tag:blogger.com,1999:blog-2214461049219354662.post8925256708446343633..comments2024-03-28T13:08:35.581+04:00Comments on Physics Reference: Physics 9702 Doubts | Help Page 107Unknownnoreply@blogger.comBlogger27125tag:blogger.com,1999:blog-2214461049219354662.post-69394862890215766232019-08-06T22:39:38.627+04:002019-08-06T22:39:38.627+04:00yes, it could also be this way. this is arbitrary....yes, it could also be this way. this is arbitrary. as long as we have the curve section, then the zero p.d. line one after the otherAdminhttps://www.blogger.com/profile/03708185681043991662noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-89659121907366237032019-08-06T15:27:43.400+04:002019-08-06T15:27:43.400+04:00Hi, In q546(c), why the new curve go up from 0V to...Hi, In q546(c), why the new curve go up from 0V to peak V for half cycle first, then becomes 0V for next half cycle? Can it be the other way round, as if a.c. supply current from X to Y first, and diode A that cannot function, causes 1st half cycle to be 0V, then 2nd half cycle where it changes direction of current from Y to X, so starts to increase from 0V to peak? ThxElinahttps://www.blogger.com/profile/15429083332480525033noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-5800751249212386572019-04-30T18:44:13.538+04:002019-04-30T18:44:13.538+04:00Thank youThank youAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-29839002934226782502019-04-28T21:51:21.966+04:002019-04-28T21:51:21.966+04:00The way the diodes are connected always make the t...The way the diodes are connected always make the top of the resistor positive.<br /><br />Consider current flowing from X (that is, X is positive). At the junction, current cannot flow through diode D as it is (here) reversed-biased (it is opposite to the direction of current). Current flows through diode A. At the next junction, current cannot flow through diode B but it goes to the resistor. So, the top is positive.<br /><br />Now, consider current flowing from Y (that is, Y is positive). At the junction, current cannot flow through diode C as it is (here) reversed-biased (it is opposite to the direction of current). Current flows through diode B. At the next junction, current cannot flow through diode A but it goes to the resistor. So, the top is again positive.<br />Adminhttps://www.blogger.com/profile/03708185681043991662noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-86657566886894215832019-04-28T20:47:33.969+04:002019-04-28T20:47:33.969+04:00can you pls tell me why the top of the load isposi...can you pls tell me why the top of the load ispositive in the rectifier questionAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-60126138614986222592019-03-07T22:39:55.424+04:002019-03-07T22:39:55.424+04:00The potential energy is proportional to the produc...The potential energy is proportional to the product of the charges. Doubling the charge would double the energy.<br /><br />Gradient = Δy / Δx <br /><br />Since the potential energy is on the y-axis, if it is doubled, the gradient also increases.<br />Adminhttps://www.blogger.com/profile/03708185681043991662noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-31673062846922380002019-03-07T13:47:31.880+04:002019-03-07T13:47:31.880+04:00In q.551 kindlyexplain the last part.. how does th...In q.551 kindlyexplain the last part.. how does the gradient doubles by doubling charge??saadia zia ulhaquehttps://www.blogger.com/profile/14095909835126113878noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-91032599859338236632019-02-24T23:23:58.905+04:002019-02-24T23:23:58.905+04:00the graph is that of potential energy against dist...the graph is that of potential energy against distance.<br /><br />so, (potential) gradient = y-axis / x-axis<br /> = potential energy / distance<br /><br />So, the gradient is proportional to to potential energy.Adminhttps://www.blogger.com/profile/03708185681043991662noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-17880297499074174082019-02-24T15:56:45.303+04:002019-02-24T15:56:45.303+04:00Why is the potential gradient proportional to the ...Why is the potential gradient proportional to the gradient of the potential energy curve in question 551? hoorhttps://www.blogger.com/profile/17127145957601988502noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-28182648778768582952019-01-26T11:33:38.238+04:002019-01-26T11:33:38.238+04:00well, the derivation shows the correct answer.
