Need help with 2 physics problems.?\nThese are the last two i need to know, i have tried to do it, but it's always wrong, i don't want to get a bad grade on this webassign. Please help me, and show the steps so i can understand it. \n7. If you were to build a pipe organ with open-tube pipes spanning the range of human hearing (20 Hz to 20 kHz), what would be the range of the lengths of pipes required?\nshortest pipe\n________ m ---> I got 8.57 and tried to submit it but it says it's wrong \nlongest pipe\n________ m \n\n\n9. In one of the original Doppler experiments, one tuba was played on a moving platform car at a frequency of 75 Hz, and a second identical one was played on the same tone while at rest in the railway station. What beat frequency was heard if the train approached the station at a speed of 13.0 m/s? \n________ Hz\nI got 5.69 but it's wrong. I found the two seperate frequencys using the dopplar effect formula then i subtracted the two and got my answer.
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Try these answers:\n\nFor the first one, shortest is 0.0086 m, longest is 8.6 m.\nFor the second one, it should be 2.95 Hz
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This problem is a conservation of momentum problem:\n\n(a) In an elastic collision, the formulas for calculating the final velocities are derived from both the conservation of momentum formula and the conservation of kinetic energy formula. Check out my source for more information, but for now just trust me that they are valid:\n\nv1 = (v1(m1 - m2) + 2m2v2)/(m1 + m2)\nv1 = (1*(4 - 7.2) + 0)/(4 + 7.2) = -0.3 m/s\n\nv2 = (v2(m2 - m1) + 2m1v1)/(m1 + m2)\nv2 = (0 + 2*4*1)/(4 + 7.2) = 0.7 m/s\n\n(b) In an inelastic collision, just use the conservation of momentum formula:\n\nm1v1' + m2v2' = (m1+ m2)v\n\n4*1 = (4+7.2)*v\nv = 4/(4+7.2) = 0.36 m/s\n\nI hope this helps!
543823
The sound intensity due to a point source radiating equally in all directions (spherically symmetric case) is proportional to 1/r^2, where r is the distance from the source of the sound:\n\nI = k/r^2, where k is a constant of proportionality.\n\nIf we know the intensity (I1) at one distance (r1), then the intensity (I2) at a second distance (r2) is given by:\n\nI2 = I1*(r1/r2)^2\n\nIn Q1, we are told that the sound intensity at 1 m is 10^-5 W/m^2, and we want to know the intensity at 5 m.  Plugging these values into the above equation, we have:\n\nI2 = 10^-5 W/m^2 * (1m/5m)^2 = (10^-5 W/m^2)/25 = 4*10^-7 W/m^2\n\nThe intensity of a sound wave is proportional to the square of the amplitude of the displacement of a particle in the medium through which the sound is propagating:\n\nI = m*x^2,\n\nwhere m is a constant of proportionality, and x is the amplitude of the displacement.  Plugging this into the equation above that relates intensity and distance, we have that:\n\n(x2)^2 = (x1)^2 * (r1/r2)^2\nx2 = x1*r1/r2\n\nThe amplitude of the displacement is inversely proportional to 1/r.\n\nFor Q1 we are told that x1  = 70 microns at 1 meter, and are asked what x2 is at 5 meters.  Plugging these values into the above equation, we have:\n\nx2 = 70 microns * 1/5 = 14 microns\n\nFor Q2, we are told that x1 = 10^-7 m at r1 = 3 m, and are asked what the amplitude is at 5 m.  Again plugging these into the above equation, we have:\n\nX2 = 10^-7 m * (3/5) = 6*10^-8 m
