#$&* course Phy 232 7/9 1030 am 006. `query 5
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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: Based on my observations, I do not believe there is a significant difference between the total KE in the x and y direction. Each direction fluxuated up and down so they were both greater than the other at certain times, and one direction wasn't greater more than the other. They both had similar numbers and nothing signifcant about the numbers showed. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. 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: The average velocities were too hard to measure, however, the red were faster than the blue. The blue particle had more mass, so therefore it was slower, and everytime something collided with it there wasn't much affect on the blue. For example, if red and blue collided, the red would speed away quickly while the blue was barely affected. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Student answer with good analogy: I did not actually measure the velocities. the red were much faster. I would assume that the blue particle has much more mass a high velocity impact from the other particles made very little change in the blue particles velocity. Similar to a bycycle running into a Mack Truck. INSTRUCTOR NOTE: : It turns out that average kinetic energies of red and blue particles are equal, but the greater mass of the blue particle implies that it needs less v to get the same KE (which is .5 mv^2) ** Your Self-Critique: OK Your Self-Critique Rating: OK ********************************************* Question: What do you think is the most likely velocity of the 'red' particle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The red wasn't as fast as the green, but it wasn't the slowest either. The green was usually at 6 or 7, so I would imagine the red was somewhere around 4. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: 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 do not believe this would ever happen. Even when I had a square 1 inch x 1 inch, there wasn't much time when there was no particles in the space. It could happen, but based on the colliding particles and the fact that they would randomly go all over the place, I don't think there will be a time when theyare all on the left side. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT ANSWER: Considering the random motion at various angles of impact.It would likely be a very rare event. INSTRUCTOR COMMENT This question requires a little fundamental probability but isn't too difficult to understand: If particle position is regarded as random the probability of a particle being on one given side of the screen is 1/2. The probability of 2 particles both being on a given side is 1/2 * 1/2. For 3 particles the probability is 1/2 * 1/2 * 1/2 = 1/8. For 100 particlles the probability is 1 / 2^100, meaning that you would expect to see this phenomenon once in 2^100 screens. If you saw 10 screens per second this would take about 4 * 10^21 years, or just about a trillion times the age of the Earth. In practical terms, then, you just wouldn't expect to see it, ever. ** Your Self-Critique: OK Your Self-Critique Rating: OK ********************************************* Question: prin phy and gen phy problem 10.36 15 cm radius duct replentishes air in 9.2 m x 5.0 m x 4.5 m room every 16 minutes; how fast is air flowing in the duct? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: ` 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?
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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: Your Self-Critique Rating: ********************************************* Question: 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Velocity is 0 at top and bottom; pressure at top is atmospheric, and if pressure in the hose was the same the water wouldn't experience any net force and would therefore remain in the hose ** Your Self-Critique: Your Self-Critique Rating: ********************************************* Question: query gen phy problem 10.43 net force on 240m^2 roof from 35 m/s wind. What is the net force on the roof? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: Your Self-Critique Rating: ********************************************* Question: gen phy which term 'cancels out' of Bernoulli's equation and why? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** because of the small density of air and the small change in y, `rho g y exhibits practically no change. ** Your Self-Critique: Your Self-Critique Rating: ********************************************* Question: `q001. Explain how to get the change in velocity from a change in pressure, given density and initial velocity, in a situation where altitude does not change. Can you tell from your expressions whether the change in velocity, for a given pressure change, is greater, less or equal when the initial velocity is greater? