A narrow beam of ultrasonic waves reflects off the liver tumor. If the speed of the wave is 11.0% less in the liver than in the surrounding medium, determine the depth of the tumor. The angle of incidence with respect to the normal is 50 degrees.
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From the following link, look for "Chapt22, Problem-24".\n\nThis problem is similar, except the speed is 10% instead of 11%. I'm sure you can use the same formulae though.\n\nhttp://www.jca.umbc.edu/~george/html/courses/phys112/2004fall/homework/04fall_phys112_hk8.pdf
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It would certainly have to be a special one.  Standard audio equipment typlically has a frequency response of up to 20 KHz.  The Ultrasound frequencies range from 1 to 20 MHz - much higher frequency.\n\nUltrasound generally works on the same principles - just at a much higher frequency.  The sensitivity is part of the difficult part - there is a lot of noise.\n\nSonar uses lower frequency ranges - anything from audible (meaning in the ~ 1-2 KHz range - to 50 KHz and 200 Khz (two standard frequencies used).\n\nUltrasound is much more "delicate", using lower amplitude waves as compared to sonar.  Some sonar systems could actually cause hearing damage if you were under water close to the system.  The ultrasound is believed to be 100% safe because of the low amplitudes used.  This also means that ultrasound would be more prone to "noise" as compared to sonar, which can blast long distances and reflect of distant surfaces much like RADAR.  \n\nFinally, on a side note, RADAR works the same way as SONAR, except Radio Waves bouncing off objects are used instead of Sound Waves.
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I am a medical sonographer and have done many liver ultrasounds.  Not sure what your question is....I hope you get good results tomorrow!
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Another person had the same basic question. This is called the Doppler Effect, which follows the equation:\nf = f0 [(S +/- So)/(S -/+ Ss)]\nWhere f is the observed frequency, f0 is the frequency of the source, S is the speed of sound, So is the speed of the observer (+ if moving towards, - if away), and Ss is the speed of the source (- if moving towards, + if away).\nIn this case, just set f = 1.05f0
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You can easily calculate the speed of the transverse wave with the given information for distance travelled and time elapsed, with the formula s=vat where s is the distance va is the magnitude of velocity and t is the time.  Rearange that to get your speed.\n\nTHe speed in strings is dependent on the tension and the linear density of the string.  The formula is Vstring= square root (Tstring/linear density of the string)  Rearange this formula to calculate your tension.  Note:  the linear desnity can be calculated with the given information of the mass of the string and its distance...I think.
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Measure the speed of sound by help of an echo, a stopwatch and a tape-measure. (depending on temperature it is about 430 m/s.\nRepeat your measurements at least 10x and have the average.\nGood luck\n\nTh
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It's an ultrasound scanner.\n\nIt works by sending out a wave of very high frequency sound waves into the body. The sound waves are reflected back from boudaries between fluid and solid matter, where the physics of sound waves affects how the wave travels. The 'echo' is detected by the scanner and displayed graphically on a screen.\n\nUsed by radiographers in the UK and ultrasonographers / sonographers in the US.
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I can think of two ways to measure the absorption and reflection coefficient of a material.  The first would only work with a polished surface.  First measure the intensity of a laser beam with a photodetector, then shoot the laser beam at the surface and aim the reflection at a photodetector.  Coefficient of reflection = transmitted intensity/initial intensity  (coefficient of absorption =1-reflection).  The other, harder way, which would work with any material, is to measure the heat gain in the target.\n\nFor this you're gonna want as small a piece of the target material you can get so that the laser still completely hits it.  Now attach a thermistor to the back of the target (again the smaller and thinner the target the better, ideal would be a foil just thick enough to be completely opaque).  Now shoot the laser at the target and measure the temperature as a function of time  It should be linear up until it starts radiating nearly as fast as it absorbs energy, so take points until it starts to level off and drop points after that.  Now find the specific heat of the thermistor-target apparatus by dipping it in water with a thermometer in it, using the initial and final temperature of the water, the volume of the water, and the initial temp of the apparatus, determine the heat it absorbed per degree it's temperature changed (this will actually give you specific heat*mass which is fine).  Now convert your temperature vs. time data to heat vs. time using the specific heat you determined (temp*(specific heat*mass)).  The slope of the line of best fit for this data will give you the rate it absorbed energy in watts.  Divide this by the lasers output in watts and you get the absorption coefficient.
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Refrective index water should be the same throught, provided that all the other conditions are the same throughout -- such as temperature, presense of impurity, etc.\n\nIts not the refrective index of water alone that determines whether light will be reflected or not or at what angle the light should be projected to be reflected. When you are talking about the surface of the water, you are really talking about the interface between water and air. When you are talking about "under the water", light is not changing media.
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READ THIS CAREFULLY\n\nCritical angle is that angle of incedence in denser medium for which the angle of refraction is 90 degrees\n\nhence\nacc to snells law\n\nsin i/sinr = 1/refractive index\n\nhence\nsin i/1 = 1/1.45 [sin 90=1]\n           =0.69\ni= sin^-1[0.69][pronounced sin inverse of 0.69]\n =43.6
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y dont u do ur homework urself .ok heres abrief intro.\n\n1.see the level of frequency changes when beam of ultra violet light rays enter to one medium from another medium(to glass from air).\n\n2.when light undergoes refraction it frequency changes its speed acc. to the medium such as(glass,air and water)\n\n3.yes there will be,as light enters to one medium from another medium.
