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Nairaland Forum / Nairaland / General / Education / UI 2018/2019 Admission Thread (333895 Views)
Olabisi Onabanjo University(oou) 2018/2019 Admission / Federal University Oye-ekiti (FUOYE) 2018/2019 Admission Thread / 2018/2019 Admission Process Thread Guide (2) (3) (4)
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Re: UI 2018/2019 Admission Thread by psalmhorah(m): 8:08am On Nov 24, 2017 |
guys pls which math textbook is recommended by jamb for this utme..I haven't given much attention to math... my fellow engineers and math gurus shld answer me pls !!! |
Re: UI 2018/2019 Admission Thread by Salarys: 9:01am On Nov 24, 2017 |
Drbabalola: Dude I saw 95 last year(2016) utme!. use of English!.. nothing is impossible ![]() |
Re: UI 2018/2019 Admission Thread by Nobody: 9:11am On Nov 24, 2017 |
Salarys:Well, as a christian I believe with God nothing shall be impossible! All I'm just saying is that one should not rely too much on use english by pursuing a prolific score at the expense of studying other subjects effectively. I also have a younger friend who had 98% in 2016 utme biology! but ended up scoring less than 50% in use of english. The major problem for use of english is that it's subject to periodic changes as directed by BBC. Since you've said you saw 95% in utme use of english: then I withdraw my previous statement. Shalom. |
Re: UI 2018/2019 Admission Thread by kidneystones(m): 11:07am On Nov 24, 2017 |
Rihsabmohd:Yh . dax right . |
Re: UI 2018/2019 Admission Thread by boyejo123(m): 1:26pm On Nov 24, 2017 |
psalmhorah:ok I'll try to get mine at dubge on Sunday. thanks |
Re: UI 2018/2019 Admission Thread by Samsonyte(m): 1:51pm On Nov 24, 2017 |
[quote author=psalmhorah post=62652049]m happy with this Invisible teacher by dele ashade ..the textbook is really revealing the secret of English language ..since I started yesterday ..I have been correcting answers in series...some series answers are always wrong esp English language...but thank God for invisible teacher... Owk you mean A1 in English |
Re: UI 2018/2019 Admission Thread by Samsonyte(m): 2:06pm On Nov 24, 2017 |
In Price Elasticity of Demand In an Inelastic Demand Increase in price of a goods will cause the quantity demanded to ________________ A. Increase alot B. Increase just a little bit C. Decrease alot D. Decrease just a little bit In an Elastic Demand Increase in price of a goods will cause the quantity demanded to _________________ |
Re: UI 2018/2019 Admission Thread by Salarys: 3:25pm On Nov 24, 2017 |
psalmhorah: Just Got mine...1200 All I can say is wow.... very in-depth Time to work ![]()
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Re: UI 2018/2019 Admission Thread by psalmhorah(m): 5:38pm On Nov 24, 2017 |
[quote author=Samsonyte post=62674531][/quote]not A1 in English pls ..A1 is specifically meant for wassce candidates ...ds 1 is invisible teacher .. |
Re: UI 2018/2019 Admission Thread by Erflog(m): 5:42pm On Nov 24, 2017 |
Please Guys...... I need your help on equilibrium of forces especially on Resolving forces in the vertical and horizontal direction. what I'm used to seeing is Horizontal forces = Fcos0 vertical forces= Fsin0 But reverse is the case in Lamad physics text .page 29 solution 1 2 3 5........I'm confused pls pardon my ignorance |
Re: UI 2018/2019 Admission Thread by felix00(m): 6:07pm On Nov 24, 2017 |
DrBESTJC: I believe the answer is correct (C) Pressure = Force/Area F=ma Recall at terminal velocity (same thing as constant velocity) acceleration is zero which means @ F=ma a=0 which means force is zero which results in pressure being 0 through force/area in which force is zero. So my point being that since pressure is zero.. I expect the viscosity to remain the same in terminal velocity cos increasing/decreasing the pressure won't have effect since the acc is nil Just my opinion.. I might be right or wrong PS I love what u guys are doing here |
Re: UI 2018/2019 Admission Thread by Pes13: 6:48pm On Nov 24, 2017 |
UNIVERSITY OF IBADAN PUTME PAST QUESTIONS PHYSICS. 1. A ball is thrown up into the air. At the highest point of its trajectory the ball: downwards (A) Is accelerating (B) has zero acceleration (C) is accelerating upwards (D) is still moving upwards Aball of mass 200 gmoving with a velocity of 8 2. Aball of mass 200 gmoving with a velocity of 8 m/s collides and sticks with another ball of mass 300 gmovingin the samedirection with velocity 4 m/s. Whatis the common velocity of the balls after the collision? (A) 5.6 ms-1 (B) 2.8 ms-1 (C) 11.2 ms-1 (D) 1.4 ms-1 3.A boy holds the end of a rubber cord from which hangs a solid metal ball, if the boy whirls the ball in a horizontal circle, keeping his hand still. If the rubber cord breaks when the ball is at a point on the circle, in what direction will the ball move? (A) towards the hand of the boy B. away from the hand of theboy C. inthe direction of the tangent to the circle at the point of break D. None of the above 4. A force of 20 Napplied parallel to the surface of horizontal table is just sufficient to make a block of mass 4 kg set for motion. Find the acceleration when the force is doubled. (A) 2 m/s (B) 4 ms-2 (C) 5 ms-2 (D) 10 ms-2 5. A man walks 1km due east and then 1 km due north. His displacement is (A) 1 kmN15°E (B) 1 Km N30°E (C) √2 km N45°E (D) 2 km N75°E |
