course Phy 202 ?????????????assignment #??z????Z????U????
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14:50:36 query introset change in pressure from velocity change. Explain how to get the change in fluid pressure given the change in fluid velocity, assuming constant altitude
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RESPONSE --> Since there is no change in h or altitude the equation pgh will not change leaving us with `dP = .5rhov^2 = .5 rho (v2^2 - v1^2) P = pressure rho = density v = velocity g = acceleration of gravity h = height or altitude
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14:52:58 ** 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 dissapear. (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). **
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RESPONSE --> I should have stated Bernoullis Equation so the statement that rho*g*h would have made more sense. (rho* g*h) + (0.5*rho*v^2) + P = 0
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14:54:15 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.
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RESPONSE --> Yes, When doing experiment 3 in kinmodel experimetn the aveKEx is 43.1 higher than the aveKEy.
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14:55:29 ** 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. **
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RESPONSE --> I believe my differed by 11% which is close to 10%. I was gauging significance by 5% which is why my answer differed.
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15:01:23 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?
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RESPONSE --> I think the KE of both balls were around 300 to 400 but the blue particle has a greater mass. So the red particle is moving at a greater velocity to achieve that KE and the blue is moving slower but has a greater mass.
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15:01:53 ** 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) **
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RESPONSE --> I should have given the KE equation to complete my answer.
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15:04:58 What do you think is the most likely velocity of the 'red' particle?
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RESPONSE --> This depends on the mass if you were to use the equation. KE = .5mv^2 so if you estimate a KE of 350 and a mass of 10 v = 8.4units/sec
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15:05:52 ** If you watch the velocity display you will see that the red particles seem to average somewhere around 4 or 5 **
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RESPONSE --> I was doing the query from memory and had forgotten about the velocity display.
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15:06:58 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?
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RESPONSE --> With that number of particles it would take a significant amount of time for the particles to all be on the left side of the screen. It would most likely be over an hour.
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15:07:54 ** 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. **
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RESPONSE --> I was a little to optomistic with my answer. I did not work out a probability but can easily recognize that is going to be a very rare occurance.
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15:17:39 What do you think the graphs at the right of the screen might represent?
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RESPONSE --> One graph is the distribution of velocities and the other is the frequency distribution of KE. They have similiar shapes because velocity is a component of KE. I had already received instructor comments on this equation from my prelim observations.
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15:17:48 ** One graph is a histogram showing the relative occurrences of different velocities. Highest and lowest velocities are least likely, midrange tending toward the low end most likely. Another shows the same thing but for energies rather than velocities. **
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RESPONSE --> ok
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15:36:09 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?
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RESPONSE --> Volume = 9.2 m * 5.0 m * 4.5 m = 207 m^3 16 mins * 60 sec/min = 960 seconds A = pi r^2 A = pi ((.15m)^2) A = .07 m^2 207 m^3 / .07m^2 = 2928.45 m 2928.45 m/ 960sec = 3 m/s
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15:36:19 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.
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RESPONSE --> Ok
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15:37:59 prin phy and gen phy problem 10.40 gauge pressure to maintain firehose stream altitude 15 m ......!!!!!!!!...................................
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RESPONSE --> P = ? h = 15 m P = rho g h P = (1.0 * 10^3 kg/m^3) (9.8 m/s^2) (15 m) P = 147000 N/m^2
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15:39:25 ** 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. **
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RESPONSE --> I didn't remember the - sign.
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15:42:20 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?
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RESPONSE --> The pressure then remains the same and cancels out of the equation.
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15:43:43 ** 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 **
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RESPONSE --> Ok
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15:49:54 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?
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RESPONSE --> (1.29 kg/m^3)(240m^2)(35m/s) = 10836 kg/s 10836 kg/s /(9.8m/s^2) = 1105.7 N
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15:52:32 ** 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. **
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RESPONSE --> I didn't think of the change in velocity of 0 to 35. That would have made Bernoullis equation applicable. I just tried to make the units work out. After using Bernoulli's `dP = -`d(.5rhov^2) to find the P you mulitply by the area to get the force on that area.
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15:53:07 gen phy which term cancels out of Bernoulli's equation and why?
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RESPONSE --> h because it is neglible
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15:53:45 ** because of the small density of air and the small change in y, `rho g y exhibits practically no change. **
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RESPONSE --> Ok.. I should have explained better it is not simply h but the whole density*g*h because the small density and the small change in h
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15:53:57 univ phy problem 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?
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RESPONSE --> not assigned
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15:54:03 ** 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). **
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RESPONSE --> k
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???????????y?? assignment #019 019. `query 9 Physics II 08-06-2007
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22:17:59 Query introductory set 6, problems 1-10 explain how we know that the velocity of a periodic wave is equal to the product of its wavelength and frequency
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RESPONSE --> The frequency is the number of peaks that occur in a given time and the wavelength measures the length of from peak to peak. So when multiplied it gives you the length in a given time so that is the velocity.
