#$&* course Phy 122 2/10 8 pm 006. `query 5
.............................................
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: I used the same varaibles that were used in the video notes, so I am not sure which one to use. I believe that I did this correctly, but I did not carry out the simplest form as the given solution did. Your Self-Critique Rating:3 ********************************************* 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: I don't think it there was really a significant difference between the total KE in the x direction than that in the y direction. It just seemed pretty much the same. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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. This difference is not statistically significant, so we conclude that the total KE is statistically the same in bot directions. ** Your Self-Critique:OK Your Self-Critique Rating:OK ********************************************* Question: 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: I believe the red balls had a higher velocity than the blue balls. I would think it was due to the blue particled having a greater mass. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** Student answer with good analogy: I did not actually measure the velocities. the red were much faster. I would assume that the blue particle has much more mass a high velocity impact from the other particles made very little change in the blue particles velocity. Similar to a bycycle running into a Mack Truck. INSTRUCTOR NOTE: : It turns out that average kinetic energies of red and blue particles are equal, but the greater mass of the blue particle implies that it needs less v to get the same KE (which is .5 mv^2) ** Your Self-Critique:OK Your Self-Critique Rating:OK ********************************************* Question: What do you think is the most likely velocity of the 'red' particle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: I saw it mostly around 5. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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:OK Your Self-Critique Rating:OK ********************************************* 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: I don't think it could ever really happen. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT ANSWER: Considering the random motion at various angles of impact.It would likely be a very rare event. INSTRUCTOR COMMENT This question requires a little fundamental probability but isn't too difficult to understand: If particle position is regarded as random the probability of a particle being on one given side of the screen is 1/2. The probability of 2 particles both being on a given side is 1/2 * 1/2. For 3 particles the probability is 1/2 * 1/2 * 1/2 = 1/8. For 100 particlles the probability is 1 / 2^100, meaning that you would expect to see this phenomenon once in 2^100 screens. If you saw 10 screens per second this would take about 4 * 10^21 years, or just about a trillion times the age of the Earth. In practical terms, then, you just wouldn't expect to see it, ever. ** Your Self-Critique:OK Your Self-Critique Rating:OK ********************************************* Question: prin phy and gen phy problem 10.36 15 cm radius duct replentishes air in 9.2 m x 5.0 m x 4.5 m room every 16 minutes; how fast is air flowing in the duct? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Since we have the information to do so, I would say we needed to find the area of the room first. V = 9.2 m * 5.0 m * 4.5 m = 207 m^3 Next we can convert the minutes to seconds. 16 minutes * (60 s / 1 minute) = 960 s Now we can find the rate it replentishes air in m^3/s. 207 m^3 / 960 s = 0.22 m^3/s Now I am not sure where to go from here. We know that the radius of the duct is 15 cm, which to me, implies that it is round. But I am still not sure of how to complete this problem. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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: I see how to do this now. It makes sense to see it wrote out, but going from the first step and arriving at the last step is tricky for me for this particular problem. ` Your Self-Critique Rating: ********************************************* 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: I assume here that this means that we want h to stay constant therefore making pg stay constant as well. We are looking for P2, I think, so I am just going to give this my best shot. 0.5 p (v1)^2 + pgh1 + P1 = 0.5 p (v2)^2 + pgh2 + P2 We can take out the constants of pgh since pgh1 = pgh2 Therefore that leaves: 0.5 p (v1)^2 + P1 = 0.5 p (v2)^2 + P2 Therefore (if I am correct in looking to solve for P2) we subtract 0.5 p (v2)^2 from both sides leaving: P2 = 0.5 p (v1)^2 + P1 - 0.5 p (v2)^2 I don't know if I was supposed to fill in the numberical values first, but I would assume so since now I have no h variable to plug in 15 m into. I guess I did this wrong, but I didn't know how to do it any other way. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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: I used the equation that relates to energy conservation here because I thought we needed to have a P1 and P2, but I was wrong. This makes sense to me, but I definitely need more practice on it. Your Self-Critique Rating:3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!