the two particles that I estimated.
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14:23:12 What do you think is the most likely velocity of the 'red' particle?
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RESPONSE --> Again, it seemed that the most commonly observed velocity for the red particle was 4.
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14:23:21 ** 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 --> I answered this problem correctly.
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14:25:42 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 --> The chance that a particle would be on the left side of the screen is 50% because it is half the screen. The chance that all one hundred particles would be on only the left side is .50^100 = 7.9*10^-31. This is a VERY small probability and therefore, I conclude that I would never observe all 100 particles on only the left side of my screen.
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14:26:51 ** 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 --> I answered this problem correctly, but my explanation may not be quite as detailed.
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14:29:18 What do you think the graphs at the right of the screen might represent?
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RESPONSE --> The two graphs at the right look somewhat similar to bell curves, so I imagine they represent some kind of distribution. They also look consistently almost identical, so whatever values they represent must be positively correlated. Since this experiment deals with kinetic energy, I would say that one represents KE of the particles and the other represents their velocities because I know the two are directly proportional.
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14:31:16 ** 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 --> I guess you could call my answer an ""educated guess."" My answer agrees with this solution, but I did not explain the details of the of the graphs such as, ""Highest and lowest velocities are least likely, midrange tending toward the low end most likely.""
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14:35:47 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 --> First I calculated the flow rate of air into the room: Every 16 minutes the duct fills the whole room's volume with air. 16min * 60sec/min = 960 sec 9.2m*5.0m*4.5m = 207m^3 So 207m^3 is filled every 960 seconds. The rate of flow = 207m^3 / 960sec = 0.216 m^3/sec Now we can find the velocity of the air in m/sec by dividing the rate of air flow by the area of the duct. Rate = 0.216 m^3/sec Area = pi*r^2 = pi(.15cm)^2 = 0.0707m^2 velocity = (0.216m^3/sec) / (0.0707m^2) = 3.05 m/s
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14:35:59 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 --> I answered this problem correctly.
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14:38:34 prin phy and gen phy problem 10.40 gauge pressure to maintain firehose stream altitude 15 m ......!!!!!!!!...................................
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RESPONSE --> For this problem I simply used the equation from the text that reads: change in pressure = density*grav. accel* change in altitude `dP = rho*g*`dh rho = 1000kg/m^3 g = 9.8m/s^2 `dh = 15m `dP = (1000kg/m^3)(9.8m/s^2)(15m) = 147,000 N/m^2
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14:39:59 ** 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 obtained the correct answer, but I don't really understand why the solution gives a more round-about process for solving. I understand how you could use Bernoulli's, but is it okay that I simply used the other equation: `dP = rho*g*`dh?
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14:40:08 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 --> I am not a general physics student
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14:40:11 ** 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 --> ok
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14:40:21 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 am not a general physics student
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14:41:07 ** 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 --> This is similar to chapter 10, problem 40 in my text. I understand how to solve it.
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14:42:02 gen phy which term cancels out of Bernoulli's equation and why?
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RESPONSE --> I am not a general physics student
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14:42:07 ** 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 --> Ok.
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14:42:16 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 --> I am not a university physics student
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14:42:40 ** 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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RESPONSE --> ok
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14:42:52 univ phy What are the meanings of the limits as f approaches 0 and 1?
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RESPONSE --> I am not a university physics student
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14:42:56 ** 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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RESPONSE --> Ok
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