tag:blogger.com,1999:blog-6403799362875544461.post1214340919478694356..comments2024-03-21T12:42:53.267+08:00Comments on Mr Sai Mun's Blog: SPM Physics 2016 Paper 1 Answersmr sai munhttp://www.blogger.com/profile/17212160178411112789noreply@blogger.comBlogger23125tag:blogger.com,1999:blog-6403799362875544461.post-57709024364580692552021-06-02T12:35:13.880+08:002021-06-02T12:35:13.880+08:00im sorry but that is accelerataion-time graph righ...im sorry but that is accelerataion-time graph right?The question is velocity-time graph<br /><br /><br />Anonymoushttps://www.blogger.com/profile/05785116670911240547noreply@blogger.comtag:blogger.com,1999:blog-6403799362875544461.post-5832166631986908242019-10-09T22:32:24.217+08:002019-10-09T22:32:24.217+08:00I agree with youI agree with youmrsm studenthttps://www.blogger.com/profile/10666325814642688485noreply@blogger.comtag:blogger.com,1999:blog-6403799362875544461.post-89101689775138673392019-10-09T22:31:18.722+08:002019-10-09T22:31:18.722+08:00Use the formula infront which is s=ut+1/2at^2 oh a...Use the formula infront which is s=ut+1/2at^2 oh and btw the english translation of the question is wrong in the book,refer the malay translation mrsm studenthttps://www.blogger.com/profile/10666325814642688485noreply@blogger.comtag:blogger.com,1999:blog-6403799362875544461.post-3693053227778504112019-03-28T13:54:17.915+08:002019-03-28T13:54:17.915+08:00can someone explain to me how to get answer for no...can someone explain to me how to get answer for no. 10? Thank you.Anonymoushttps://www.blogger.com/profile/06611736887827313668noreply@blogger.comtag:blogger.com,1999:blog-6403799362875544461.post-92041180350525484652016-11-30T16:02:18.827+08:002016-11-30T16:02:18.827+08:00nope,from question,we need to produce a big,uprigh...nope,from question,we need to produce a big,upright(virtual) image,if the answer is B,the image will be a upright(virtual) but deminished image,still D is the correct answerAnonymoushttps://www.blogger.com/profile/00129520714030197328noreply@blogger.comtag:blogger.com,1999:blog-6403799362875544461.post-83257382199866695692016-11-25T13:21:48.594+08:002016-11-25T13:21:48.594+08:00No. 23 is convex, isn't it? It's BNo. 23 is convex, isn't it? It's B Anonymoushttps://www.blogger.com/profile/10662271436338956633noreply@blogger.comtag:blogger.com,1999:blog-6403799362875544461.post-35219996048142031812016-11-24T13:39:47.796+08:002016-11-24T13:39:47.796+08:00To be honest, regarding on Question 4.
It is test...To be honest, regarding on Question 4. <br />It is testing your knowledge on terminal velocity. <br /><br />As the gradient is positive, it is called acceleration.<br />As the gradient is negative, it is called deceleration.<br /><br />Regarding my comments in this question, I would say a badly made question and should be removed and not considered in digital marking.iSean1997https://www.blogger.com/profile/18209674306465538891noreply@blogger.comtag:blogger.com,1999:blog-6403799362875544461.post-62112149090441732542016-11-24T01:17:29.024+08:002016-11-24T01:17:29.024+08:00Cn i get science...? Cn i get science...? Anonymoushttps://www.blogger.com/profile/02751539869795275254noreply@blogger.comtag:blogger.com,1999:blog-6403799362875544461.post-32866416051926486362016-11-24T01:17:18.135+08:002016-11-24T01:17:18.135+08:00Cn i get science...? Cn i get science...? Anonymoushttps://www.blogger.com/profile/02751539869795275254noreply@blogger.comtag:blogger.com,1999:blog-6403799362875544461.post-865423150079535562016-11-23T18:50:07.864+08:002016-11-23T18:50:07.864+08:00There is no such rule that an acceleration makes a...There is no such rule that an acceleration makes a speed time graph a straight line. In fact, it is almost impossible irl for a uniform acceleration, due to many factors such as gravity, air drag, friction etc. You might be a little misleaded on this, as this graph only states that the velocity is changing in a not uniform rate, but as the graph starts from 0 and moves upwards, it can be assumed Anonymoushttps://www.blogger.com/profile/13921904376175405700noreply@blogger.comtag:blogger.com,1999:blog-6403799362875544461.post-66094761927293047482016-11-23T18:46:40.673+08:002016-11-23T18:46:40.673+08:00According to this post Can I have some help with t...According