course PHY 202 ???????????assignment #007?????€???????Physics II
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10:44:53 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 --> Ave force is change in momentum divided by timer imterval. The time interval is 2L/v... average force is m*v^2/L if there were several particles M*mv^2/L
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10:44:59 ** 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 --> right
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10:46:54 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 --> energy that goes in equals the energy that comes out and the work done by the system
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10:47:03 ** 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 --> right it is conserved
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10:47:49 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 --> work done divided by energy input.
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10:48:12 ** 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 --> right work done is thermal energy removed added to the amount of work done to get the input
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10:50:16 Query video experiment 8 **** for each water height in the tube, give that height and the length of the air column at the sealed end of the tube and the pressure in the sealed end
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RESPONSE --> p = 1.013 x 10^5 + 1000*9.8*0.2 = 103, 260 Pa p = 1.013 x 10^5 + 1000*9.8*0.44 = 105, 512 Pa p = 1.013 x 10^5 + 1000*9.8*0.68 = 107, 964 Pa p = 1.013 x 10^5 + 1000*9.8*0.2 = 110, 316 Pa LO/L = 1.02 1.05 1.08 1.09
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10:50:22 ** STUDENT SOLUTION: Distance(cm) Height(cm) 0 0 1 20 2 44 3 68 4 92 water at top of tube is at pressure 1 atm = 1.013 x 10^5 Pascals. Bernoulli's Equation applied to the top of the tube and to the inside of the container, between which we have the water column, tells us that the pressures at these heights are p = 1.013 x 10^5 Pa + (rho)(g)(dh), giving us pressures p = 1.013 x 10^5 + (1000)(9.8)(0.2) = 103, 260 Pa when the water column is .2 m high, p = 1.013 x 10^5 + (1000)(9.8)(0.44) = 105, 512 Pa when the water column is .44 m high, p = 1.013 x 10^5 + (1000)(9.8)(0.68) = 107, 964 Pa when the water column is .68 m high and p = 1.013 x 10^5 + (1000)(9.8)(0.92) = 110, 316 Pa when the water column is .92 m high. The pressure in the sealed end is L0 / L atmospheres, where L0 is the original length of this air column. The observed lengths gave me L0 / L results of 1.02 atm for the 20 cm water height, 1.05 atm for the 44 cm water height, 1.08 atm for the 68 cm water height and 1.11 atm for the 1.09 atm height. **
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RESPONSE --> right
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10:50:46 describe your graph of pressure difference vs. air pressure (in multiples of the original pressure) in the sealed end of the tube and give its slope
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RESPONSE --> the graph is increasing at an constant rate
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10:51:05 ** CONTINUED STUDENT SOLUTION: Pressure difference vs. air pressure points were (1.02 atm, 103260 Pa), (1.05 atm, 105512 Pa), (1.08 atm, 107964 Pa) and (1.11 atm, 110316 Pa). The graph is increasing almost at a constant rate. Slope is (110316 Pa - 103260 Pa) / (1.11 atm - 1.02 atm) = 7056 Pa / (.9 atm) = 78000 Pa / atm, approximately. INSTRUCTOR COMMENT: Your pressure differences based on water column heights are only accurate to about 2 significant figures, and your air column length differences are certainly not significant to more than two significant figures, probably only to one significant figure. So your estimate of the slope is probably best expressed at 80,000 Pa / atm. Also note that your results should have been based on a best-fit straight line, not on the first and last data points. **
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RESPONSE --> the slope is 80,000Pa/atm
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10:52:29 explain why the slope of your graph should equal atmospheric pressure if we assume that the original pressure was atmospheric pressure
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RESPONSE --> slope is water pressure / gas pressure which is equal to kPa of presure for atmosphere
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10:52:36 ** The slope of the graph is a ratio of water pressure in kPa to gas pressure in multiples of atmospheric, which should have slope equat to the number of kPa of pressure per atmosphere. **
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RESPONSE --> right
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10:54:21 explain why the motion of the column in the second part of the experiment is strongly influenced by the slope of the tube
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RESPONSE --> the more vertical the tube the more pressure that has to be done to the water.
