tag:blogger.com,1999:blog-2214461049219354662.post168656856952278346..comments2024-03-28T13:08:35.581+04:00Comments on Physics Reference: Physics 9702 Doubts | Help Page 143Unknownnoreply@blogger.comBlogger22125tag:blogger.com,1999:blog-2214461049219354662.post-88378602126468764462020-08-23T23:02:59.938+04:002020-08-23T23:02:59.938+04:00the explanation has been updatedthe explanation has been updatedAdminhttps://www.blogger.com/profile/03708185681043991662noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-31830782875411519122020-08-23T14:32:43.749+04:002020-08-23T14:32:43.749+04:00can you please help me understand why for question...can you please help me understand why for question 708 solution (c) (i) above why the answer is 'from bottom to top'? [june 2011 paper 41 question 5 c(i)].. <br />thank youDrewhttps://www.blogger.com/profile/04590736321387982341noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-636752563741457252019-12-04T22:59:21.660+04:002019-12-04T22:59:21.660+04:00go to
http://physics-ref.blogspot.com/2017/12/9702...go to<br />http://physics-ref.blogspot.com/2017/12/9702-june-2015-paper-41-43-worked.htmlAdminhttps://www.blogger.com/profile/03708185681043991662noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-52719587281906853872019-12-04T14:31:55.970+04:002019-12-04T14:31:55.970+04:00where can i find the answer to may-june, 2015, pap...where can i find the answer to may-june, 2015, paper 41 question 6 b (ii) ?hmmmmhttps://www.blogger.com/profile/05979039611598352333noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-1509333979346044722019-10-16T18:29:57.309+04:002019-10-16T18:29:57.309+04:002015 may/june 9702/42 ...question number 5?
2015 may/june 9702/42 ...question number 5?<br />Anonymoushttps://www.blogger.com/profile/15359844870767650795noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-23575738066252243222019-03-07T22:34:29.323+04:002019-03-07T22:34:29.323+04:00From Fig 5.2, we can observe that the path is NOT ...From Fig 5.2, we can observe that the path is NOT the arc of a circle as in Fig 5.1. <br />We are 2 options for the direction of the path: either from right to left or from bottom to top of the diagram. <br />As the charged particle passes through the coil, it loses some kinetic energy. Its speed v decreases and so, the radius of the path also decreases (since r = mv / Bq). In the diagram, this reduction can be observed as we pass through the foil from the bottom to top and NOT from right to left along the path.<br />Adminhttps://www.blogger.com/profile/03708185681043991662noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-77504060508793571362019-03-07T14:00:41.569+04:002019-03-07T14:00:41.569+04:00why is the direction from top to bottom, and not i...why is the direction from top to bottom, and not its reverse?saadia zia ulhaquehttps://www.blogger.com/profile/14095909835126113878noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-60900567966692248652017-09-13T21:08:56.170+04:002017-09-13T21:08:56.170+04:00Consider 2 forces: 10N and 7N both opposing each o...Consider 2 forces: 10N and 7N both opposing each other.<br />THe resultant would be 3N in the same direction as the 10N force.<br /><br />SImilarly, the centripetal is the resultant of the 2 other forces and there is no problem for it to be in the same direction as one of them. Here the centripetal force is in the same direction as the force of gravity.Adminhttps://www.blogger.com/profile/03708185681043991662noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-59536909259715682322017-09-13T15:39:43.202+04:002017-09-13T15:39:43.202+04:00Pls try to reply asap....
