#$&* course PHY 202 2/8/15 at 12:05AM 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: ********************************************* 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: This is impossible to tell without doing specialized statistics, but it does appear that the KEx mean (1390) is quite lower than the KEy mean (1483). 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. My means differ by 93, which is lower than the 10% necessary to suggest statistical significance, so KEx is not statistically different than KEy. Your Self-Critique Rating: ********************************************* 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 was approximately 4 times slower than the green particle, so if the green particle had an average speed of 5m/s, then the red particle has an average speed of 1.25m/s. The blue particle had a speed of half of that of the green particle, so its average speed would be 2.5m/s. The blue particle is more massive, which leads to a lower velocity if energy is the same. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Student answer with good analogy: I did not actually measure the velocities. the red were much faster. I would assume that the blue particle has much more mass a high velocity impact from the other particles made very little change in the blue particles velocity. Similar to a bycycle running into a Mack Truck. INSTRUCTOR NOTE: : It turns out that average kinetic energies of red and blue particles are equal, but the greater mass of the blue particle implies that it needs less v to get the same KE (which is .5 mv^2) ** Your Self-Critique: OK Your Self-Critique Rating: ********************************************* Question: What do you think is the most likely velocity of the 'red' particle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The red particle appears to travel at 1-2m/s most frequently. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: If the green particles range from 0 to 10m/s, then their average speed would be 5m/s. The red particle is clearly slower (it takes approximately 4 times as long for the red particle to cross the screen as it does the green particle, so it is 4 times slower), so I do not see why the average speed of the red particle is around 4 or 5. Also, I am unable to see the velocity displayed for the red particle in the simulation. 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: The default number of particles is 43, and it already seems unlikely to have all particles appear on half of the screen, so if you more than double this number to obtain 100 particles, then it will be a very brief, very rare occurrence to see such an event. It might take days for this rare moment to 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: Wow, the actual statistics blow my guess of “days” away. The probability is definitely simpler to calculate than I thought, so now I know for the future how to calculate this. Your Self-Critique Rating:3
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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: ********************************************* 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: Gauge Pressure = P2 - P1 v1 = v2 P1 + rho*g*h1 + 1/2 *rho*v1^2 = P2 + rho*g*h2 + 1/2 *rho*v2^2 P1 + (1000kg/m^3)(9.8m/s^2)(0m) + 1/2*(1000kg/m^3)*v1^2 = P2 + (1000kg/m^3)(9.8m/s^2)(15m) + 1/2*(1000kg/m^3)*v2^2 Cancel out the KE terms on either side because v is the same for both. P1 = P2 + 147000 P2 - P1 = -147000 Pa The pressure inside the hose must be 147000 Pa lower than atmospheric pressure.
