Ultrasound waves and the kidney

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Homework Help Overview

The discussion revolves around a problem involving ultrasound waves used in a kidney study, specifically focusing on the calculation of the kidney's width based on reflected ultrasound signals and their respective travel times through different tissues.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of the kidney's width by analyzing the time differences of ultrasound reflections. Questions are raised about the reasoning behind the calculations and the interpretation of the travel paths of the ultrasound waves.

Discussion Status

Some participants are attempting to clarify their reasoning and the steps taken in their calculations. There is an acknowledgment of a potential misunderstanding regarding the time intervals and how they relate to the distances being calculated. Guidance is being sought on the interpretation of the problem and the calculations involved.

Contextual Notes

Participants are working under the constraints of the problem statement and the provided data, with some expressing confusion over the correct interpretation of the time intervals and the necessary calculations to determine the kidney's width.

Kolika28
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Homework Statement


In a kidney study, ultrasound was used with frequency 3.5 MHz. In normal body tissue, the speed of the ultrasound is 1.50 km / s, while the speed in the kidney tissue is 1.55 km / s. Reflected signals came as follows: after 20 microseconds from the skin, after 75 microseconds from the front of the kidney, and after 152 microseconds from the back of the kidney.
Find out how long the kidney is across.

Homework Equations


S=v*t

The Attempt at a Solution


(152-75-20)/2=28,5
S=v*t=(1,55*1000)*(28,5*10^(-6)=0,044175

But the answer is 6 cm
 
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Kolika28 said:

The Attempt at a Solution


(152-75-20)/2=28,5
S=v*t=(1,55*1000)*(28,5*10^(-6)=0,044175
Please elaborate your reasoning; can you explain in words what your calculation is trying to accomplish?

Have you made a sketch of the sound paths?
 
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Kolika28 said:

Homework Statement


In a kidney study, ultrasound was used with frequency 3.5 MHz. In normal body tissue, the speed of the ultrasound is 1.50 km / s, while the speed in the kidney tissue is 1.55 km / s. Reflected signals came as follows: after 20 microseconds from the skin, after 75 microseconds from the front of the kidney, and after 152 microseconds from the back of the kidney.
Find out how long the kidney is across.

Homework Equations


S=v*t

The Attempt at a Solution


(152-75-20)/2=28,5
S=v*t=(1,55*1000)*(28,5*10^(-6)=0,044175

But the answer is 6 cm

Explain to yourself the path of the ultrasound, broken down into stages with the time for each.
 
Ok, I can try to explain. 152 microseconds is the time it takes to the back of the kindney and back, therefore I divide with 2, because I only need one way. I don't need to find the length from the skin to the back of the kidney, so therefore I subtract 75 and 20. Then I multiply seconds I'm left with with the speed through the kidney tissue. I don't see what I have done wrong. Could someone tell me?
 
My bad, I should have read the question better. It should be (152-75)/2. Thank's for your help!
 

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