tag:blogger.com,1999:blog-2214461049219354662.post4182109707700938053..comments2024-03-28T13:08:35.581+04:00Comments on Physics Reference: Physics 9702 Doubts | Help Page 20Unknownnoreply@blogger.comBlogger12125tag:blogger.com,1999:blog-2214461049219354662.post-90803288515080075472019-11-10T23:04:48.232+04:002019-11-10T23:04:48.232+04:00go to
http://physics-ref.blogspot.com/2018/10/9702...go to<br />http://physics-ref.blogspot.com/2018/10/9702-june-2016-paper-13-worked.htmlAdminhttps://www.blogger.com/profile/03708185681043991662noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-10631700939283469202019-11-10T18:22:30.125+04:002019-11-10T18:22:30.125+04:009702/13/M/J/16 question 35 my exam is on the 15th,...9702/13/M/J/16 question 35 my exam is on the 15th, would be really helpful if I can get the explanation beforehand, thank you. Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-51148440187488978942019-04-10T21:45:45.275+04:002019-04-10T21:45:45.275+04:00go to
http://physics-ref.blogspot.com/2018/09/9702...go to<br />http://physics-ref.blogspot.com/2018/09/9702-june-2016-paper-12-worked.htmlAdminhttps://www.blogger.com/profile/03708185681043991662noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-52424701887272005662019-04-10T21:17:05.247+04:002019-04-10T21:17:05.247+04:00problem on Q.33 9702/12/MJ/2016
problem on Q.33 9702/12/MJ/2016<br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-59263431245093934692019-03-17T22:22:35.742+04:002019-03-17T22:22:35.742+04:00Note that it is NOT millimetre.
unit of h = m
Unit...Note that it is NOT millimetre.<br />unit of h = m<br />Unit of g = m s^-2<br /><br />So, unit of hg = m . m s^-2 <br /><br />For the equation to be homogeneous <br />the overall base unit on the left hand side should be the same as the overall unit on the right hand side. <br /><br />To make the unit equal to that of speed (m s^-1), the powers of 2 should be reduced to one. This is true if then value of n is ½. Adminhttps://www.blogger.com/profile/03708185681043991662noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-72496952338484689202019-03-17T03:01:50.356+04:002019-03-17T03:01:50.356+04:00I don't understand why choose your in millimet...I don't understand why choose your in millimeter and how you got 1/2<br />Anonymoushttps://www.blogger.com/profile/01971684386791420954noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-75608126846364350612019-03-07T20:17:48.728+04:002019-03-07T20:17:48.728+04:00thanks
for
qn 108
thanks<br /> for<br /> qn 108 <br />7th_Intellectualhttps://www.blogger.com/profile/06797999955160410650noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-1058073933843209602016-03-30T00:32:16.700+04:002016-03-30T00:32:16.700+04:00Thank you .Thank you .Anonymoushttps://www.blogger.com/profile/02204003080621355122noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-52912743145257882742015-05-01T14:52:28.377+04:002015-05-01T14:52:28.377+04:00From Ohm’s law: I = V / R
Increasing the resistanc...From Ohm’s law: I = V / R<br />Increasing the resistance R causes the current induced in the circuit to decrease.<br /><br />Power dissipated = (I^2)R. <br /><br />The power dissipated depends on the square of current I. So, as I decreases, the power dissipated also decreases, even if the resistance R increases. The dependence of current I is higher than on the resistance.<br /><br />So, less energy is loss. Since this energy is derived from the oscillation of the magnet, less energy is now taken from the oscillation. So, there is less damping.<br />Adminhttps://www.blogger.com/profile/03708185681043991662noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-87242186950560352862015-05-01T13:30:15.685+04:002015-05-01T13:30:15.685+04:00For question 106 b), why is there less damping as ...For question 106 b), why is there less damping as resistance of resistor R is increased ?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-58713387049105144312015-04-27T09:33:38.042+04:002015-04-27T09:33:38.042+04:00Time for 1 revolution = period of 1 pulse = 20ms
...Time for 1 revolution = period of 1 pulse = 20ms<br /><br />For A,<br />the setting is chosen such that 1cm (1 square on the screen) represents a time of 1s. But 1 pulse takes 20ms. In 1 second, there are 1/0.02 = 50 pulses. 20ms = 0.02s<br /><br />For B,<br />1cm represents 10ms. The period is 20ms. So, in every 2cm (2 squares), a pulse is seen. Since the screen is 10cm, 5 pulses will be seen. This is good enough to analyse the pulses.<br /><br /><br /><br />Calculating in seconds seemed easier. And actually, I just started solving without even thinking about whether it's better to calculate in cm or second. But, now I think that it's really better to calculate in second.Adminhttps://www.blogger.com/profile/03708185681043991662noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-5347720281867315822015-04-22T08:41:31.194+04:002015-04-22T08:41:31.194+04:00In question 105 why didn't you calculate in ce...In question 105 why didn't you calculate in centimetres? how did you do A and B?Anonymoushttps://www.blogger.com/profile/09821574713784543809noreply@blogger.com