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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: I’m not sure why the negative gauge pressure means that the pressure inside the hose must be higher than atmospheric??? Your Self-Critique Rating: ********************************************* Question: Openstax: The heart of a resting adult pumps blood at a rate of 5.00 L/min. Convert this to cm^3 /s . What is this rate in m^3 /s ? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: 1000 L = 1 m^3 1.0 L = 0.001 m^3 1 m^3 = 1,000,000 cm^3 1.0 L = (1,000,000 * 0.001) = 1000 cm^3 5.0 L/min = 5000 cm^3/min = 83.3 cm^3/sec 5.0 L/min = 0.005 m^3/min = 8.3*10^-5 m^3/min confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: A cube 10 cm on a side is a liter. It takes 10 layers each of 10 rows each of 10 such cubes to fill a one-meter cube. It follows that l liter is a volume of 10 cm * 10 cm * 10 cm = 1000 cm^3, and a cubic meter is a volume equivalent to 10 * 10 * 10 liters = 1000 liters. 5 L / min is therefore the same as 5 L * (1000 cm^3 / L) / min = 5000 cm^3 / min. Your Self-Critique: OK Your Self-Critique Rating: ********************************************* Question: Openstax: Every few years, winds in Boulder, Colorado, attain sustained speeds of 45.0 m/s (about 100 mi/h) when the jet stream descends during early spring. Approximately what is the force due to the Bernoulli effect on a roof having an area of 220 m^2 ? Typical air density in Boulder is 1.14 kg/m^3 , and the corresponding atmospheric pressure is 8.89×104 N/m^2 . (Bernoulli’s principle as stated in the text assumes laminar flow. Using the principle here produces only an approximate result, because there is significant turbulence.) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: P1 + rho*g*h1 + 1/2 *rho*v1^2 = P2 + rho*g*h2 + 1/2 *rho*v2^2 8.89*10^4 Pa + (1.14kg/m^3)(9.8m/s^2)(0) + 1/2*(1.14kg/m^3)(0m/s)^2 = P2 + (1.14kg/m^3)(9.8m/s^2)(0) + 1/2*(1.14kg/m^3)(45.0m/s)^2 8.89*10^4 Pa = P2 + 1154.25 Pa P2 = 87745.8 Pa F = P*A = 87745.8Pa * 220m^2 F = 1.93*10^7 N confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Bernoulli's Equation says that rho g y + 1/2 rho v^2 + P = constant. On one side of the roof v = 0 (that would be the inside). On the other side v = 45 m/s. The difference in y from one side of the roof to the other is small, so that the change in rho g y from one side to the other is negligible. Thus 1/2 rho v^2 + P is constant, and we conclude that 1/2 rho v_1^2 + P_1 = 1/2 rho v_2^2 + P_2, where v_1 and P_1 are on one side of the roof (let's say that's the inside) and v_2 and P_2 are on the other. The pressure change from inside to outside is thus P_2 - P_1 = 1/2 rho v_2^2 - 1/2 rho v_1^2 = 1/2 rho ( v_2^2 - v_1^2) = 1/2 * 1.14 kg/m^3 ( (45 m/s)^2 - (0 m/s)^2) = 1400 kg m^2 / (s^2 m^3) = 1400 kg / (m s^2) = 1400 N / m^2, or 1400 Pa. Exerted on a roof area of 220 m^2 this pressure will thus exert a force of F = P * A = 1400 Pa * 220 m^2 = 1400 N/m^2 * 220 m^2 = 300 000 Newtons, very approximately. Your Self-Critique: I see that I made an error in substituting Denver’s atmospheric pressure (8.89*10^4 Pa) for P1. The atmospheric pressure would be the same on either side of the roof in absence of wind, and the wind would result in a pressure difference that would be discovered by subtracting P1 from P2. So, in the future, I should ignore the atmospheric pressure and solve for P2-P1 instead of P2. The rest of my math is correct based on my calculation of P2. In fact, if I subtract atmospheric pressure from my calculation for P2, I still get the correct answer, so the atmospheric pressure is indeed where I went wrong. Your Self-Critique Rating:3
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Given Solution: Your Self-Critique: OK Your Self-Critique Rating: ********************************************* Question: Openstax: A sump pump (used to drain water from the basement of houses built below the water table) is draining a flooded basement at the rate of 0.750 L/s, with an output pressure of 3.00×10^5 N/m^2 . (a) The water enters a hose with a 3.00-cm inside diameter and rises 2.50 m above the pump. What is its pressure at this point? (b) The hose goes over the foundation wall, losing 0.500 m in height, and widens to 4.00 cm in diameter. What is the pressure now? You may neglect frictional losses in both