#$&* course Phy 202 Friday, July 6, 2010, between 2:10 AM and 2:15 AMPlease note that I was unable to answer some of the questions related to the kinmodel.exe program because it does not work on Windows Vista (or at least my copy of Windows Vista). If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: Bernoulli's Equation can be written * 1/2 rho v1^2 + rho g y1 + P1 = 1/2 rho v2^2 + rho g y2 + P2 If altitude is constant then y1 = y2, so that rho g y1 is the same as rho g y2. Subtracting this quantity from both sides, these terms disappear, leaving us * 1/2 rho v1^2 + P1 = 1/2 rho v2^2 + P2. The difference in pressure is P2 - P1. If we subtract P1 from both sides, as well as 1/2 rho v2^2, we get * 1/2 rho v1^2 - 1/2 rho v2^2 = P2 - P1. Thus * change in pressure = P2 - P1 = 1/2 rho ( v1^2 - v2^2 ). Caution: (v1^2 - v2^2) is not the same as (v1 - v2)^2. Convince yourself of that by just picking two unequal and nonzero numbers for v1 and v2, and evaluating both sides. ALTERNATIVE FORMULATION Assuming constant rho, Bernoulli's Equation can be written 1/2 rho `d(v^2) + rho g `dy + `dP = 0. If altitude is constant, then `dy = 0 so that 1/2 rho `d(v^2) + `dP = 0 so that `dP = - 1/2 rho `d(v^2). Caution: `d(v^2) means change in v^2, not the square of the change in v. So `d(v^2) = v2^2 - v1^2, not (v2 - v1)^2. STUDENT SOLUTION: 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 disappear. (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). ** Your Self-Critique: OK Your Self-Critique Rating: OK ********************************************* Question: 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: ???? Are we asking about the KinModel program's objects or the ""real-world"" situation that they represent? If the latter, are the x and y directions both horizontal (""bird's-eye view"") or vertical? My guess is that they'd both be horizontal because we don't see any projectile motion (even slight), but I don't know. ???? - SOLUTION: Assuming that we're asking about the model alone, there appear to be stretches of time over which KE in the x direction dominates and stretches of time over which KE in the y direction dominates. (In other words, if one samples at fairly frequent intervals, there will be large groups of consecutive readings where the x-value is greater and large groups where the y-value is greater.) These stretches appear to be approximately equal in length, such that everything would balance out over time. In other words, when one graphs absval(Value:X - Value:Y) over time and calculates its mean value when measured at consistent intervals, the difference will be present and significant, but when one graphs (Value:X - ValueY) or (Value:Y - Value:X) over time, although the value will be highly volatile, its mean will be near zero.
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Given Solution: ** 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. ???? Do you mean that the sum of the x-values over 30 readings and the sum of the y-values over 30 readings are less than 10% different from each other, or do you mean that over 30 readings the average of the respective readings' differences between x- and y-values will be less than 10%? ???? This difference is not statistically significant, so we conclude that the total KE is statistically the same in both directions. **
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Given Solution: ** Student answer with good analogy: I did not actually measure the velocities. the red were much faster. ???? Using the Windows version [billiard_simulation_0006b.exe] and your default settings, the green ones are the fast, low-mass ones; the red ones are the slow, high-mass ones. ????? 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) ** Your Self-Critique: 3 to the extent that I was able to find something that matches your description; indeterminate otherwise ???? When trying to remember what if anything I'd learned about Maxwell-Boltzmann distributions in previous courses, the visual similarity with Graham's Law reminded me of it, and upon looking it up, I realized that my above guess is basically a (mis)application of Graham's Law having to do with velocities in a mixture rather than effusion rates. ????
