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UI 2018/2019 Admission Thread - Education (10) - Nairaland

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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:
Lols! You mean scoring 90% in utme english? It's practically impossible, unless you want to be reading only use of english at the expense of other subjects. My brother, the more you understand the 'invisible teacher', the more confused you'll be when you get to utme hall; though your construction will improve significantly. Read the book normally, read your novels and newspapers as well to learn tactics of passing english exams. God bless you.

Dude I saw 95 last year(2016) utme!. use of English!.. nothing is impossible cheesy
Re: UI 2018/2019 Admission Thread by Nobody: 9:11am On Nov 24, 2017
Salarys:


Dude I saw 95 last year(2016) utme!. use of English!.. nothing is impossible cheesy
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:
i thought that heterogenous mixture is when the mixture in nt uniform while homogenous is otherwise
Yh . dax right .
Re: UI 2018/2019 Admission Thread by boyejo123(m): 1:26pm On Nov 24, 2017
psalmhorah:
i bought it #1250 @ beere,Ibadan.
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:
go to a very good bookshop ..d textbook is very essential for you to score minimum of 90% in utme English...

m currently enjoying it :Pgo to a very good bookshop ..d textbook is very essential for you to score minimum of 90% in utme English...grin

m currently enjoying it grin grin

Just Got mine...1200

All I can say is wow.... very in-depth
Time to work smiley

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:
PHYSICS.

In an experiment to determine terminal velocity, as the pressure increases, the viscosity of a liquid...
(A) Increases then decreases
(B) Decreases
(C) Remains constant
(D) Increases.

Please I need explanations.


>> The answer they gave to the question above is (C).Take note of the 'terminal velocity'. <<


...but let a guru explain this...


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:
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

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 KillerFrost(f): 9:36pm On Nov 24, 2017
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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:



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

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:
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?
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:
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
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:

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

u solved it 100%.
Re: UI 2018/2019 Admission Thread by Towbaba500(m): 9:30am On Nov 25, 2017
boyejo123:
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
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:


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
Re: UI 2018/2019 Admission Thread by samuelizz(m): 2:19pm On Nov 25, 2017
Erflog:
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


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:
[/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.
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:

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.
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:
it would be nice if we could av a study group here or on watsapp
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:
The last answer y sin ,y not cos or tan (getting the angle)


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:

kindly create the group and share the link or inform us to submit our number.
am eager to join self

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:


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.
Seen . it helps . thanks.

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