tag:blogger.com,1999:blog-2214461049219354662.post8783481903703789151..comments2024-03-28T13:08:35.581+04:00Comments on Physics Reference: Physics 9702 Doubts | Help Page 84Unknownnoreply@blogger.comBlogger12125tag:blogger.com,1999:blog-2214461049219354662.post-7569589327081795912019-05-07T15:35:06.874+04:002019-05-07T15:35:06.874+04:00see the updated link added abovesee the updated link added aboveAdminhttps://www.blogger.com/profile/03708185681043991662noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-4068825683060627882019-05-07T15:34:34.420+04:002019-05-07T15:34:34.420+04:00the solution has been updated with much more detai...the solution has been updated with much more details.<br /><br />see the link aboveAdminhttps://www.blogger.com/profile/03708185681043991662noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-1095635983474118192019-05-07T14:32:07.383+04:002019-05-07T14:32:07.383+04:00Q443).
Why did u take
10^-2. When considering the...Q443).<br />Why did u take <br />10^-2. When considering the thickness of 2X. Because data is given in meters?<br />So my answer must also in metres and then I convert into cm but I don't get the answer when doing so. <br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-88699420923185956302019-05-06T12:35:33.278+04:002019-05-06T12:35:33.278+04:00I still dont undertand this. so we say that instan...I still dont undertand this. so we say that instantly after the reflection t the fat muscle boundary, the intensity was .018I. and at S it was .012I. so basically it attenuated from .018I to .012I after travelling a distance x through the fat. it did NOT travel through x twice. it traveled straight from the interface to S. then why is the distance 2x?<br />imranhttps://www.blogger.com/profile/16195872191077742475noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-40415287109844910232019-03-25T22:21:32.229+04:002019-03-25T22:21:32.229+04:00The ultrasound moves through a distance x of fat t...The ultrasound moves through a distance x of fat to the boundary where some of waves are transmitted and others reflected. The reflected wave would also need to move a distance x in the fat to reach surface S where it is detected.Adminhttps://www.blogger.com/profile/03708185681043991662noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-53304676125808895022019-03-25T15:23:30.598+04:002019-03-25T15:23:30.598+04:00Since we're taking the reflected intensity as ...Since we're taking the reflected intensity as the incident intensity, it means we're only considering the reflection of ultrasound from boundary to surface X. Then why double the distance x?<br /> Anonymoushttps://www.blogger.com/profile/00738361939316421844noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-44968568122645288882019-02-13T18:02:34.581+04:002019-02-13T18:02:34.581+04:00α = IR / I = 0.018
So, 0.018 of the intensity has...α = IR / I = 0.018<br /><br />So, 0.018 of the intensity has been reflected at the interface. <br /><br />Now, this reflected intensity will move through the fat towards surface S.<br /><br />Considering this intensity from the boundary towards the surface S, we can take this reflected intensity as the new incident intensity. The ultrasound will move from the boundary towards surface S.<br /><br />After being transmitted in the fat, the intensity at surface S is given to be 0.012 I. <br /><br />So, the intensity was 0.018 I at the boundary, and as the ultrasound moves through the fat, it got attenuated and the new intensity at the surface S is now 0.012 I.<br />Adminhttps://www.blogger.com/profile/03708185681043991662noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-8251579515768222952019-02-12T19:09:55.362+04:002019-02-12T19:09:55.362+04:00hi for question 443, why Incident intensity is 0.0...hi for question 443, why Incident intensity is 0.018?<br />Anonymoushttps://www.blogger.com/profile/05554585538662322233noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-85418895200444910572016-12-03T18:05:17.090+04:002016-12-03T18:05:17.090+04:00the ultrasound is first incident in the fat. It tr...the ultrasound is first incident in the fat. It travels a distance x. Upon reflection, it returns along the fat itself towards the detector. This is also a distance x.<br /><br />Total distance travelled = x+x = 2xAdminhttps://www.blogger.com/profile/03708185681043991662noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-49744517917846155892016-11-30T07:39:04.814+04:002016-11-30T07:39:04.814+04:00Sir, for question 443, why is the thickness 2x X 1...Sir, for question 443, why is the thickness 2x X 10^-2?Anonymoushttps://www.blogger.com/profile/11606677351807860244noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-62941312902423124252015-09-04T18:58:44.961+04:002015-09-04T18:58:44.961+04:00I diagram has been added. Read the explanation aga...I diagram has been added. Read the explanation again and see if you understand.Adminhttps://www.blogger.com/profile/03708185681043991662noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-34952799803337381542015-09-03T18:46:04.114+04:002015-09-03T18:46:04.114+04:00Sir I completely don't understand Question 440...Sir I completely don't understand Question 440. It looks to me like A is going up and C is going down. I understand about the speed but I don't understand the upward/downward directions. It only makes sense that C is going down if we consider the wave to be the white part of the diagram, not the gray part. If a surfer were on this wave of water, and he was at point A, wouldn't he be going upwards, towards the crest of the wave? asaakihttps://www.blogger.com/profile/03891564210444946590noreply@blogger.com