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Given Solution: ** If you watch the velocity display you will see that the red particles seem to average somewhere around 4 or 5 ** Your Self-Critique: Indeterminate: ???? The billiard_simulation_0006b.exe program does not have a velocity display. ???? Your Self-Critique Rating: Indeterminate ********************************************* Question: 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: ???? Again, I'm answering with respect to the billiard_simulation_0006b.exe program and its default settings. ???? - SOLUTION: There's a slight chance that I would see such a screen before the Sun swells up into a red giant and burns the computer to a crisp. In other words, pretty [expletive] long. If the left-hand side of the screen is about 15 ""billiards"" wide and the screen is square, then the left-hand side of the screen is big enough for only 2(15*15) = 450 billiards anyway, so it will be incredibly rare that all of them make it over there without bumping into each other and sending at least one back to the right side of the screen. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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. ????? Doesn't this estimate assume that each 50/50 probability re: a given particle is independent of all the others, i.e., that the particles neither gravitationally attract each other (negligible here) nor get in each other's way (nonnegligible here). Because the billiards can get in each other's way (they make contact at their edges, not their centers), the presence of a billiard on the left side of the screen reduces by a nonzero factor the chances of the next billiard making it over there. Therefore, I think that the odds would be even less than this. ????? In practical terms, then, you just wouldn't expect to see it, ever. ** Your Self-Critique: OK/3 (it didn't occur to me to do a formal probabilistic analysis, but I did go beyond the given solution [I think] by considering an additional factor that's important at least in this model, so we'll call it a wash or close.) Your Self-Critique Rating: OK
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Given Solution: ** 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. ** Your Self-Critique: N/A Your Self-Critique Rating: N/A
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Given Solution: 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. Your Self-Critique: OK Your Self-Critique Rating: OK ********************************************* Question: prin phy and gen phy problem 10.40 What gauge pressure is necessary to maintain a firehose stream at an altitude of 15 m? `....................................... YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: - SOLUTION: Pressure differential = +147,000 Pa, i.e., approx. 1.47 ""bars"" or just over 1.45 ""standard atmospheres"" in excess of atmospheric pressure - - Gauge pressure = difference in pressure between ""cis"" (hose) and ""trans"" (air) sides of nozzle = pressure in excess of atmospheric pressure - - Pressure = fluid density * gravity * fluid column height = 1000 (kg [m^-3]) * 9.8 (m [s^-2]) * 15 m = 147,000 Pa confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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. ** Your Self-Critique: OK/3 (I used only Pascal's simpler equation re: density * gravity * height, in part because the absence of other data didn't prompt me to think of Bernoulli's full equation, but in context it was good enough, and in any event this simplified version of the equation is the one that I used in the lab experiments re: raising water and measuring atmospheric pressure.) Your Self-Critique Rating: OK
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Given Solution: ** 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 ** Your Self-Critique: OK Your Self-Critique Rating: OK ********************************************* Question: 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: - SOLUTION: If ""net force"" in this problem = net effect of wind alone, then net force = 191,100 N; if ""net force"" means force from wind - force due to atmospheric pressure, then treating upward direction as positive (such that atmospheric pressure = -101,325 Pa) and assuming height of flowing air mass to be negligible: - - Assuming net pressure to involve offsetting of atmospheric pressure, net pressure = (.5 * air density * velocity^2) + [-](atmospheric pressure) = .5 * (1.3 kg [m^-3]) * ([35 m/s]^2) - 101,325 Pa = 796.25 Pa - ' ' ' = -100,528.75 Pa - - - Therefore, force = pressure * area = (796.25 Pa * 240 [m^2]) - (100,528.75 Pa * 240 [m^2]) = 191,100 N - 24,126,900 N = -23,935,800 N confidence rating #$&*:ntellectually, 2.5 psychologically (I'm a bit worried by how to interpret the term ""net force."") ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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. ** Your Self-Critique: OK Your Self-Critique Rating: OK ********************************************* Question: gen phy which term 'cancels out' of Bernoulli's equation and why? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: There may actually be two terms that ""cancel out"" of the equation: In any event, because the airstream is of negligible thickness, the ""density * gravity * height"" term cancels out; moreover, assuming from the previous Given Solution that net force does *not* include accounting for ambient vertical (atmospheric) pressure, there are actually two terms that ""cancel out"" of Bernoulli's equation here: Because ambient vertical pressure (here, atmospheric pressure) is the same before and after the wind blows, ambient vertical pressure (generally notated as the first term with symbol ""P-sub-0"" [Pinitial] or ""P-sub-1"" [Pfinal]) cancels out of the equation. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** because of the small density of air and the small change in y, `rho g y exhibits practically no change. ** Your Self-Critique: OK Your Self-Critique Rating: OK ********************************************* Question: univ phy problem 12.77 / 14.75 (11th edition 14.67): prove that if weight in water if f w then density of gold is 1 / (1-f). Meaning as f -> 0, 1, infinity. Weight of gold in water if 12.9 N in air. What if nearly all lead and 12.9 N in air? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: [Omitted as marked for University Physics students only] confidence rating #$&*:A] ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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). ** Your Self-Critique: [N/A] Your Self-Critique Rating: [N/A] ********************************************* Question: univ phy What are the meanings of the limits as f approaches 0 and 1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: [Omitted as marked for University Physics students only] confidence rating #$&*:A] ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** GOOD STUDENT ANSWER: f-> 0 gives (density of crown)/(density of water) = 1 and T=0. If the density of the crown equals the density of the water, the crown just floats, fully submerged, and the tension should be zero. When f-> 1, density of crown >> density of water and T=w. If density of crown >> density of water then B is negligible relative to the weight w of the crown and T should equal w. **" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: univ phy What are the meanings of the limits as f approaches 0 and 1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: [Omitted as marked for University Physics students only] confidence rating #$&*:A] ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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. **" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!