Force of gravitation is...Pls try to reply asap....<br /><br />Force of gravitation is towards the centre and F(centripetal) is also centre directed so how ????? F(normal) becomes F(gravity)-F(centripetal).<br /><br />Normal reaction is balancing both the forces by acting in opposite direction....<br />or Fn should have been Fn= Fg + FcAnonymoushttps://www.blogger.com/profile/09241325219450184999noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-23025098826080349612016-10-07T21:26:23.332+04:002016-10-07T21:26:23.332+04:00Acceleration = Rω^2 where R is the perpendicular d...Acceleration = Rω^2 where R is the perpendicular distance from the axis. At the poles, the value of R is zero. So, the acceleration = 0Adminhttps://www.blogger.com/profile/03708185681043991662noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-74958209266765534302016-10-07T19:46:17.045+04:002016-10-07T19:46:17.045+04:00why is the acceleration 0 at the axis??why is the acceleration 0 at the axis??Anonymoushttps://www.blogger.com/profile/14115447598963293624noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-54541815802561025762016-08-31T10:44:14.402+04:002016-08-31T10:44:14.402+04:00could you read again. It seems that the explanatio...could you read again. It seems that the explanation is already available above - (b)(i) in red.Adminhttps://www.blogger.com/profile/03708185681043991662noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-17868109784853477372016-08-31T07:08:18.327+04:002016-08-31T07:08:18.327+04:00Why when mrw^2 =0 then normal reaction =0? Isn'...Why when mrw^2 =0 then normal reaction =0? Isn't the normal reaction at the pole= GMm/r^2 - mRw^2? So the normal reaction will be equal to the gravitational force isn't it๑ YUSHAN ♥https://www.blogger.com/profile/12381172128563255594noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-49661612394222664682015-11-26T13:56:55.531+04:002015-11-26T13:56:55.531+04:00By accelerating the electrons through different p....By accelerating the electrons through different p.d.`s, electrons with different de Broglie wavelength are produced. Now, if one of these electrons is diffracted by a crystal, it means that the value of its de Broglie wavelength (this can be calculated as above) is similar to the separation of the atoms (this is inferred due to the diffraction taking place). Thus, diffraction allows us to predict the separation of the atoms in the crystal.Adminhttps://www.blogger.com/profile/03708185681043991662noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-33357282463424786782015-11-25T18:54:20.598+04:002015-11-25T18:54:20.598+04:00Solution 705 (b)(ii)
Could you please explain the ...Solution 705 (b)(ii)<br />Could you please explain the answer because I'm not sure what the question or answer is trying to specify...Anonymoushttps://www.blogger.com/profile/00889217957296109314noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-13254777335726201642015-08-26T10:54:39.727+04:002015-08-26T10:54:39.727+04:00The path is already shown in the diagram - it'...The path is already shown in the diagram - it's the curved line. You only need to insert the head of an arrow to indicate its direction. This is from the bottom to the top along the path. <br /><br />There could only be 2 directions - either the one stated above, or its reverse - both are along the path shown - the curved line.Adminhttps://www.blogger.com/profile/03708185681043991662noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-41051581137017674772015-08-25T22:06:49.131+04:002015-08-25T22:06:49.131+04:00For Q.708 Please show the path of the particle in ...For Q.708 Please show the path of the particle in part (c)(i)Anonymoushttps://www.blogger.com/profile/11459103661460827376noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-51663676760464373652015-08-24T17:27:33.090+04:002015-08-24T17:27:33.090+04:0041/M/J/11 Q.5(c)(i) Can you please show the path o...41/M/J/11 Q.5(c)(i) Can you please show the path of the particle?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-41907024588932000132015-06-03T17:11:38.789+04:002015-06-03T17:11:38.789+04:00Check solution 803 at
http://physics-ref.blogspot....Check solution 803 at<br />http://physics-ref.blogspot.com/2015/06/physics-9702-doubts-help-page-161.htmlAdminhttps://www.blogger.com/profile/03708185681043991662noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-55179966986169103192015-06-03T12:07:35.775+04:002015-06-03T12:07:35.775+04:00can you please explain oct/nov 2012, variant -11, ...can you please explain oct/nov 2012, variant -11, (p-1) , Q.21? :)Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-31084680289742842942015-05-17T14:55:05.029+04:002015-05-17T14:55:05.029+04:00For 12/O/N/10 Q.33, go to
http://physics-ref.blogs...For 12/O/N/10 Q.33, go to<br />http://physics-ref.blogspot.com/2014/11/9702-november-2010-paper-12-worked.htmlAdminhttps://www.blogger.com/profile/03708185681043991662noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-11938292917578817432015-05-17T00:42:28.314+04:002015-05-17T00:42:28.314+04:0012/O/N/10 Q.33
11/M/J/11 Q.9,1012/O/N/10 Q.33<br />11/M/J/11 Q.9,10Anonymoushttps://www.blogger.com/profile/11459103661460827376noreply@blogger.com