course Phy 122 W旀OB}Stassignment #006
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20:29:57 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 --> Using this constant formula: .5 `rho v^2 + P = 0 d`(.5 `rho v^2) = d`(-P) d`(P) = `d(-1)(.5 `rho v^2)
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20:30:07 ** 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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20:32:05 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 --> On average there is not a significant different between the KE x & y because the average in my trials were so close. Since it is a random simulation anything can happen. The difference would be a very small percentage.
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20:32:22 ** 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 --> okay
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20:41:13 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 KE of the red and blue particles appear to be the same. However, the particles are not themselves identical. One of the particles is probably greater in their mass.
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20:45:52 ** 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 The blue particle has greater mass so the velocity of the blue particle is less to create an equivalent KE = .5mv^2.
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20:46:20 What do you think is the most likely velocity of the 'red' particle?
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RESPONSE --> About 4.
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20:46:27 ** 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 --> okay
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20:47:55 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 --> (1/2)^100 = 7.8 E -31
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20:48:22 ** 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 --> alright Small chance.
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20:49:11 What do you think the graphs at the right of the screen might represent?
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RESPONSE --> They show how the velocities and kinetic energy changes over time.
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20:49:18 ** 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 --> okay
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21:36:10 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 --> Volume = 9.2 m * 5.0 m * 4.5 m = 207 m^3 rate of replinishment = 207 m^3 / (16 min * 60 s/min) = .215 m^3 / s duct = area of circle = pi r^2 = pi ( .15 m)^2 = .071 m^2 rate of volume flow = area * speed speed = rate of volume flow / area speed = .22 m^3 / s / (.071 m^2) = 3.1 m/s
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21:39:12 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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21:43:04 prin phy and gen phy problem 10.40 gauge pressure to maintain firehose stream altitude 15 m ......!!!!!!!!...................................
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RESPONSE --> altitude = 15 m `d(rho * g * h ) = 1000 kg/m^3 * 9.8 m/s^2 * 15 m `d(rho * g * h) = 147,000 N/m^2 `dP = -`d(rho * g * h) = -147,000 N/m^2 P is 147,000 N/m^2 higher than atmospheric P.
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21:44:03 ** 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 --> okay
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21:44:06 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 -->
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21:44:09 ** 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 -->
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ƟΑۿǚđ냝 assignment #008 008. `query 7 Physics II 09-23-2008
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22:26:44 Query set 5 problems 16-20 explain how we calculate the average force exerted by a particle of mass m moving at speed v which collides at a time interval `dt elastically and at a right angle with a solid wall
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RESPONSE --> The particle's change in momentum is equivalent to the impulse given on that particle. force = `d(momentum) / time interval force = (-2 m v ) / (2*L / v) force = m v^2 / L For x number of particles the force is x * m*v^2 / L
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22:28:08 ** The impulse exerted on a particle in a collision is the change in the momentum of that particle during a collision. The impulse-momentum theorem says that the change in momentum in a collision is equal to the impulse, the average force * the time interval between collisions. The average force is thus change in momentum / time interval; the time interval is the round-trip distance divided by the velocity, or 2L / v so the average force is -2 m v / ( 2L / v) = m v^2 / L If there were N such particles the total average force would be N * m v^2 / L If the directions are random we distribute the force equally over the 3 dimensions of space and for one direction we get get 1/3 the force found above, or 1/3 N * m v^2 / L. This 3-way distribution of force is related to the fact that for the average velocity vector we have v^2 = vx^2 + vy^2 + vz^2, where v is average magnitude of velocity and vx, vy and vz the x, y and z components of the velocity (more specifically the rms averages--the square root of the average of the squared components). **
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RESPONSE --> Force is distributed over 3-D space, therefore one direction is 1/3 x * m*v^2 / L. The sum of the average velocity vectors is related to 3-way distribution of force.
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22:30:08 Summarize the relationship between the thermal energy that goes into the system during a cycle, the work done by the system during a cycle, and the thermal energy removed or dissipated during the cycle.