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first range of time:\n***************************************\ninitial values: v0 = 0 m/s; t - t0 = 12.0 s; s0 = 0 m\na = 1.80 m/s²\n\nfinal velocity v:\nv = v0 + a(t-t0)\nv = 0 m/s + (1.80 m/s²)(12.0 s)\nv = 21.6 m/s\n\nfinal position s:\ns = s0 + v0*(t-t0) + 0.5(a)(t-t0)²\ns = 0 m + (0 m/s)(12.0 s) + 0.5(1.80 m/s²)(12.0 s)²\ns = 129.6 m\n\nsecond range of time: (final values of first range are initial values of second range)\n***************************************\ninitial values: v0 = 21.6 m/s; t - t0 = 8.0 s; s0 = 129.6 m\na = 0 m/s²\n\nfinal velocity v:\nv = v0 + a(t-t0)\nv = 21.6 m/s + (0 m/s²)(8.0 s)\nv = 21.6 m/s\n\nfinal position s:\ns = s0 + v0*(t-t0) + 0.5(a)(t-t0)²\ns = 129.6 m + (21.6 m/s)(8.0 s) + 0.5(0 m/s²)(12.0 s)²\ns = 302.4 m\n\nthird range of time: (final values of second range are initial values of third range)\n***************************************\ninitial values: v0 = 21.6 m/s; t - t0 = 10.0 s; s0 = 302.4 m\nfinal values: v = 0 m/s\na = ? m/s²\n\nacceleration a:\nv = v0 + a(t - t0)\na = (v - v0)/(t - t0)\na = (0 m/s - 21.6 m/s)/(10.0 s)\na = -2.16 m/s²\n\nfinal position s:\ns = s0 + v0*(t-t0) + 0.5(a)(t-t0)²\ns = 302.4 m + (21.6 m/s)(10.0 s) + 0.5*(-2.16 m/s²)(10.0 s)²\ns = 410.4 m
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hi chuck, \n\n5300 lbs of pressure?  yikes!  hate to be working near that thing if it should blow!\n\nanyway the point of a bladder accumulator is to expand/contract thereby absorbing transient fluctuations in the pressure, right?  so i would expect you would want it to be somewhere near the middle of its range - too empty and it won't be full enough to compensate for pressure dips, and too full and it won't be able to expand to handle pressure spikes.  so somewhere near halfway i.e. 60% sounds reasonable to me.  \n\nthis is one of those things where a calculation of the pressure depends on a LOT of weird variables like the elastic constant of the bladder, the shape of it etc.  your best bet to get maximum damping is probably to just do an empirical trial and error sort of thing.\n\n200 Hz is well within the audible range of the human ear.  G3 on the piano (find middle A, go down an octave plus one white key - that's low G) is supposed to be tuned to 196 Hz, so your 200 cps hum should sound almost like low G.  depending on the amplitude of the pulses you could probably just listen to the pipes, or use a stethoscope if it's too quiet, and then vary the charge pressure looking for a minimum in the hum.\n\ngood luck!
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When you copy and paste these, why don't you convert the "W"s into "ohms"?\n\nHey!  The guy who did your homework for you ("The Great One") gave three WRONG answers (as of 1930 MST 11-14.  He'll probably redo it after he reads this)...heh, heh, heh...
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The sound energy is being dissipated uniformly in all directions.\nThe intensity of the sound energy decreases inversely to\nthe distance squared.\n\nSo from a location 78m versus 22m, the decrease in intensity\nis a factor of (22/78)^2.  And teh actual intenisty is then\n\n3x10^-4 x (22/78)^2.
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1.\nPhases:\nf1 = (t/T1)*2*pi, where T1 = 1 sec\nf2 = (t/T2)*2*pi, T2 = 1 sec\nIn phase: f1 = f2 + 2*pi*n, where n is integer (0, +-1, +-2, ...)\n(t/T1)*2*pi = (t/T2)*2*pi + 2*pi*n\nt(1/T1 + 1/T2) = n\nT2 = 0.98 = 98/100 = 49/50, 1/T2 = 50/49\nt(49/49 - 50/49) = n\n-t/49 = n\nt = 49 seconds, n = -1\nNumber of oscillations if second, shorter pendulum (shorter pendulum has shorter period): N = t/T2 = 49/(49/50) = 50\n\n2.\nCoordinate: y(t) = 3*sin(20*pi*t)\nSpeed: v(t) = y'(t) = 20*pi*3*cos(20*pi*t)\nMean position: y(t) = 0. Take t0 = 0.\nMaximum displacement: y(t1) = amplitude = 3, speed = v(t1) = 0\nv(t1) = 0\ncos(20*pi*t1) = 0\n20*pi*t1 = pi/2\nt1 = 1/(2*20) = 1/40
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the velocity of fluid flow is proportional to area.\nfind the total area of one 4 mm tube compared to 2  mm tubes.  Here is a hint, they wont be the same.