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High intensity focused ultrasound\n\nHIFU (high intensity focused ultrasound) (sometimes FUS or HIFUS) is a highly precise medical procedure using high-intensity focused ultrasound to heat and destroy pathogenic tissue rapidly. It is one modality of therapeutic ultrasound, and although it induces hyperthermia it should not be confused with this technique which heats much less rapidly and to much lower therapeutic temperatures (generally < 45oC).\n\nThis is typically under computerized MRI guidance, when it is sometimes called Magnetic Resonance guided Focused Ultrasound, often shortened to MRgFUS. Magnetic resonance imaging (MRI) is used to identify tumors or fibroids in the body, before they are destroyed by the ultrasound. MRgFUS is currently used in the US, Israel, Europe, and Asia to treat uterine fibroids. Current clinical trials are underway, examining the possible use of the technique in the treatment of cancers of the brain, breast, liver, and bone.\n\nTherapeutic ultrasound is a minimally invasive or non-invasive method to deposit acoustic energy into tissue. Applications include tissue ablation (HIFU) (for tumor treatments, for example), hyperthermia treatments (low-level heating combined with radiation or chemotherapy), or the activation or enhanced delivery of drugs.\n\nThe Foundation for Focused Ultrasound Research is promoting research into medical uses of high intensity focused ultrasound.\n\nThe International Society of Therapeutic Ultrasound may be found at: www.itsu.org\n\nAiming\n\nThe ultrasound beam can be focused in these ways:-\n\nGeometrically, for example with a lens or with a spherically curved transducer. \nElectronically, by adjusting the relative phases of elements in an array of transducers (a "phased array"). By dynamically adjusting the electronic signals to the elements of a phased array, the beam can be steered to different locations, and aberrations due to tissue structures can be corrected. \n\nHow HIFUS works\n\nAs an acoustic wave propagates through the tissue, part of it is absorbed and converted to heat. With focused beams, a very small focus can be achieved deep in tissues. When hot enough, the tissue is thermally coagulated. By focusing at more than one place or by scanning the focus, a volume can be thermally ablated. At high enough acoustic intensities, cavitation (microbubbles forming and interacting with the ultrasound field) can occur. Microbubbles produced in the field oscillate and grow (due to factors including rectified diffusion), and eventually implode (inertial or transient cavitation). During inertial cavitation, very high temperatures inside the bubbles occur, and the collapse is associated with a shock wave and jets that can mechanically damage tissue. Because the onset of cavitation and the resulting tissue damage can be unpredictable, it has generally avoided in clinical applications. However, cavitation is currently being investigated as a means to enhance HIFU ablation and for other applications.\n\nMethod of use\n\nIn HIFU therapy, ultrasound beams are focused on diseased tissue, and due to the significant energy deposition at the focus, temperature within the tissue rises to 65° to 85°C, destroying the diseased tissue by coagulation necrosis. Each "sonication" of the beams treats a precisely defined portion of the targeted tissue. The entire therapeutic target is treated by moving the applicator on its robotic arm in order to juxtapose multiple shots, according to a protocol designed by the physician. This technology can achieve precise "ablation" of diseased tissue, therefore being called HIFU surgery. Because it destroys the diseased tissue non-invasively, it is also known as "Non-invasive HIFU surgery". Anesthesia is not required. The treatment can be combined with radiotherapy or chemotherapy.\n\nUses\n\nUterine fibroids\n\nDevelopment of this therapy significantly broadened the range of treatment options for patient
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Ultrasound is the diagnostic procedure of choice for imaging the hepatobiliary system in patients who present with acute right upper quadrant pain. The sensitivity of ultrasound for detecting gallstones is >95%, and other adjacent structures can be simultaneously evaluated. CT scanning is not as accurate for detecting gallbladder wall abnormalities or gallstones; cholesterol stones can have the same density as bile.\n\n-random physician
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The last answer is correct.  The equation:\n\nA*sin(bx+c)\n\nCan be graphed.  A is the amplitude of the wave (how high along the y axis).  b is the frequency (how fast do you need to complete one wavelength).  c is a wavelength shift, how far away from x=0 do you have to start the wavelength.
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Check out this website. They discuss both direct and reverb loudness. It seems the steady-state reverb level depends on surface area and absorption coefficient while direct is 1/R^2 dependent. Unfortunately the symbols in their equations are not always defined right below the equation but they will be somewhere on the page. If you see a symbol that looks like 2 vertical sides and the top of a square, it's pi. Also they're not too careful about parentheses. One eq. you'll notice says\nWreverb = Wsource * (1 - a / Sa) but I'm betting it should be (1 - a) / Sa. (Otherwise they're using the fraction a/Sa which reduces to 1/S with 'a' irrelevant). Good luck. Maybe you'll find a more readable site.\nP.S. Of course one major effect you'll notice if your program material extends into the deep bass is a falloff of low-frequency response below about 22 Hz. From the ref., Fcutoff = c / 2 * longest dimension, should be c / (2 * longest dimension).
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simple  for smooth surfaces, the angle of incidence is equal to the angle of reflection\nso the angle of reflection is going to be the same angle mentioned in the question
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a)\n\n(change in frequency)/(recorded frequency) = (Speed of emitter)/(wave speed).\n\nChange in frequency = 22 Hz\nRecorded frequency = 820 Hz\nSpeed of emitter = what you are trying to find.\nWave speed = speed of sound = 340 m/s\n\n(22 / 820) x 340 = 9.12 m/s\n\nb)\n\nThe frequency of the echo is (I'm guessing for this one sorry) double the regular frequency, which is 1684 Hz.\n\nc) \n\nThe frequency heard by the man is 820 Hz by definition isn't it? Thats the frequency as the boy passes anyway.\n\n[EDIT] Wait I just realised your whole question makes no sense. Due to the limited info in the question I was assuming you meant it falls to 820 Hz from the emission frequency at the point where the boy passes the man. But why would the frequency fall? The point where the boy passes the man is the point where the frequency experienced by the man is highest.\n\nOh well I hope my answer helped a bit anyway.