#$&* course Phy 232 If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
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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: We can safely assume that the majority of the time, the kinetic energy in the x direction is equivalent to the kinetic energy in the y direction. They may be some slight difference but they are not far apart compared to how close most of the cases are. 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 kinetic energy for the blue particles and for the red particle are equivalent. The blue particle having the higher mass means that somewhere something must compensate for the red and the blue to be equal. Looking at the KE equation, (1/2)m(v^2), we see that to balance to things out for something with a greater mass, the velocity is the only thing left that can change. This implies that while the blue particles weight more, they move at a slower velocity than the red particles to ensure that everything balances out in the end. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Student answer with good analogy: I did not actually measure the velocities. the red were much faster. I would assume that the blue particle has much more mass a high velocity impact from the other particles made very little change in the blue particles velocity. Similar to a bycycle running into a Mack Truck. INSTRUCTOR NOTE: : It turns out that average kinetic energies of red and blue particles are equal, but the greater mass of the blue particle implies that it needs less v to get the same KE (which is .5 mv^2) ** Your Self-Critique: OK Your Self-Critique Rating: OK ********************************************* Question: What do you think is the most likely velocity of the 'red' particle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: During the experiment, the velocity of the red particle is shown. It seems to be constantly around 4.5. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: Breaking down this problem into individual probabilities is the best method to approach it. There is a 50/50 chance at any moment of one particle being on the left or on the ride side of the screen. 50/50 is also written as (1/2) probability. Therefore, if you multiply the probability of one particle being on either side by the total number of 100 particles, you would get (1/2)*(1/2)*(1/2)… and so on. The final equation for finding the probability is (1/2)^100 = 7.89*10^-31. This probably could not get closer to 0, therefore the changes of all the particles ending up on one side of the screen at once will not occur. 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: I didn’t end up calculating the number of days or years but I now remembered and see how to calculate this given the probabilities. 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: First we must calculate the volume of the room. This is done by: volume = (9.2)*(5.0)*(4.5) = 207 m^3. With the air flowing in every 16 minutes, we can find the rate at which the air is flowing into the room in m/s. 16 minutes*60 seconds = 960 seconds. Take the volume and divide it by the rate in seconds. This comes out to be, (207 / 960) = .22 m^3/second. This is the rate at which air flows into the room. We must now find the area of the duct, which can be done by the given radius of .15 m and by using the formula, pi*(r^2). Therefore, the area of the duct comes out to be, pi*(.15^2) = .071 m^2. To find the speed that the air is flowing into the room can be done by using the equation, volume / area. Therefore, speed of air flow = (.22) / (.071) = 3.1 m/s. 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: 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?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: To find the pressure, we can use Bernoulli’s equation. The first step to take is to establish that the velocity of the water is zero inside of the hose and this must be assumed. Also, the velocity of water at its peak height of 15 m is also zero. Given those two we can properly use Bernoulli’s equation. The altitude of the water stream is given at 15 m; therefore the h is 15 m. Since gravity plays a role in the water falling back down, we can use the formula, rho*(g)*(h), to determine the change in pressure. Plugging in the numbers, we get the equation, pressure change = (1,000)*(9.8)*(15) = 147,000 N/m^2. This means that the pressure changes of total of 147,000 N/m^2 after exiting the hose and reaching a height of 15 m. 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 Your Self-Critique Rating: OK ********************************************* 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: The velocity term is the one that cancels out of Bernoulli’s equation. This is because the velocity of the water remains 0 inside of the hose before exiting and 0 at the peak of the 15 m altitude. These velocities being 0 made the equation that I used in the previous problem accurate. Velocity is the only factor that remains constant throughout this process. Pressure and altitude both are changing. 