course Phy 202
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11:00:40 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 --> GIVEN: change in fluid velocity (rho) constant altitude Bernoulli's Law... .5*'rho*v^2+'rho*g*h+P =0 The g here is the acceleration of gravity which when you multiply that by y you get the acceleration of gravity from that specific Altitude. The velocity is v^2. When you solve for this you get... (""rho*g*h+P) / (.5*rho)= v^2
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11:02:23 ** 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 --> .5 * rho* v1^2 + P = .5*rho*v2^2 +P2 Well I was getting close. We known the change which gives us two quantities. P2-P1 = .5*rho*v2^2-.5*rho*v1^2 =.5*rho(v2^2-v1^2) Okay.
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11:03:19 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 --> Well sometimes the y was bigger than x. This occurred less frequently but it still occurred. This leads me to believe that the difference in KE is not significant.
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11:03:34 ** 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 --> Less than 10%. That definitely proves the point.
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11:05:01 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 --> When the blue particle hit into other particles it really affected them more than the red. It is probably significantly bigger than the red, but perhaps not moving much faster.
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11:05:19 ** 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 --> The are equal. Okay. Less v to get same KE.
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11:05:47 What do you think is the most likely velocity of the 'red' particle?
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RESPONSE --> Approximately 4?
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11:05:58 ** 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 --> 4 or 5 okay.
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11:06:47 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 do not believe this would occur often if ever.
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11:07:17 ** 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 --> For each particle added the probability gets less and less.
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11:09:05 What do you think the graphs at the right of the screen might represent?
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RESPONSE --> I am fairly certain one graph represents velocity. I think this because it has a bulk of occurence and that is probably accounted for by the close average velocities.
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11:09:24 ** 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 --> The second graph is energies. I thought that is my initial KINMODEL obesevation.
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11:11:57 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 --> KNOWN: 15 cm radius time 16 minutes volume 207 m^3 (9.2 * 5 * 4.5) 16 min = 960 seconds Velocity is 207 m^3 / 960 seconds = .216 m^3/s
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11:13:05 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 --> speed of flow = rate of volume flow/ cross sectional area I only found the rate of volume flow which is not the speed of flow. pi*r^2 = pi * .15 m= .71 m^2 .22 m^3 /s / .071 m^2 = 3.1 m/s Okay.
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11:14:25 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 1000 kg/m^3 * 9.8 m/s^2 * 15 m = 147,000 N/m^2
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11:15:36 ** 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 --> Gravity is the acceleration and water is 1000 kg/m^3. Zero at both points would give a change in pressure of - 147,000 N/m^2
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11:16:25 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 --> Well the velocity of the water is 0 at both ends. To allow water flow there must be pressure.
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11:16:49 ** 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 --> If pressure in the hose was same as the outside the water would not move. Net force remains the same. Okay.
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11:19:21 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 --> v1= 0 m/s v2= 35 m/s What is the density of air? it would be .5* density of air * 35m/s^2 - 0 = XXX n/m^2
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11:20:36 ** 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 --> 1.29 kg/m^3 .5 * 1.29 kg/m^3 8 35 m/s^2 - 0 = 790 N/m^2 delta P = -790 N/m^2 790 N/m^2 * 240 m^2 = 190,000 N Okay.
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11:21:15 gen phy which term cancels out of Bernoulli's equation and why?
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RESPONSE --> The rho cancels out because of the little effect it has on the initial problem.
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11:21:36 ** 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 --> Oh and there is a small change in y okay. rho*g*y would see no change. okay.
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11:21:40 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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11:21:42 ** 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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11:21:45 univ phy What are the meanings of the limits as f approaches 0 and 1?
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11:21:47 ** 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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