to this post Can I have some help with this? http://iob.imgur.com/uYGg/M0YkKp8cxy many agree that the answer for number 4 is C. Could you provide an explanation as to why you think the answer is D? Still, thanks for the post! Much appreciatedAnonymoushttps://www.blogger.com/profile/13921904376175405700noreply@blogger.comtag:blogger.com,1999:blog-6403799362875544461.post-18528661929780979472016-11-23T18:11:16.581+08:002016-11-23T18:11:16.581+08:00no matter it's a straight line or curve,if the...no matter it's a straight line or curve,if the gradient is positive,it's acceleration,their difference is just uniform and ununiform accelerationAnonymoushttps://www.blogger.com/profile/00129520714030197328noreply@blogger.comtag:blogger.com,1999:blog-6403799362875544461.post-79275588325201007742016-11-23T17:27:03.836+08:002016-11-23T17:27:03.836+08:00no 4 motion OP is a curve,and its gradient is posi...no 4 motion OP is a curve,and its gradient is positive(so,accelerates).<br />Just looking at the curve, as the time increases ,the velocity increases,how can it be a deceleration?Anonymoushttps://www.blogger.com/profile/00129520714030197328noreply@blogger.comtag:blogger.com,1999:blog-6403799362875544461.post-45359351470114671452016-11-23T13:22:12.709+08:002016-11-23T13:22:12.709+08:00This comment has been removed by the author.Anonymoushttps://www.blogger.com/profile/16048310326170278759noreply@blogger.comtag:blogger.com,1999:blog-6403799362875544461.post-54786930602895680512016-11-23T12:57:56.277+08:002016-11-23T12:57:56.277+08:00According to success book, it's still accelera...According to success book, it's still acceleration, just not uniform.Anonymoushttps://www.blogger.com/profile/06186565885195120644noreply@blogger.comtag:blogger.com,1999:blog-6403799362875544461.post-86838767058782804772016-11-23T11:19:32.999+08:002016-11-23T11:19:32.999+08:00Have science paper 1 ans?Have science paper 1 ans?喵喵 ニャンニャンhttps://www.blogger.com/profile/11320765913807708501noreply@blogger.comtag:blogger.com,1999:blog-6403799362875544461.post-34668784361760820772016-11-22T23:43:48.262+08:002016-11-22T23:43:48.262+08:00But still quite a lot of people say is C. Their re...But still quite a lot of people say is C. Their reason is according to success, it says accelerate to achieve maximum velocity basihttps://www.blogger.com/profile/11442686235405434641noreply@blogger.comtag:blogger.com,1999:blog-6403799362875544461.post-86089860529729470082016-11-22T23:13:00.555+08:002016-11-22T23:13:00.555+08:00no 4 OP is a curve graph, if acceleration must be ...no 4 OP is a curve graph, if acceleration must be exact straight line, curve is deceleration<br />Anonymoushttps://www.blogger.com/profile/12201984481488451010noreply@blogger.comtag:blogger.com,1999:blog-6403799362875544461.post-19147644737445052252016-11-22T21:50:17.658+08:002016-11-22T21:50:17.658+08:00for question 4,motion at OP is not decreasing acce...for question 4,motion at OP is not decreasing acceleration?<br /><br />Kumpulan STEM 1https://www.blogger.com/profile/13407863067622983977noreply@blogger.comtag:blogger.com,1999:blog-6403799362875544461.post-47331556841644777882016-11-22T21:10:01.962+08:002016-11-22T21:10:01.962+08:00May I know why it's D??? I've put that as ...May I know why it's D??? I've put that as my answer tho but quite a number of ppl also say C ..:(Anonymoushttps://www.blogger.com/profile/09762657171088594920noreply@blogger.comtag:blogger.com,1999:blog-6403799362875544461.post-41378012694238441662016-11-22T20:58:53.009+08:002016-11-22T20:58:53.009+08:00Airil izzah the answer for the fourth question is ...Airil izzah the answer for the fourth question is D.. Refer the bookAnonymoushttps://www.blogger.com/profile/13195835615631484085noreply@blogger.comtag:blogger.com,1999:blog-6403799362875544461.post-678254248279003492016-11-22T20:47:10.032+08:002016-11-22T20:47:10.032+08:00Could you please recheck the answer for question 4...Could you please recheck the answer for question 4. Some said the answer is CAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6403799362875544461.post-49269856852862410512016-11-22T17:31:27.751+08:002016-11-22T17:31:27.751+08:00hv sc kertas1 answer?hv sc kertas1 answer?Anonymoushttps://www.blogger.com/profile/08829903586353050000noreply@blogger.com