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RESPONSE --> In an isolated system, thermal energy put into the system has to equal the work done by the system and thermal energy removed from it, thus work and energy is conserved.
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22:32:37 ** Work-energy is conserved within an isolated system. So the thermal energy that goes into the system must equal the total of the work done by the system and the thermal energy removed from the system. What goes in must come out, either in the form of work or thermal energy. **
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RESPONSE --> ok
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22:42:43 If you know the work done by a thermodynamic system during a cycle and the thermal energy removed or dissipated during the cycle, how would you calculate the efficiency of the cycle?
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RESPONSE --> cycle efficiency = work done / input of energy input of energy = thermal energy removed + work done
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22:43:09 ** STUDENT SOLUTION: Efficiency is work done / energy input. Add the amount thermal energy removed to the amount of work done to get the input. Then, divide work by the energy input. **
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RESPONSE --> work / energy input = -thermal energy removed
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22:58:29 prin phy and gen phy problem 15.2, cylinder with light frictionless piston atm pressure, 1400 kcal added, volume increases slowly from 12.0 m^3 to 18.2 m^3. Find work and chagne in internal energy.
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RESPONSE --> `dW = P * `dV `dW = 1 atm * (18.2 m^3 - 12 m^3) `dW = 6.2 m^2 * atm 1 atm = 101.3 * 10^3 N/m^2 `dW = 6.2 m^2 * 101.3 * 10^3 N/m^2 `dW = 6.3 * 10^5 J 1400kcal * 4200J = 5.9 * 10^6 J `dU = `dQ - `dW `dU = 5.9 * 10^6 J - 6.3 * 10^5 J `dU = 5.3 * 10^6 J
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22:58:41 Work done at constant pressure is P `dV, so the work done in this situation is `dW = P `dV = 1 atm * (18.2 m^3 - 12 m^3) = (101.3 * 10^3 N/m^2) * (6.2 m^3) = 630 * 10^3 N * m = 6.3 * 10^5 J. A total of 1400 kcal = 1400 * 4200 J = 5.9 * 10^6 J of thermal energy is added to the system, the change in internal energy is `dU = `dQ - `dW = 5.9*10^6 J - 6.3 * 10^5 J = 5.9 * 10^6 J - .63 * 10^6 J = 5.3 * 10^6 J. It is worth thinking about the P vs. V graph of this process. The pressure P remains constant at 101.3 * 10^3 J as the volume changes from 12 m^3 to 18.2 m^3, so the graph will be a straight line segment from the point (12 m^3, 101.3 * 10^3 J) to the point (18.2 m^3, 101.3 * 10^3 J). This line segment is horizontaland the region above the horizontal axis and beneath the segment is a rectangle whose width is 6.2 * 10^3 m^3 and whose altitude is 101.3 * 10^3 N/m^2; its area is therefore the product of its altitude and width, which is 6.3 * 10^5 N m, or 6.3 * 10^5 J, the same as the word we calculated above.
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RESPONSE --> okay
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23:05:14 prin phy and gen phy problem 15.5, 1.0 L at 4.5 atm isothermally expanded until pressure is 1 atm then compressed at const pressure to init volume, final heated to return to original volume. Sketch and label graph.