Fo...well, the derivation shows the correct answer.<br /><br />For square roots, we cannot simply add numbers inside the root directly for addition. If that was multiplication, then that would have been possibleAdminhttps://www.blogger.com/profile/03708185681043991662noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-65030179912223174172019-01-26T11:09:27.683+04:002019-01-26T11:09:27.683+04:00WHY CANT THE ANSWER BE "D" IN Q.549?WHY CANT THE ANSWER BE "D" IN Q.549? saadia zia ulhaquehttps://www.blogger.com/profile/14095909835126113878noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-40271966206392551172018-02-08T22:16:44.645+04:002018-02-08T22:16:44.645+04:00divided by timedivided by timeAdminhttps://www.blogger.com/profile/03708185681043991662noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-89128487360881530692018-02-08T05:50:19.003+04:002018-02-08T05:50:19.003+04:00This comment has been removed by the author.leenah jawaathhttps://www.blogger.com/profile/09576944185545616791noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-69019151329839003242017-12-03T22:09:03.406+04:002017-12-03T22:09:03.406+04:00The question has been solved at
http://physics-ref...The question has been solved at<br />http://physics-ref.blogspot.com/2017/12/a-digital-voltmeter-with-three-digit.htmlAdminhttps://www.blogger.com/profile/03708185681043991662noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-11956097199981481632017-12-01T16:34:56.522+04:002017-12-01T16:34:56.522+04:00Plz post
june 2010 paper 23 question 1 solution
...Plz post <br /> june 2010 paper 23 question 1 solution<br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-47745351536967619762017-05-07T14:14:18.236+04:002017-05-07T14:14:18.236+04:00No, most of the resistance is due to the voltmeter...No, most of the resistance is due to the voltmeter. From V = IR, it will have a p.d. of 2V across it as the other resistance is negligible compared to that of the voltmeterAdminhttps://www.blogger.com/profile/03708185681043991662noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-41254458598211514002017-05-07T07:31:22.420+04:002017-05-07T07:31:22.420+04:00Shouldn't it be zero for voltmeter reading? As...Shouldn't it be zero for voltmeter reading? As the voltmeter is incorrectly installed, it should be connected in parallel to the part of the circuit you wish to measure. It's true that resistance of the voltmeter is so high that it 'takes up' all the p.d but it shouldn't be giving any reading isn't it so?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-18861414194208100852015-11-25T13:52:26.206+04:002015-11-25T13:52:26.206+04:00That's the formula for electric potential, not...That's the formula for electric potential, not electric potential energy. For formula in the explanation above is the correct one - it's the product of the electric potential with the charge.Adminhttps://www.blogger.com/profile/03708185681043991662noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-32657007704762121182015-11-24T23:37:48.540+04:002015-11-24T23:37:48.540+04:00I'm sorry, but I don't get question 551.
I...I'm sorry, but I don't get question 551.<br />Isn't the formula for electric potential energy: V = k · q1 / r? So, why do we have to multiply both charges: q1 and q2? In any case, what I would do it would be to add both potentials (scalar quantity). Any help will be very appreciatedAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-27597477134920375122015-05-29T22:13:59.576+04:002015-05-29T22:13:59.576+04:00Yeah, but here we are given a graph - so we should...Yeah, but here we are given a graph - so we should consider the momentum as displayed in the graph.<br /><br />+ we want the magnitude.<br />With the direction, the change would be (consider the example using numbers) -6 - 8 = -14<br />same value but with a negative signAdminhttps://www.blogger.com/profile/03708185681043991662noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-14413962784566171742015-05-29T22:02:22.205+04:002015-05-29T22:02:22.205+04:00but shouldn't the change be final minus initia...but shouldn't the change be final minus initial, p2-p1?<br />qdkhanhttps://www.blogger.com/profile/18294886518542671162noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-9887414945508996962015-05-29T17:25:11.161+04:002015-05-29T17:25:11.161+04:00Consider the following example involving number fo...Consider the following example involving number for better understanding.<br />Let’s say momentum changes from +8 to -6. On a graph of momentum against time, (we will ignore the time here because it is obvious), the ‘+8’ would be above the time-axis (x-axis) and the ‘-6’ would be below the time-axis. The time-axis would be drawn at ‘momentum = 0’.<br /><br />The MAGNITUDE of the change is as follow:<br />From +8 to 0, we have a magnitude of 8<br />From 0 to -6, we have a magnitude of 6<br />The total magnitude of the change is 8 + 6 = 14<br /><br />Now, let’s try to write it in a single equaltion:<br />Magnitude of change = 8 – (-6) = 14<br /><br />If we compare with the question, p1 = 8 and p2 = -6<br />So, magnitude of change = p1 – p2<br /><br />Here the question asks for magnitude, so we consider only the magnitude.<br />Adminhttps://www.blogger.com/profile/03708185681043991662noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-79003196308125662572015-05-28T23:05:52.634+04:002015-05-28T23:05:52.634+04:00bt how is the momentum change P1-P2? can you pleas...bt how is the momentum change P1-P2? can you please explain ASAP with the directions of the changes etc?<br />qdkhanhttps://www.blogger.com/profile/18294886518542671162noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-9226999869639238312015-05-27T16:53:30.390+04:002015-05-27T16:53:30.390+04:00Sorry, that was a mistake. I have already correctl...Sorry, that was a mistake. I have already correctly it.<br />ThanksAdminhttps://www.blogger.com/profile/03708185681043991662noreply@blogger.comtag:blogger.com,1999:blog-2214461049219354662.post-44764291980192286392015-05-27T15:53:47.896+04:002015-05-27T15:53:47.896+04:00In 550, it says 'the voltmeter reading will be...In 550, it says 'the voltmeter reading will be zero' I am confused Just 'Me'https://www.blogger.com/profile/09742741295678489085noreply@blogger.com