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Wave length and frequency are inversions.\n\nThe average speed of sound is 343 meters per second.\n\nA 1 meter wavelength then would be a frequency of 1/343 hertz.\n\nA tube will produce most easily a wavelength equal to it's length.  It can also produce a wave length of twice it's length.\n\nAll other tones will be some division of these lengths.\n\nA 12 meter tube would produce wave lengths of 24m (twice the tube) 12m (equal), 6m (half), 4m (a third), 3m (a quarter), 2m (a sixth) etc.  They will all will be completely divisible into the tube.\n\nThat makes frequencies of 12/343 Hz, etc.
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Hello,\n\nIn this link you will have what you need:\n\nhttp://www.bsu.edu/web/fseec/pie/Lessons2/Physics/Potato%20Battery.doc\n\nI believe now will be OK.
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lol \nYou have homework that you don't know the answer to.  \n\nWhy would someone ask these here out of interest? Like you have these specific parameters and conditions that you are just raring to want to know just out of the blue.\n\n Quit wasting time writing and typing questions out waiting here for answers when you could be the next engineering graduate who actuallly UNDERSTANDS this stuff by doing it yourself.
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FORMULA: speed = frequency * wavelenght (sorry cant get symbols to work on yahoo)\n\n(1) 0.2= 1/0.5 * w   ...(as f=1/T)\n=> 0.2=2w\n=> w=0.1\n\n(2) s= 25*(10^8) * 0.12    ...(12cm= 0.12m)\n=> s= 3*10^8\n=> s= 300000000 m/s\n\n(3) 2= f * 0.5\n=> f= 2/0.5\n=> f= 8\n\n8 waves strike dock each second.\n\npretty easy questions.
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L+ e = C/4N\n\nL is the resonance length.\n\ne is the end correction \n\nC is the speed of sound in air\n\nN is the frequency of the fork.\n\nThe above equation can be written as \n\n4NL + 4Ne = C.\n\n\nFor first fork, 4N = 1024.\n\n\n1024 x 0.32 + 1024 e = C --------------- (1)\n\nFor the next fork, 4N = 1536\n\n1536 x 0.208 + 1536 e = C----------------- (2).\n\nEquating (2) and (1) \n\n1536 x 0.208 + 1536 e = 1024 x 0.32 + 1024 e \n\n512 e = 327.68 - 319.49 = 8.19 \n\ne = 0.016 m = 1.6 cm.
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Now there's some strange questions. I'll help as best I can. I hope I've understood the questions properly.\n\nQuestion 1) Bob will always win, assuming they play perfectly.\nIf Ann takes 1 matchstick, Bob should take 2. That leaves 3 left. Now whether Ann takes 1 or 2 from the remaining 3, Bob is left with 1 or 2, so he can take what's left and wins.\nIf Ann takes 2 matchsticks,  Bob should take 1. That also leaves 3 left, so again regardless of how many Ann takes from the 3 remaining ones, Bob can take what's left and win again.\n\nQuestion 2) The answer is 5.52pm.\nLets say the slow train moves at speed X. That means the fast train moves at speed 2.75X. This is because the fast one moves 2.75 times faster than the slow one (since it travels the same distance in 8 hours compared to the slow time of 22 hours, so it must be going 22/8 times faster).\nIn the time taken for the trains to meet, between them they have travelled the full distance. Therefore you can add their speeds together to calculate the relative approaching speed of the trains, which would be 2.75X + X = 3.75X.\nNow, the slow train travels at speed X and takes 22 hours to go the same distance. Therefore we can say that the time taken for the two trains to meet is 22X/3.75X, or 22/3.75, which is 5.8666.\nTo convert 0.8666 of an hour to minutes is simple - just multiply by 60. This gives 52.\n\nSo the total time taken is 5 hours 52 mins. \n\nTo put it another way:\nSpeed of train 1 = 2.75 times faster than train 2. So the speeds of the two trains together = 3.75 times faster than the slow train.\nThe slow train takes 22 hours to go the distance. Therefore the two trains together take 22/3.75 to do the same distance, which is 5.8666.\n\nI hope this helps :)
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Let f1 = 35.0 kHz\nvbat = 12 m/s\nvsound = 330 m/s\n\na) f2  = vsound*f1/(vsound-vbat) = 330*35/(330-12) = 36.32kHz\n\nb) For bat f = (vsound+vbat)*f2/vsound = 342*36.32/330 = 37.64kHz\n\nc) It will obviously affect the answers
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Sorry, but with the time you spent typing these questions you could of looked them up in your physics book or googled them.  Sometimes getting others to do your leg work isn't always the best answer to a problem.  Good Luck in your new class!