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RESPONSE --> P V = n R T with constant P and V P = 4.5 atm V = 1 L P * V = 4.5 atm * 1 L = 4.5 atm * L 4.5 atm * L = n R T 1 atm = 101.3 kPa = 101.3 * 10^3 Pa = 101.3 * 10^3 N/m^2 4.5 atm = 4.5 * 101.3 * 10^3 Pa = 4.6 * 10^3 Pa = 4.6 * 10^3 N/m^2
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23:05:22 When a confined ideal gas is expanded isothermally its pressure and volume change, while the number of moled and the temperature remain constant. Since PV = n R T, it follows that P V remains constant. In the initial state P = 4.5 atm and V = 1 liter, so P V = 4.5 atm * 1 liter = 4.5 atm * liter (this could be expressed in standard units since 1 atm = 101.3 kPa = 101.3 * 10^3 N/m^2 and 1 liters = .001 m^3, but it's more convenient to first sketch and label the graph in units of atm and liters). During the isothermal expansion, therefore, since P V remains constant we have P V = 4.5 atm liters. At a pressure of 1 atm, therefore, the volume will be V = 4.5 atm liter / P = 4.5 atm liter / (1 atm) = 4.5 liters. The graph follows a curved path from (1 liter, 4.5 atm) to (4.5 liters, 1 atm). At the gas is compressed at constant pressure back to its initial 1 liter volume, the pressure remains constant so the graph follows a horizontal line from (4.5liters, 1 atm) to (1 liter, 1 atm). Note that this compression is accomplished by cooling the gas, or allowing it to cool. Finally the gas is heated at constant volume until its pressure returns to 4.5 atm. The constant volume dictates that the graph follow a vertical line from (1 liter, 1 atm) back to (4.5 liters, 1 atm). The graph could easily be relabeled to usestandard metric units. 1 atm = 101.3 kPa = 101.3 * 10^3 Pa = 101.3 * 10^3 N/m^2, so 4.5 atm = 4.5 * 101.3 * 10^3 Pa = 4.6 * 10^3 Pa = 4.6 * 10^3 N/m^2. 1 liter = .001 m^3 so 4.5 liters = 4.5 m^3. Since P V = 4.5 atm liters, P = 4.5 atm liters / V. This is of the form P = c / V, with c a constant. For positive values of V, this curve descendsfrom a vertical asymptote with the vertical axis (the V axis) through the point (1, c) then approaches a horizontal asymptote with the horizontal axis. For c = 4.5 atm liters, the curve therefore passes through the point (1 liter, 4.5 atm). As we have seen it also passes through (4.5 liters, 1 atm).
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23:05:26 gen phy problem 15.12, a-c curved path `dW = -35 J, `dQ = -63 J; a-b-c `dW = - 48 J gen phy how much thermal energy goes into the system along path a-b-c and why?
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23:05:28 ** I'll need to look at the graph in the text to give a reliably correct answer to this question. However the gist of the argument goes something like this: `dQ is the energy transferred to the system, `dW the work done by the system along the path. Along the curved path the system does -35 J of work and -63 J of thermal energy is added--meaning that 35 J of work are done on the system and the system loses 63 J of thermal energy. If a system gains 35 J of energy by having work done on it while losing 63 J of thermal energy, its internal energy goes down by 28 J (losing thermal energy take internal energy from the system, doing work would take energy from the system so doing negative work adds energy to the system). So between a and c along the curved path the system loses 28 J of internal energy. In terms of the equation, `dU = `dQ - `dW = -63 J -(-35 J) = -28 J. It follows that at point c, the internal energy of the system is 28 J less than at point a, and this will be the case no matter what path is followed from a to c. Along the path a-b-c we have -48 J of work done by the system, which means that the system tends to gain 48 J in the process, while as just observed the internal energy goes down by 28 Joules. The system therefore have `dQ = `dU + `dW = -28 J + (-48 J) = -76 J, and 76 J of internal energy must be removed from the system.**
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23:05:32 gen phy How are the work done by the system, the thermal energy added to the system and the change in the internal energy of the system related, and what is this relationship have to do with conservation of energy?
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23:05:36 ** If a system does work it tends to reduce internal energy, so `dW tends to decrease `dU. If thermal energy is added to the system `dQ tends to increase `dU. This leads to the conclusion that `dU = `dQ - `dW. Thus for example if `dW = -48 J and `dU = -28 J, `dQ = `dU + `dW = -28 J + -48 J = -76 J. **
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23:05:38 gen phy How does the halving of pressure caused a halving of the magnitude of the work, and why is the work positive instead of negative as it was in the process a-b-c?
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