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22:18:09 ** we know how many wavelength segments will pass every second, and we know the length of each, so that multiplying the two gives us the velocity with which they must be passing **
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RESPONSE --> Ok
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22:20:04 explain how we can reason out that the period of a periodic wave is equal to its wavelength divided by its velocity
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RESPONSE --> Because velocity is rate of speed that the wave is passes a particular point if you divide this by the wave length it gives you the time. m/s / m = s
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22:23:25 ** If we know how far it is between peaks (wavelength) and how fast the wavetrain is passing (velocity) we can divide the distance between peaks by the velocity to see how much time passes between peaks at a given point. That is, period is wavelength / velocity. **
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RESPONSE --> It is easier explained when you say it is the distance between the peaks divided by the velocity giving you how much time passes between peaks.
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22:24:48 explain why the equation of motion at a position x along a sinusoidal wave is A sin( `omega t - x / v) if the equation of motion at the x = 0 position is A sin(`omega t)
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RESPONSE --> I don't feel confident I can explain how the equation is derived.
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22:27:11 ** the key is the time delay. Time for the disturbance to get from x = 0 to position x is x / v. What happens at the new position is delayed by time x/v, so what happens there at clock time t happened at x=0 when clock time was t = x/v. In more detail: If x is the distance down the wave then x / v is the time it takes the wave to travel that distance. What happens at time t at position x is what happened at time t - x/v at position x=0. That expression should be y = sin(`omega * (t - x / v)). } The sine function goes from -1 to 0 to 1 to 0 to -1 to 0 to 1 to 0 ..., one cycle after another. In harmonic waves the motion of a point on the wave (think of the motion of a black mark on a white rope with vertical pulses traveling down the rope) will go thru this sort of motion (down, middle, up, middle, down, etc.) as repeated pulses pass. If I'm creating the pulses at my end, and that black mark is some distance x down in rope, then what you see at the black mark is what I did at time x/v earlier. **
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RESPONSE --> From 0 to point x the time disturbance is x/v. So t = x/v beacuse what happens at clock time t happened at x = 0 so it all depends on the time lag.
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22:31:08 Query introductory set six, problems 11-14 given the length of a string how do we determine the wavelengths of the first few harmonics?
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RESPONSE --> only problems 1-10 were assigned in 19
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22:33:31 ** As wavelength decreases you can fit more half-waves onto the string. You can fit one half-wave, or 2 half-waves, or 3, etc.. So you get 1 half-wavelength = string length, or wavelength = 2 * string length; using `lambda to stand for wavelength and L for string length this would be 1 * 1/2 `lambda = L so `lambda = 2 L. For 2 wavelengths fit into the string you get 2 * 1/2 `lambda = L so `lambda = L. For 3 wavelengths you get 3 * 1/2 `lambda = L so `lambda = 2/3 L; etc. } Your wavelengths are therefore 2L, L, 2/3 L, 1/2 L, etc.. **
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RESPONSE --> ok. so it can be found by n * 1/2 lamba = L where n is the number of wavelengths lamaba = wavelength L = string length
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22:38:29 Given the wavelengths of the first few harmonics and the velocity of a wave disturbance in the string, how do we determine the frequencies of the first few harmonics?
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RESPONSE --> Wavelength = m/peak velocity = m/s velocity divide by wavelength = peak/s which is the frequency which is measured in Hz
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22:38:40 ** The frequency is the number of crests passing per unit of time. We can imagine a 1-second chunk of the wave divided into segments each equal to the wavelength. The number of peaks is equal to the length of the entire chunk divided by the length of a 1-wavelength segment. This is the number of peaks passing per second. So frequency is equal to the wave velocity divided by the wavelength. **
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RESPONSE --> ok
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22:40:34 Given the tension and mass density of a string how do we determine the velocity of the wave in the string?
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RESPONSE --> It was not discussed in the question 1-10.
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22:40:57 ** We divide tension by mass per unit length and take the square root: v = sqrt ( tension / (mass/length) ). **
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RESPONSE --> v = squareroot (tension/ (mass/length))
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22:51:44 gen phy explain in your own words the meaning of the principal of superposition
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RESPONSE --> it is when two or more waves are moving in a system and they cause interference or diffraction. That is the waves either cancel each other out or build each other up
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22:52:18 ** the principle of superposition tells us that when two different waveforms meet, or are present in a medium, the displacements of the two waveforms are added at each point to create the waveform that will be seen. **
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RESPONSE --> I think i confused this with other properties discussed in the book.
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23:01:34 gen phy what does it mean to say that the angle of reflection is equal to the angle of incidence?
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RESPONSE --> ""The angle of incidence is the angle the incident ray makes with the perpendicular to the reflecting surface. While the angle of reflection is the corresponding angle for the reflected wave. So they are equal.
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23:02:41 ** angle of incidence with a surface is the angle with the perpendicular to that surface; when a ray comes in at a given angle of incidence it reflects at an equal angle on the other side of that perpendicular **
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RESPONSE --> ok
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