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10:54:27 ** At the temperature in the gas increases air expands and pressure increases. As observed, for a given temperature rise water only climbs to a certain height in a vertical tube, and if the tube is narrow this allows for negligible expansion. In a horizontal tube pressure wouldn't change and therefore the water column would vacate a significant volume as temperature increases; in a narrow tube this would correspond to the displacement of a very long column of fluid. For an upward-sloped tube the pressure increase corresponds only to the vertical displacement of the water in the tube, so in order to accomplish the same pressure increase the water has to move much further along the tube than in the vertical case. This vacates a small but significant volume, allowing some expansion of the gas. For a given temperature increase there is less pressure increase and more volume increase than in the vertical case. **
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RESPONSE --> right
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10:57:30 query 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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RESPONSE --> 63J - 35J = 28J + 48J = 76J removed from the system
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10:57:38 ** 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. Aong 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 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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RESPONSE --> right
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10:59:41 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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RESPONSE --> reduce = dW decreases dU add = dQ increase dU dU = dQ-dW
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10:59:50 ** 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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RESPONSE --> ok
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11:01:32 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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RESPONSE --> 1/ pressure for volumes so 1/2 height = half area width is final vol - initial vol
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11:02:07 ** Work is the area under the pressure vs. volume curve. If you have half the pressure between two volumes the graph has half the altitude, which leads to half the area. The 'width' of a region is final volume - initial volume. If the direction of the process is such that final volume is less than initial volume (i.e., going 'backwards', in the negative x direction) then with 'width' is negative and the area is negative. **
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RESPONSE --> right if the final volume is less than the initial volume then with width is negative and the area is negative
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11:02:22 query univ phy problem 19.56 (17.40 10th edition) compressed air engine, input pressure 1.6 * 10^6 Pa, output 2.8 * 10^5 Pa, assume adiabatic.
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RESPONSE --> ok
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11:02:25 ** For an adiabatic process in an ideal gas you know that PV = nRT and PV^`gamma is constant. You are given P1 and P2, and you want T2 > 273 K to prevent formation of frost. Assume T2 = P2 V2 / (n R) = 273 K and n R = (P2 V2) / 273 K . Then T1 = P1 V1 / (n R) = P1 V1 * 273 K / (P2 V2) = (P1 / P2) * (V1 / V2) * 273 K. Since PV^`gamma = constant it follows that V1 / V2 = (P2 / P1)^(1/`gamma) = (P1 / P2)^(-1/`gamma). Thus T1 = (P1 / P2) ( P1 / P2)^(-1/`gamma) * 273 K = (P1 / P2)^(1 - 1/`gamma) = (P1 / P2)^(1-1/1.4) * 273 K = (P1 / P2)^.29 * 273 K = 5.6^.29 * 273 K = 443 K, approx. **
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RESPONSE --> ok
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11:02:27 query univ 19.62 (17.46 10th edition) .25 mol oxygen 240 kPa 355 K. Isobaric to double vol, isothermal back, isochoric to original pressure.
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RESPONSE --> ok
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11:02:28 ** .25 mol oxygen at 240 kPa occupies about V = n R T / P = .25 mol * 8.31 J / (mol K) * 355 K / (2.4 * 10^5 N/m^2) = .003 m^3, very approximately. Doubling volume, `dV = 2 * V - V = V = .003 m^2 and P = 2.4 * 10^5 Pa so P `dv = 700 J, very approximately. During isothermal compression we have n = const and T = const so P = n R T / V. Compressing to half the volume, since PV = const, gets us to double the pressure, so max pressure is 2 * 240 kPA = 480 kPa. To get work we integrate P dV. Integral of P dV is calculated from antiderivative n R T ln | V |; integrating between V1 and V2 we have n R T ln | V2 | - n R T ln | V1 | = n R T ln | V2 / V1 |. In this case V2 = V and V1 = 2 V so V2 / V1 = 1/2 and we have `dW = n R T ln(1/2) = .25 mol * 8.31 J/(mol K) * 710 K * (-.7) = -1000 J, approx. So net work is about 700 J - 1000 J = -300 J **
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RESPONSE --> ok
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11:02:33 univ phy describe your graph of P vs. V
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RESPONSE --> ok
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11:02:36 ** The graph proceeds horizontally to the right from original P and V to doubled V, then to the left along a curve that increases at an incr rate as we move to the left (equation P = 2 P0 V0 / V) until we're just above the starting point, then vertically down to the starting pt. **
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RESPONSE --> ok
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11:02:38 univ phy What is the temperature during the isothermal compression?
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RESPONSE --> ok
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11:02:39 ** If vol doubles at const pressure then temp doubles to 710 K, from which isothermal compression commences. So the compression is at 710 K. **
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RESPONSE --> ok
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11:02:41 univ phy What is the max pressure?
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RESPONSE --> ok
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11:02:44 ** It starts the isothermal at the original 240 kPa and its volume is halved at const temp. So the pressure doubles to 480 kPa. **
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RESPONSE --> ok
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