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: 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: In this scenario, the Bernoulli’s equation has some differences, unlike in many other cases. The difference lies in the velocities of the (1/2)*(rho)*(v^2) term of the equation. The total equation to use (1/2)*(rho)*(v^2) - (1/2)*(rho)*(v^2) = 0. On one side of the roof, the velocity = 0 m/s and on the other side the velocity = 35 m/s. Plugging those velocities into the equations gives us, (1/2)*(1.3)*(35^2) - (1/2)*(1.3)*(0^2) = 0. This comes out to be a difference of 790 N/m^2. This answer is in the form of pressure, and we know that force = pressure*area. Therefore, the force on the roof is, force = (790)*(240) = 189,600 N. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** air with density around 1.29 kg/m^3 moves with one velocity above the roof and essentially of 0 velocity below the roof. Thus there is a difference between the two sides of Bernoulli's equation in the quantity 1/2 `rho v^2. At the density of air `rho g h isn't going to amount to anything significant between the inside and outside of the roof. So the difference in pressure is equal and opposite to the change in 1/2 `rho v^2. On one side v = 0, on the other v = 35 m/s, so the difference in .5 rho v^2 from inside to out is `d(.5 rho v^2) = 0.5(1.29kg/m^3)*(35m/s)^2 - 0 = 790 N/m^2. The difference in the altitude term is, as mentioned above, negligible so the difference in pressure from inside to out is `dP = - `d(.5 rho v^2) = -790 N/m^2. The associated force is 790 N/m^2 * 240 m^2 = 190,000 N, approx. ** Your Self-Critique: OK Your Self-Critique Rating: OK ********************************************* Question: gen phy which term 'cancels out' of Bernoulli's equation and why? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The terms that cancel out in this scenario are gravity and height. As previously stated, the change in altitude between both sides in this problem is so small that it is negligible. Also, gravity is cancelled out because it doesn’t play a role in this scenario and if it did, the gravity constant remains the same for all objects and problems. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: I forgot to include the density of air. I understand how and why it can be cancelled out now and fully understand all of the cancelling terms in this problem. 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: From the master equation of Bernoulli’s equation, we can show a change in volume resulting from a change in pressure as (1/2)*(rho)*(v1^2) + (rho)*(g)*(h1) - (1/2)*(rho)*(v2^2) + (rho)*(g)*(h2) = (P2-P1). We can factor out the common terms on the left side of the equation to get a final equation of (P2-P1) = (1/2)*rho*(v1^2-v2^2). This equation now shows how the change in pressure on the left side, can determine the change in volume on the right side. This shows the relationship between the two using Bernoulli’s equation. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-Critique Rating: OK
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Given Solution: ** The tension in the rope supporting the crown in water is T = f w. Tension and buoyant force are equal and opposite to the force of gravity so T + dw * vol = w or f * dg * vol + dw * vol = dg * vol. Dividing through by vol we have f * dg + dw = dg, which we solve for dg to obtain dg = dw / (1 - f). Relative density is density as a proportion of density of water, so relative density is 1 / (1-f). For gold relative density is 19.3 so we have 1 / (1-f) = 19.3, which we solve for f to obtain f = 18.3 / 19.3. The weight of the 12.9 N gold crown in water will thus be T = f w = 18.3 / 19.3 * 12.9 N = 12.2 N. STUDENT SOLUTION: After drawing a free body diagram we can see that these equations are true: Sum of Fy =m*ay , T+B-w=0, T=fw, B=(density of water)(Volume of crown)(gravity). Then fw+(density of water)(Volume of crown)(gravity)-w=0. (1-f)w=(density of water)(Volume of crown)(gravity). Use w==(density of crown)(Volume of crown)(gravity). (1-f)(density of crown)(Volume of crown)(gravity) =(density of water)(Volume of crown)(gravity). Thus, (density of crown)/(density of water)=1/(1-f). ** Your Self-Critique: OK ------------------------------------------------ Self-Critique Rating: OK ********************************************* Question: univ phy What are the meanings of the limits as f approaches 0 and 1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: As f approaches 0, the density of the crown is the same as the density in water. This means that the crown will float on top of the water. As f approaches 1, the density in the crown is greater than the density of the water. This means that the crown sinks and there is tension in rope unlike when the crown is just floating on the water. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: To achieve the maximum horizontal distance x away from the cylinder, the hole must be placed as far down on the cylinder to allow for the most water to be able to push down along with the hole not being too far down to not allow the water enough room to flow out far enough.
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