Re: UI 2018/2019 Admission Thread by Nobody: 8:19pm On Nov 24, 2017 |
Pls who can send a comprehension tip from the invisible teacher for me. I really need to improve my comprehension skill. Its just so painful that I can't find it in any bookshop around ojo in Lagos state. |
Re: UI 2018/2019 Admission Thread by Salarys: 9:19pm On Nov 24, 2017 |
Pes13: For question 2, for two bodies moving in the same direction, total momentum before impact= m1v2+m2v2 After impact, the two move with a common velocity=V(m1 +m2) Therefore, M1v1 + m2v2= V(m1+m2) by conservation principle Making V the subject formula, V=M1V1 + m2v2/m1+m2 V= 200x8 + 300x4/200+300 V=1600+1200/500 V=2800/500 V=5.6m/s For question 4, F=ma 20x2=4a 40=4a a=40/4 a=10m/s^2 For question 5, using Pythagoras theorem Hyp^2= opp2+adj2 Hyp2=1^2+1^2 Hyp2=/2 I.e(root2) To get the angle, sin TITA= opp/hyp Sin TITA= 1/root2 SinTita= 0.7071 TITA= sin- 0.7071 TITA= 45 degrees Therefore root 2km N45 degreesE 3 Likes |
Re: UI 2018/2019 Admission Thread by Nobody: 9:36pm On Nov 24, 2017 |
. |
Re: UI 2018/2019 Admission Thread by Hardebaryor(m): 9:56pm On Nov 24, 2017 |
Bosses of these thread please what is the importance attached to using the syllabus to read since you have gotten the right books to read. I just can't see the meaning of using syllabus to read. Can someone explain please? |
Re: UI 2018/2019 Admission Thread by overcomer90: 12:10am On Nov 25, 2017 |
felix00: Sorry this is wrong because viscosity increases with increase in pressure. Check this out http://www.viscopedia.com/basics/factors-affecting-viscometry/ What u mentioned above is for when terminal velocity has been achieved then viscosity remains constant . observe the question well, u will see that terminal velocity hasn't been achieved yet. |
Re: UI 2018/2019 Admission Thread by overcomer90: 12:13am On Nov 25, 2017 |
Hardebaryor:After reading ur textbook confirm from ur syllabus to know if u have actually covered all that is required. |
Re: UI 2018/2019 Admission Thread by Towbaba500(m): 8:21am On Nov 25, 2017 |
morning guys,i was unable to solve this question so i decided to bring it over to the gurus here.it's a physics question.it goes thus- An engine pumps water from a river 10m below its own level and discharges it through a nozzle of diameter 10cm with a speed of 50ms-¹.Find the power required assuming: (a)no losses (b)70% efficiency. (density of water=10³kgm–³,g=10ms–²) source: okeke |
Re: UI 2018/2019 Admission Thread by boyejo123(m): 9:11am On Nov 25, 2017 |
Towbaba500:i can provide in sight help of formula suitable for application let's say power-work/time work-force*distance I.e power=force*distance/time therefore power=force*velocity we have been given velocity but no force, I think Force=pressure*Area pressure=height (10)*density (1000)*g(10) then find area of nozzle, circle at the mouth, pie*r² then divide diameter/2 and solve for force, then multiply with speed. let the gurus come |
Re: UI 2018/2019 Admission Thread by Pes13: 9:25am On Nov 25, 2017 |
boyejo123: u solved it 100%. |
Re: UI 2018/2019 Admission Thread by Towbaba500(m): 9:30am On Nov 25, 2017 |
boyejo123:big ups to you bro,thanks so much |
Re: UI 2018/2019 Admission Thread by kidneystones(m): 2:10pm On Nov 25, 2017 |
The last answer y sin ,y not cos or tan (getting the angle) Salarys: |
Re: UI 2018/2019 Admission Thread by samuelizz(m): 2:19pm On Nov 25, 2017 |
Erflog: That Fy = Fsin∅ and Fx = Fcos∅ is not constant. It all depends on the diagram, the part at which the angle is facing. Fx could also be Fsin∅.. It is advisable to always draw the diagram first before revolving. Shalom! |
Re: UI 2018/2019 Admission Thread by mercy288(f): 4:26pm On Nov 25, 2017 |
[/color][color=#000000] Good evening everyone, I've been following dis thread since it started and I noticed dat there are no Art students here or perhaps they aren't active....Pls I need to know if dis thread Is for science students only or d Art students aren't just interested. |
Re: UI 2018/2019 Admission Thread by Nobody: 5:50pm On Nov 25, 2017 |
![]() why is thread not moving any longer na |
Re: UI 2018/2019 Admission Thread by boyejo123(m): 6:07pm On Nov 25, 2017 |
mercy288:I think it's for everyone, but art students don't always have problem when reading, you know you guys calculate less,so science students bring much tough questions to be tackled by the ogas. |
Re: UI 2018/2019 Admission Thread by mercy288(f): 6:55pm On Nov 25, 2017 |
boyejo123:it would be nice if we could av a study group here or on watsapp |
Re: UI 2018/2019 Admission Thread by boyejo123(m): 7:25pm On Nov 25, 2017 |
mercy288:kindly create the group and share the link or inform us to submit our number. am eager to join self |
Re: UI 2018/2019 Admission Thread by Salarys: 7:59pm On Nov 25, 2017 |
kidneystones: Because we’re looking for the angle of displacement.. therefore to get the angle.. we must use any from SOHCAHTOA which includes the hypotenus(value for the displacement) If u use SOH or CAH you’ll get same answer. Hope it helps. |
Re: UI 2018/2019 Admission Thread by Salarys: 8:00pm On Nov 25, 2017 |
boyejo123: There is one already. Check the previous pages for the link |
Re: UI 2018/2019 Admission Thread by kidneystones(m): 8:22pm On Nov 25, 2017 |
Salarys:Seen . it helps . thanks. |
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