parts of the problem. The speed of the water in the hose satisfies the continuity equation A_cs * v = rate of volume flow. A_cs is the cross-sectional area of the hose, which is about 7 cm^2, so v = rate of volume flow / A_cs = .750 L/s / (7 cm^2) = .000750 m^3 / s / (.0007 m^2) = 1.1 m/s, approx.. The water enters at rest. At the point 2.50 m above the pump its velocity is 1.1 m/s. Since change in 1/2 rho v^2 + change in rho g y + change in pressure = 0 we have (1/2 rho v_2^2 - 1/2 rho v_1^2) + (rho g y_2 - rho g y_1) + (P_2 - P_1) = 0 with v_1 = 0, v_2 = 1.1 m/s, y_1 = 0, y_2 = 2.5 m and P_1 = 3 * 10^5 N/m^2. We get 1/2 * 1000 kg/m^3 * (1.1 m/s)^2 + 1000 kg/m^3 * 9.8 m/s^2 * 2.5 m + P_2 - 3 * 10^5 N/m^2 = 0 so that P_2 = 2.74 * 10^5 N/m^2, approx.. With a .500 m loss of height the quantity rho g y will change by - 5000 Pa. With the diameter increase to 4.00 cm the velocity will reduce to about .6 m/s so 1/2 rho v^2 will reduce to about 180 Pa, which is pretty much negligible. The pressure at this point will therefore be about 2.79 * 10^5 Pa. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: a) Rate of volume flow = c.s. area * flow rate 7.5*10^-4 m^3/s = (pi*(0.015m)^2) * flow rate Flow rate = 1.06 m/s P1 + rho*g*h + 1/2*rho*v1^2 = P2 + rho*g*h + 1/2*rho*v2^2 3*10^5 Pa + (1000kg/m^3)(9.8m/s^2)(0m) + 1/2*(1000kg/m^3)(0m/s)^2 = P2 + (1000kg/m^3)(9.8m/s^2)(2.5m) + 1/2*(1000kg/m^3)(1.06m/s)^2 3*10^5 Pa - 24500 - 561.8 = P2 P2 = 274938.2 Pa b) A1v1 = A2v2 (pi*(0.015m)^2)*(1.06m/s) = (pi*(0.02m)^2)*v2 7.49*10^-4 m^2 = 1.26*10^-3 m^2 * v2 v2 = 0.59 m/s 2.75*10^5 Pa + (1000kg/m^3)(9.8m/s^2)(0) + 1/2*(1000kg/m^3)(1.06m/s)^2 = P2 + (1000kg/m^3)(9.8m/s^2)(-0.5m) + 1/2*(1000kg/m^3)(0.59m/s)^2 2.75*10^5 + 561.8 = P2 - 4900 + 174.05 P2 = 280287.8 Pa confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Your Self-Critique: OK 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: For the velocity to move at 0 velocity, the pressure must be the same. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Velocity is 0 at top and bottom; pressure at top is atmospheric, and if pressure in the hose was the same the water wouldn't experience any net force and would therefore remain in the hose ** Your Self-Critique: Your Self-Critique Rating: ********************************************* Question: query gen phy problem 10.43 net force on 240m^2 roof from 35 m/s wind. What is the net force on the roof? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: P1 + rho*g*h1 + 1/2*rho*v1^2 = P2 + rho*g*h2 + 1/2*rho*v2^2 101300Pa + (1.29kg/m^3)(9.8m/s^2)(0m) + 1/2*(1.29kg/m^3)(0m/s)^2 = P2 + (1.29kg/m^3)(9.8m/s^2)(0m) + 1/2*(1.29kg/m^3)(35m/s)^2 P2 = 100509.9 Pa = N/m^2 F = P*A = 100509.9 N/m^2 * 240m^2 F = 24122376 N down onto roof F = P*A = 101300 N/m^2 * 240m^2 F = 24312000 N up from house Fnet = Fhouse - Froof = 24312000 N - 24122376 N Fnet = 1.9*10^5 N confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: ********************************************* Question: gen phy which term 'cancels out' of Bernoulli's equation and why? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The PE term cancels out of the equation because there is a negligible difference in height on either side of the roof. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** because of the small density of air and the small change in y, `rho g y exhibits practically no change. ** Your Self-Critique: OK Your Self-Critique Rating: ********************************************* Question: `q001. Explain how to get the change in velocity from a change in pressure, given density and initial velocity, in a situation where altitude does not change. Can you tell from your expressions whether the change in velocity, for a given pressure change, is greater, less or equal when the initial velocity is greater? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: If the pressure change is known, the known values for pressure, density, and initial velocity can be substituted into Bernoulli’s equation, which can be solved for v2 when the PE terms are canceled out for the zero change in altitude. P1 + rho*g*h1 + 1/2*rho*v1^2 = P2 + rho*g*h2 + 1/2*rho*v2^2 P1 + rho*g*0 + 1/2*rho*v1^2 = P2 + rho*g*0 + 1/2*rho*v2^2 P1 - P2 + 1/2*rho*v1^2 = 1/2*rho*v2^2 [2(P1 - P2 + 1/2*rho*v1^2)]/g = v2^2 v2 = sqrt([2(P1 - P2 + 1/2*rho*v1^2)]/g) The change in velocity is greater when initial velocity is greater. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ 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. 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: P1 + rho*g*h1 + 1/2*rho*v1^2 = P2 + rho*g*h2 + 1/2*rho*v2^2 P1 + (1000kg/m^3)(9.8m/s^2)(0m) + 1/2(1000kg/m^3)(4m/s)^2 = P2 + (1000kg/m^3)(9.8m/s^2)(0m) + 1/2(1000kg/m^3)(8m/s)^2 P2 - P1 = 1/2(1000kg/m^3)[(8m/s)^2 - (4m/s)^2] P2 - P1 = 24000 Pa If the stream is traveling at 8m/s at an angle of 45 degrees, we need to find the velocity in the x direction and the velocity in the y direction to analyze the kinematics of this stream. Cos 45 = v_ix/(8m/s) Where v_ix is initial velocity in x direction v_ix = 5.66m/s Sin 45 = v_iy/(8m/s) Where v_iy is initial velocity in y direction v_iy = 5.66m/s We split the parabolic motion of the stream in half - the trip up and the trip down TRIP UP: v_ix = 5.66m/s v_fx = 5.66m/s a_x = 0m/s^2 t = ? dx = ?? v_iy = 5.66m/s v_fy = 0m/s a_y = -9.8m/s^2 dy = ?? v_fy = v_iy + 2*a_y*dy 0m/s = 5.66m/s + 2*(-9.8m/s^2)dy dy = 1.63m v_fy - v_iy = at 0m/s - 5.66m/s = (-9.8m/s^2)t t = 0.58s dx = vt dx = (5.66m/s)(0.58s) dx = 3.28m TRIP DOWN v_ix = 5.66m/s v_fx = 5.66m/s a_x = 0m/s^2 t = 0.58s (equal to time on trip upward) dx = ?? v_iy = 0m/s v_fy = -5.66m/s a_y = 9.8m/s^2 dy = -1.63m Since the trip downward is identical in time, altitude, and velocities as the trip upward, the distance in the x direction is identical to the distance traveled in the x direction for both halves of the parabolic motion. So, the dx found for trip upward can be multiplied by two to find total distance traveled in the x direction. 3.28m * 2 = 6.57m Therefore, the stream of water travels approximately 6.6m horizontally before the stream returns to the altitude of the nozzle.
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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: ------------------------------------------------ Self-Critique Rating: ********************************************* Question: univ phy What are the meanings of the limits as f approaches 0 and 1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** GOOD STUDENT ANSWER: f-> 0 gives (density of crown)/(density of water) = 1 and T=0. If the density of the crown equals the density of the water, the crown just floats, fully submerged, and the tension should be zero. When f-> 1, density of crown >> density of water and T=w. If density of crown >> density of water then B is negligible relative to the weight w of the crown and T should equal w. ** ------------------------------------------------ Self-Critique Rating: ********************************************* Question: `q003. Water exits a large tank through a hole in the side of a cylindrical container with vertical walls. The water stream falls to the level surface on which the tank is resting. The tank is filled with water to depth y_max. The water stream reaches the level surface at a distance x from the side of the container. Without doing any calculations, explain why there must be at least one vertical position at which the hole could be placed to maximize the distance x. Explain also why there must be distances x that could be achieved by at least two different vertical positions for the hole. Give all the possible vertical levels of the hole. What is the maximum possible distance x at which it is possible for a water stream to reach the level surface, and where would the hole have to be to achieve this? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-Critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!