#$&* course Phy 242 006. `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: No, I don't think there is a significant difference. If you watch the simulation for long enough, at some points it seems that the KE in the x direction is higher than the y and other times it seems that the KE in the y direction is greater than that in the x. However, this is constantly changing so I think it sort of balances itself out, making the difference not very significant. 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 'red' particle seems to be moving at a 1 or 2 speed. Since it shows the mass of the red particle to be 64, I assume it moves so slow because the force of the lighter particles aren't having much effect on it. The light blue particles seem to move at a speed of about 3 or 4. The light blue particles (there are only two of them) are the second heaviest in weight and move a little faster than the red but not much faster. They seem to be effected more by the impact of the green and dark blue particles than the red particle is. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Student answer with good analogy: I did not actually measure the velocities. the red were much faster. I would assume that the blue particle has much more mass a high velocity impact from the other particles made very little change in the blue particles velocity. Similar to a bycycle running into a Mack Truck. INSTRUCTOR NOTE: : It turns out that average kinetic energies of red and blue particles are equal, but the greater mass of the blue particle implies that it needs less v to get the same KE (which is .5 mv^2) ** Your Self-Critique: I think this solution must be from an old simulation. Because the mass of my red particle is greater than the mass of my two blue particles.
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Given Solution: ** If you watch the velocity display you will see that the red particles seem to average somewhere around 4 or 5 ** Your Self-Critique: I don't know if this solution is from a different simulation but mine seems to be moving a lot slower than 4 or 5. It seems to be moving at around 1 or 2. Your Self-Critique Rating: ********************************************* Question: If the simulation had 100 particles, how long do you think you would have to watch the simulation before a screen with all the particles on the left-hand side of the screen would occur? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: I don't think this would ever occur. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 think of setting it up as a probability. But I understand why this is useful and it supports my idea that this would never happen. Your Self-Critique Rating: 3 ********************************************* 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 #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The volume of the room is 9.2 m * 5.0 m * 4.5 m = 210 m^3. This air is replentished every 16 minutes, or at a rate of 210 m^3 / (16 min * 60 sec/min) = 210 m^3 / (960 sec) = .22 m^3 / second. The cross-sectional area of the duct is pi r^2 = pi * (.15 m)^2 = .071 m^2. The speed of the air flow and the velocity of the air flow are related by rate of volume flow = cross-sectional area * speed of flow, so speed of flow = rate of volume flow / cross-sectional area = .22 m^3 / s / (.071 m^2) = 3.1 m/s, approx. Your Self-Critique: ` Your Self-Critique Rating: ********************************************* Question: prin phy and gen phy problem 10.40 What gauge pressure is necessary to maintain a firehose stream at an altitude of 15 m?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: 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 #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** because of the small density of air and the small change in y, `rho g y exhibits practically no change. ** Your Self-Critique: Your Self-Critique Rating: ********************************************* Question: `q001. Explain how to get the change in velocity from a change in pressure, given density and initial velocity, in a situation where altitude does not change. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Looking at Bernoulli's Law that states the sum of 0.5*rho*v1^2+rho*g*h+P= 0.5*rho*v2^2+rho*g*h+P, you can see that if the altitude does not change, it will cancel out. Then the equation will look like this: 0.5*rho*v1^2+P1= 0.5*rho*v2^2+P2 So, knowing the change in pressure, you can solve for the change in velocity. Shown below: 0.5*rho*v1^2-0.5*rho*v2^2= P1-P2 You can see that the change in velocity will be equal but opposite signs of the change in pressure. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-Critique Rating: ********************************************* 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: The stream would be expected to rise to a height of 3.265 meters. I solved this by treating the water as a projectile. Explanation below: Since the hose is pointed straight upward these are the velocities, acceleration, and distance in the y direction (f & i stand for final & initial). Vf^2= Vi^2+2a(yf-yi) 0^2= 8^2+2*-9.8*(yf-0) The final velocity is zero because this is where the water stops going up and starts going down, the y initial is zero because this is where the water is first exiting the hose, the initial velocity is 8 m/s because you are told the water exits the nozzle at 8 m/s. Now solving for yf, the height that the water travels up, you get: -64= -19.6(yf) -64/-19.6= yf So you get the height the water rises (yf) to be 3.26 meters. To solve for the horizontal distance the stream would travel if it were directed at a 45 degree angle you start by finding the velocities in the x and y direction since you're given the initial velocity. You draw a right triangle and the given velocity (8 m/s) is the hypotenuse and the horizontal leg is the x direction, the vertical leg is the y direction and the angle is theta. Using sin and cosine you can find the velocity initial in the x direction (8cos45= 5.66 m/s) and the velocity initial in the y direction (8sin45= 5.66 m/s). You also know the velocity in the y direction when it comes back to the nozzle because it is equal and opposite to the velocity at which it left the nozzle. So the final velocity (the velocity that the water gets back to the nozzle) in the y direction is equal to -5.66 m/s. Using the kinematics equation below (for the y components) you can find the time the water travels (which is the same for both x and y components) and use it to solve for the distance x: Vf= Vi+a*t 5.66= -5.66+9.8*t t= 1.155 seconds Now plugging in the time to the kinematics equation for the x components you can find the distance in x the water travels: x= Vi*t+1/2*a*t The acceleration in the x direction is zero so you're left with: x= Vi*t You know the initial velocity in the x direction (Vi) & you just solved for t so you have for x: x= 5.66*1.155 x= 6.54 meters. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-Critique Rating: ********************************************* Question: univ phy problem 12.77 / 14.75 (11th edition 14.67): prove that if weight in water if f w then density of gold is 1 / (1-f). Meaning as f -> 0, 1, infinity. Weight of gold in water if 12.9 N in air. What if nearly all lead and 12.9 N in air? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: T=f*w Buoyant force plus tension in the rope is equal to the weight of the object. f*w+ density of water*g*volume= w where volume equals: V= w/(density of the crown*g) f*w=w-[density of water*g*(w/density of crown*g)] Now dividing out the w's and canceling the g's: f=1-(density of water/density of crown) Relative density is the ratio between density of the object and density of water: Relative density= density of the crown/density of the water= 1/(1-f) The relative density of gold is 19.3 and the weight is 12.9 N so the apparent weight is f*w=(1-1/19.3)*(12.9 N )= 12.2 N For lead in part C Fw=(1-1/11.3)*(12.9 N)= 11.8 N confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: When f approaches 0 the specific gravity approaches 1 and the crown would float. When f approaches 1 the buoyant force would be negligible and the crown would sink. 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: 1. The distance x at which the water hits the ground is dependent upon two things, the velocity at which it exits the hole and the height between water level and the hole so there's got to be a position that has the best combination of height (which means an increase in time it takes to hit the level surface) and initial velocity.