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: This can be done by using Bernoulli's equation: 1/2(density)v1^2 + (density)(g)(h1) + P1 = 1/2(density)v2^2 + (density)(g)(h2) + P2 If the altitude does not change, then (density)(g)(H) will be the same on each side, so they will cancel out, leaving us: 1/2(density)v1^2 + P1 = 1/2(density)v2^2 + P2 Change of pressure = P2 - P1. If we change around the above equation we get the change of pressure = P2 - P1 = 1/2(density)v1^2 - 1/2(density)v2^2 So, 1/2(density)v2^2 = 1/2(density)v2^2 - change of pressure v2 = sqrt([1/2(density)v2^2 - change of pressure]/[1/2(density)]) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-Critique Rating: OK ********************************************* Question: `q002. Water is moving inside a garden hose at 4 meters / second, and it exits the nozzle at 8 meters / second. What is the change in pressure as the water moves from the end of the hose out into the surrounding air? 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: We can use bernoullis equation for this. Since altitude seems to be constant, we can use: change of pressure = P2 - P1 = 1/2(density)v1^2 - 1/2(density)v2^2 change of pressure = (1/2)(1000 kg/m^3)(4 m/s)^2 - (1/2)(1000 kg/m^3)(8 m/s)^2 = 24000 Pa The height of the stream if pointed straight upwards would be found by: v = vo + at 0 = (8 m/s) + (-9.81 m/s^2)(t^2) t = 0.9 s y = y1 + vy(t) + 1/2a(t)^2 y = 0 + (8 m/s)(0.9 s) + (1/2)(-9.81 m/s^2)(0.9)^2 = 3.2 m To figure out how far the stream would travel horizonally: 0 = 0 + (8 m/s)(sin(45))(t) + (1/2)(-9.81)(t)^2 t = 1.2 s x = 0 + (8 m/s)(cos(45))(1.2 s) x = 6.8 m So the horizonal distance is 6.8 meters confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-Critique Rating: OK ********************************************* Question: univ phy problem 12.77 / 14.75 (11th edition 14.67): prove that if weight in water if f w then density of gold is 1 / (1-f). Meaning as f -> 0, 1, infinity. Weight of gold in water if 12.9 N in air. What if nearly all lead and 12.9 N in air? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: As stated in the problem, Tension = fw. Also, F = T - mg = 0 with mg = w. Since it is under water, we also have to take buoancy into account, so T + B - w = 0 Bouancy = (density of water)(volume of object)(gravity) So taking this into mind, the equation can turn into: Fw + (density of water)(volume of object)(gravity) - w = 0 (density of water)(volume of object)(gravity) = w - fw (density of water)(volume of object)(gravity) = w(1 - f) Bouancy is equal to the weight of the fluid displaced by an object, so w = (density of object)(volume of object)(gravity) (density of water)(volume of object)(gravity) = (density of object)(volume of object)(gravity)(1 - f) Relative density is the proportion of the density of the object to the density of water, so: density of object / density of water = 1/(1-f) Since the crown is made out of solid gold, its relative density to water would be 19.3 So 19.3 = 1/(1-f) 19.3 - 19.3 f = 1 f = 18.3/19.3 = 0.948 Since T = fw, and Tension can be the apparent weight, the apparent weight immersed in water would equal 0.948 x 12.9 N = 12.2 N The relative density of lead is 11.3. So repeating the process: 11.3 = 1/(1-f) 11.3 - 11.3 f = 1 f = 10.3/11.3 = 0.91 Apparent weight = 0.91 x 12.9 N = 11.8 N confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: OK ------------------------------------------------ Self-Critique Rating: OK ********************************************* Question: univ phy What are the meanings of the limits as f approaches 0 and 1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: If f approached 0, then relative density = 1/(1-0) = 1 and Tension = (0)w = 0. Since relative density is 1, then the density of the object is equal to water and it will float, with no tension. If f approached 1, then relative density = 1/(1-1) = 1/0 = DNE and Tension = (1)w = w. The density of the object will be greater than water. And then tension will be equal to the objects weight. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 Rating: OK ********************************************* 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 at least one vertical position at which the hole could be placed to maximize the distance x because if it was not completely vertical, the water would flow out at an angle, which would change where the water falls. In a vertical position, the water will come out faster because of water pressure, and will therefore go the furthest.