course PHY 202 ͤ怅kЮꇀassignment #015
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19:46:53 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 --> The answer will be based on the formula, for the change in KE which is equal to: W=>(.5)mvf^2-(.5)mvo^2 To get the unknown variables of mass and work we have to work backwards, starting with Force=AP, where A is area, of P=Pressure, solving for Force. Once you get Force, use the formula, W=F(D), where W=work, F=Force, D=Distance. Once have work, you only need one more unknown variable which is mass. Mass can be obtained by rho=m/V, where rho=density of substance, m=mass(our unknown), V=volume. Then plug into change in KE formula to arrive at velocity change.
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19:50:38 ** 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 --> Bernouli's equation is simpler, and it is based on the same theory that I used. Depending on the given variables in the problem, this is a more direct and easier way to solve the problem. P is like Work done .5rho(v^2) is like KE rho(g)(y) is like PE
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19:53:06 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. I observed it 30 different times and averaged the results. The average of KE in the x direction was 357.21 and in the y direction it was 376.78. I imagine that if took subsequent trials my answers would be very similar.
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19:53:24 ** 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 --> ok
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19:55:04 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 --> I think the average velocity of the red particle would be faster than the average velocity of the blue particle. Because the blue particle is 10 times the mass of the red atom, I beleive it would be slower on average than the red particle.
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19:57:31 ** 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 was going in the right direction but this answer was based on F=MA, which is good way to explain it further. It will require more force to move a particle with higher mass, as the analogy explains.
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20:01:25 What do you think is the most likely velocity of the 'red' particle?
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RESPONSE --> I've watched the simulation a few times, but I'm not sure what to base the velocity of the red particle on. I'm sure you can calculate it with the help of the KE values, but I'm not sure where to start.
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20:03:17 ** 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 --> Use the two numbers given at the top of the screen. The velocity that changes more often is obviously the red particle, because we established that it's moving at a faster rate. The other velocity number refers to the blue particles.
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20:07:37 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 think that would ever happen, because I think that would require more change of KE in the x direction vs the y. Since the particle movement in the x direction and the y direction are so similiar, I think that woudn't happen.
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20:10:52 ** 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 --> My answer was correct, but my reasoning could have been clearer. The explanation here is pretty easy to understand.
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20:12:47 What do you think the graphs at the right of the screen might represent?
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RESPONSE --> The graphs on the right of the screen represent probablitity distributions. One shows relative probabilities of velocities, the other the relative probablities of different kinetic energies.
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20:12:56 ** 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 --> ok
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20:52:46 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 --> The answer is based on a form of the equation of continuity or v2=V2/t(A2)= where V2= rate of flow in the air duct, V2=Volume of air in the room, t=time in seconds of air in room, and A2=area of room. Thus: v2=207/960(pi)(r^2)=207/67.82=3.05 m/s
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20:53:04 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 --> ok
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21:18:01 prin phy and gen phy problem 10.40 gauge pressure to maintain firehose stream altitude 15 m ......!!!!!!!!...................................
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RESPONSE --> Start with Bernoulli's equation of P1+.05rhov^2+rho(g)y=P2 +.05rhov^2+rho(g)(y) Where P1=Pressure at point 1 .05rhov^2=velocity squared(density) at points 1 and 2 respectively rho(g)y=density(gravity)(height) at pts 1 and 2 respectively. In this equation, I would assume that both of the velocity cross out because the water in the firehouse is moving at a constant velocity. Rho(g)(y) for the first point will cross out because we are starting at an arbitrary height of zero. This leaves, P1=P2(rho)(g)(y). I think that P2 is atmospheric pressure, thus 1.013E5(1.00E3)(9.8)(15)=1.5E5.
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21:20:33 ** 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 --> Our reasoning is the same, the answer is a little different, but this may be due to rounding.
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21:22:49 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 is constant between the inside of the hose and the top of the stream, so .05(rho)(velocity^2) will cancel out of the equation.
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21:25:28 ** 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 --> oh...that' s very true. Pressure does supply the net force or does the work required to move the water out of the hose.
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21:35:41 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 would use the equation of P=F/A, where P=Atmospheric pressure, F=Force, A= Area, so 1.013E5/240=422.08. But I see that the velocity is 35 m/s. I'm not quite sure where I would need the velocity in this problem, however, I am sure it is used.
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21:38:29 ** 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 --> Ok..I was supposed to use bernoulli's equation again. I have to remember that when different values are given at different points, I must use bernoulli's equation to account for each value at each point in the problem. I was right to use F=PA, but only after I determined the correct Pressure by using Bernoulli's equation.
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21:39:28 gen phy which term cancels out of Bernoulli's equation and why?
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RESPONSE --> In the previous problem, the term that cancelled out what rho(g)(y) because we assumed that there wasn't going to be a significant height difference from one part of the roof to the other.
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21:39:37 ** 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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KZ}ݧʳR`~h assignment #015 015. `query 5 Physics II 07-09-2007"