#$&* course phy 242 june 16 around 11:30 pm 1)Question: query introset change in pressure from velocity change.
.............................................
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 2) 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: Kinetic energy is x and y direction do not have significant differences between them but they do different as seen in the experiment using kinmodel simulation. majority of the differences between the two directions were within 12 to 20 percent. 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 3) 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: The average velocity of the red particles is about 3 m/s compared to blue particles which are about 4m/s. The reason that the velocity of the blue particles is greater than that of the red is that its mass is lower than that of the red particle, making it move faster than the red with a collisions occurs. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** Student answer with good analogy: I did not actually measure the velocities. the red were much faster. I would assume that the blue particle has much more mass a high velocity impact from the other particles made very little change in the blue particles velocity. Similar to a bycycle running into a Mack Truck. INSTRUCTOR NOTE: : It turns out that average kinetic energies of red and blue particles are equal, but the greater mass of the blue particle implies that it needs less v to get the same KE (which is .5 mv^2) ** Your Self-Critique: didn’t mention kinetic energy. Your Self-Critique Rating: ok 4) Question: What do you think is the most likely velocity of the 'red' particle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The most likely velocity of the red particle is between 4 or 5 m/s. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ok ** 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 5) 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: It would take a very long time given the ransom movements and collisions of the particles. And a specific time cannot be given, but it can be estimated through using probability: since its half of the screen ˝ is incorporated and the 10 particles being in that side is also , (1/2)^100 = 7.9 *10^-31. The probably of that happening is 7.9 *10^-31. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: ** STUDENT ANSWER: Considering the random motion at various angles of impact.It would likely be a very rare event. INSTRUCTOR COMMENT This question requires a little fundamental probability but isn't too difficult to understand: If particle position is regarded as random the probability of a particle being on one given side of the screen is 1/2. The probability of 2 particles both being on a given side is 1/2 * 1/2. For 3 particles the probability is 1/2 * 1/2 * 1/2 = 1/8. For 100 particlles the probability is 1 / 2^100, meaning that you would expect to see this phenomenon once in 2^100 screens. If you saw 10 screens per second this would take about 4 * 10^21 years, or just about a trillion times the age of the Earth. In practical terms, then, you just wouldn't expect to see it, ever. ** Your Self-Critique: did not mention the specific time Your Self-Critique Rating: ok 6) Question: `q001. Explain how to get the change in velocity from a change in pressure, given density and initial velocity, in a situation where altitude does not change. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: P1+1/2*rho*V1^(2) + rho*g*h1 = P2+1/2*rho*V2^(2) + rho*g*h2 We are looking for v2-v1: Since the altitude is constant and initial velocity given we have: P1+1/2*rho*V1^(2) = P2+1/2*rho*V2^(2) V1 = C P1+1/2*rho*C^(2) = P2+1/2*rho*V2^(2) P2-P1 =1/2*rho(C^(2)- V2^(2)) (P2-P1)/( 1/2*rho) = (C^(2)- V2^(2)) [(P2-P1)/( 1/2*rho)] - C^2 = -v2^2 C^2 -[(P2-P1)/( 1/2*rho)] = v2^2 Sqrt(C^2 -[(P2-P1)/( 1/2*rho)] = V2 Sqrt(C^2 -[(P2-P1)/( 1/2*rho)] - C = V2 - V1 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-Critique Rating: ok 7) Question: `q002. Water is moving inside a garden hose at 4 meters / second, and it exits the nozzle at 8 meters / second. Neglecting the effect of air resistance: How high would the stream be expected to rise if the hose was pointed straight upward? How far would the stream travel in the horizontal direction before falling back to the level of the nozzle, if directed at 45 degrees above horizontal? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: 1/2*rho*V1^(2) + rho*g*h1 = 1/2*rho*V2^(2) + rho*g*h2 .5*(1000kg/m^3)*(4m/s)^2 + (1000kg/m^3 * 9.8m/^2 *0 m) = .5*(1000kg/m^3)*(8m/s)^2 + (1000kg/m^3 * 9.8m/s^2 *h2 ) 8000 pa = 32000pa+9800*h2 -24000 pa = 9800*h2 -2.45 m= h2 Abs (-2.45 ) = 2.45 m H2 = x*tan45
.............................................
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: ok ------------------------------------------------ Self-Critique Rating: ok 9) Question: univ phy What are the meanings of the limits as f approaches 0 and 1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: First if the f approaches 0 then we get density of the gold to be 1 and weight of water of water to be 0 which means the tension is 0 therefore the crown will float since there is no tension. When f approaches 1 then the density of gold will be infinity which will be greater than water’s density and the water weight is w which is tension = w, therefore the crown will sink. . confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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 Rating: ok 10) Question: `q003. Water exits a large tank through a hole in the side of a cylindrical container with vertical walls. The water stream falls to the level surface on which the tank is resting. The tank is filled with water to depth y_max. The water stream reaches the level surface at a distance x from the side of the container. Without doing any calculations, explain why there must be at least one vertical position at which the hole could be placed to maximize the distance x. Explain also why there must be distances x that could be achieved by at least two different vertical positions for the hole. Give all the possible vertical levels of the hole. What is the maximum possible distance x at which it is possible for a water stream to reach the level surface, and where would the hole have to be to achieve this? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: There must be one vertical place where the hole must be placed to maximize its distance x and that place is halfway between the tank and the surface of the water. The reason for having a vertical position for the hole which maximizes the x direction is that in that particular hole there is a higher pressure and so it projects the water at a higher speed resulting in a further distance. There must be two different vertical positions for the hole that can achieve the distance x , because while it takes a longer time on the whole near the surface of the water , in the bottom that time is eliminated while the distance is elongate and so therefore there time manipulates the system for it to achieve the same distances within two different placements of the holes. Below the halfway mark, the water will be projected as a higher speed but smaller distance than that of the maximized, above the halfway mark, the water will be in the air for longer time along with lower speed. Halfway between the tank and the surface of the water. The maximum possible distance for x can be found through the following equation: X=2* sqrt(2*y1^2 - y1^2) , where y1 is height of the hole.