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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 sirgalahad26(m): 9:02am On Jul 26, 2018 |
Pes13:Big Boss!!! Looking forward to seeing you too! 1 Like |
Re: UI 2018/2019 Admission Thread by sirgalahad26(m): 9:04am On Jul 26, 2018 |
AASOO: Like there is something wrong with that link!!! |
Re: UI 2018/2019 Admission Thread by Damayogenius18: 9:10am On Jul 26, 2018 |
U guys are trying o......eben..drbest,Teflon,ntaliban.....I should be back 2moro....... |
Re: UI 2018/2019 Admission Thread by Damayogenius18: 9:13am On Jul 26, 2018 |
Damayogenius18:....Eben,grace sent ur question to me but I don't know it was dat important cos a lot of ppl sent question to me den on what's app and my phone is not working well....I will repair it today by Christ grace...I don't think it is hard though....DAT was y I did not attempt it then.....maybe I will look at it later....m so sori.... |
Re: UI 2018/2019 Admission Thread by Damayogenius18: 9:25am On Jul 26, 2018 |
Maybe I should give u some warming up b4 i come back....A 3kg ball moving at 8m/s strike a 2kg ball at rest.... If the collision is elaestic,what is the speed of each ball after collision ...its simple o..... |
Re: UI 2018/2019 Admission Thread by freshsteph: 10:37am On Jul 26, 2018 |
Damayogenius can you please add me to your group on Whatsapp . i'm Stephen 09069015773 |
Re: UI 2018/2019 Admission Thread by freshsteph: 10:39am On Jul 26, 2018 |
Damayogenius18: Velocity of each ball is 4.8m/s |
Re: UI 2018/2019 Admission Thread by Eben331: 12:18pm On Jul 26, 2018 |
The correct answers are marked with * Eben331: |
Re: UI 2018/2019 Admission Thread by Eben331: 12:19pm On Jul 26, 2018 |
Damayogenius18:no prob boss...welcome back!! |
Re: UI 2018/2019 Admission Thread by Eben331: 12:31pm On Jul 26, 2018 |
Eben331: |
Re: UI 2018/2019 Admission Thread by nessie17(f): 1:18pm On Jul 26, 2018 |
DKrown:Well, last year both jamb score and post utme score was used... Jamb score/8 + post utme score/2 Last two years and three years ago, only post utme was used I do not know how they will do it this year Just prepare very well and try to score high for your preferred course |
Re: UI 2018/2019 Admission Thread by nessie17(f): 1:21pm On Jul 26, 2018 |
If u av still not done your post utme registration U can contact me for a stress free registration and it is affordable price, quality service Registration closes on 10th of August 2018 |
Re: UI 2018/2019 Admission Thread by Eben331: 2:39pm On Jul 26, 2018 |
Orezy5:D...Am I right? |
Re: UI 2018/2019 Admission Thread by Eben331: 2:53pm On Jul 26, 2018 |
Damayogenius18:M1=3kg, U1=8m/s, M2=2kg, U2=0 (at rest), V1=?, V2=? Using law of conservation of momentum: M1U1+ M2U2=M1V1+M2V2 3*8+2*0=3*V1+2*V2 In elastic collision, the relative velocity of the balls will be unchanged in magnitude but reversed in direction; so V1=V2=V in magnitude 24+0=3*V1+2*V2 >>> 24=(3+2)V >>> 24=5V >>> V=24/5 >>> V=4.8m/s. Therefore V1=4.8m/s and V2=4.8m/s. |
Re: UI 2018/2019 Admission Thread by Pr0ton: 3:30pm On Jul 26, 2018 |
Hello everyone. This is for incoming students who are yet do do the Post UTME registration, YINO does post UTME registration for just 3500NGN total pay for online payment, registration and printing. Contact Yino on 08128200178(whatsapp) or 08100269515 |
Re: UI 2018/2019 Admission Thread by esivue007(m): 3:55pm On Jul 26, 2018 |
Eben331:i think the motion of conduction is vibratory not oscillatory. |
Re: UI 2018/2019 Admission Thread by esivue007(m): 3:57pm On Jul 26, 2018 |
Eben331:no its ficsher's projection |
Re: UI 2018/2019 Admission Thread by MikeSpokes(m): 4:06pm On Jul 26, 2018 |
Pr0ton: LMao, this guy. whats up? |
Re: UI 2018/2019 Admission Thread by Eben331: 4:38pm On Jul 26, 2018 |
esivue007:It's vibratory I choose not oscillatory. |
Re: UI 2018/2019 Admission Thread by Eben331: 4:40pm On Jul 26, 2018 |
esivue007:ok, noted...thanks. You can as well help me with the English questions. |
Re: UI 2018/2019 Admission Thread by BiafranDel: 4:47pm On Jul 26, 2018 |
How much sodium hydroxide is required to make 25cm3 of a 0.1 molar solution? ( na=23, O=16, H=1) A4g B2g C1g D5g |
Re: UI 2018/2019 Admission Thread by AASOO: 5:34pm On Jul 26, 2018 |
message me via Whatsapp, I will send you the link. 07033539277 |
Re: UI 2018/2019 Admission Thread by AASOO: 5:38pm On Jul 26, 2018 |
sirgalahad26: message me via Whatsapp, I will send you the link. 07033539277 |
Re: UI 2018/2019 Admission Thread by AASOO: 5:39pm On Jul 26, 2018 |
AASOO: message me via Whatsapp, I will send you the link. 07033539277 |
Re: UI 2018/2019 Admission Thread by Damayogenius18: 5:57pm On Jul 26, 2018 |
Eben331:....u tried bro...not correct sha |
Re: UI 2018/2019 Admission Thread by Damayogenius18: 5:59pm On Jul 26, 2018 |
freshsteph:...thanks for attempting it...but its not correct |
Re: UI 2018/2019 Admission Thread by Eben331: 6:11pm On Jul 26, 2018 |
Damayogenius18:chai!!! pls show the correction. |
Re: UI 2018/2019 Admission Thread by Eben331: 6:29pm On Jul 26, 2018 |
BiafranDel:C=0.1mol/dm^3, V=250cm^3, m.m of NaOH =23+16+1=40g/mol Using n=CV ( n=m/m.m) >>> m/m.m=CV >>> m/40=0.1*250/1000 >>> m=1g there is a mistake in d question, the volume supposed to be 250cm^3 and nt 25cm^3 according to the options. |
Re: UI 2018/2019 Admission Thread by DRZAY(m): 6:43pm On Jul 26, 2018 |
Eben331 pes13 please who is able to login after C.O.I because I did mine on 17 of this month. |
Re: UI 2018/2019 Admission Thread by DRZAY(m): 6:44pm On Jul 26, 2018 |
Eben331 pes13 please who is able to login after C.O.I because I did mine on 17 of this month. And I can't login. |
Re: UI 2018/2019 Admission Thread by freshsteph: 8:08pm On Jul 26, 2018 |
Re: UI 2018/2019 Admission Thread by esivue007(m): 8:09pm On Jul 26, 2018 |
DRZAY:infact dude we many like |
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