UI 2018/2019 Admission Thread - Education (91) - Nairaland
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| Re: UI 2018/2019 Admission Thread by Eben331: 6:51pm On Jul 23, 2018 |
The correct answers are marked with * Eben331: |
| Re: UI 2018/2019 Admission Thread by Eben331: 7:11pm On Jul 23, 2018 |
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| Re: UI 2018/2019 Admission Thread by Eben331: 7:12pm On Jul 23, 2018 |
@Aceed. I am using a press-button phone and u knw hw strenuous it is to type with it. If I am using a touchscreen phone, it will be much easier for me to show workings to calculation questions. So bear with me. I will try my possible best to solve ur questions. |
| Re: UI 2018/2019 Admission Thread by DrBESTJC(m): 7:44pm On Jul 23, 2018 |
Eben331:Thanks for this......really helpful.......more of it please.. |
| Re: UI 2018/2019 Admission Thread by Eben331: 11:10pm On Jul 23, 2018*. Modified: 11:37pm On Jul 23, 2018 |
(1) The tendon in a mass leg is 10cm long and 1cm in diameter. How much will it be stretched by a force of 157N if the Young Modulus for the tendon is 2*10^8N/m^2 (2) A steel bar of length 4m and of rectangular section 1cm by 2cm supports a load of 10kg. By how much is the bar stretched? take Young Modulus=2*10^11N/m^2 (3) A metal rod of diameter 7mm, length 100mm is made from steel of Young Modulus 2*10^11N/m^2. Calculate the force constant of the rod. PLS SHOW YOUR WORKINGS!! |
| Re: UI 2018/2019 Admission Thread by sirkhaleb47: 11:42pm On Jul 23, 2018 |
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| Re: UI 2018/2019 Admission Thread by Nltaliban(m): 4:16am On Jul 24, 2018 |
Aceed:Why did you add more questions? ![]()
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| Re: UI 2018/2019 Admission Thread by nessie17(f): 6:20am On Jul 24, 2018*. Modified: 4:45pm On Aug 02, 2018 |
Hello |
| Re: UI 2018/2019 Admission Thread by Eben331: 6:35am On Jul 24, 2018 |
DrBESTJC:You also use post more questions frequently bro. You should endeavour to also show corrections to them and workings if possible, to help one another on this thread. |
| Re: UI 2018/2019 Admission Thread by Specyano(m): 10:29am On Jul 24, 2018 |
cant move pass d upload of passport and signature...pls anyone having same issues and how will i go about it...pls quote me asap |
| Re: UI 2018/2019 Admission Thread by Orezy5(m): 10:30am On Jul 24, 2018 |
Eben331:SOLUTION .
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| Re: UI 2018/2019 Admission Thread by Hardebaryor(m): 10:54am On Jul 24, 2018 |
Mirabel2032:I don't know. |
| Re: UI 2018/2019 Admission Thread by Eben331: 10:57am On Jul 24, 2018 |
@Orezy5. Thanks but I will prefer you typing the answers out because i am using a 'small' phone. I'm sorry for any inconvenience this may give you. |
| Re: UI 2018/2019 Admission Thread by Orezy5(m): 11:13am On Jul 24, 2018 |
Eben331:All right I'll do that |
| Re: UI 2018/2019 Admission Thread by Aceed: 1:19pm On Jul 24, 2018 |
Eben331:length = 10cm=0.1m diameter = 1cm = 0.01m radius = diameter/2 = 0.005m force = 157N Young modulus = 2 x 10^8 Nm^-2 Young modulus = stress/strain stress = force/area; strain = extension/length Young modulus = F/A ÷ e/l YM = Fl/Ae make e subject of formula e = Fl/YA (Area = pie x r^2) therefore, e = (157 x 0.1) ÷ (2*10^8 x 3.14 x 0.005 x 0.005) e = 15.7 ÷ 15700 e = 0.001m or 1x10^-3m (2) A steel bar of length 4m and oflength = 4m load = 10kg force = load x g = 100N Area (of rectangle) = L x B = 2 x 1 = 2cm^2 = 0.0002m extension = ? Young modulus = 2x10^11 Nm^-2 extension e = Fl/YA e = (100 x 4) ÷ (2x10^11 x 0.0002) e = 1 x 10^-5m (3) A metal rod of diameter 7mm,length = 100mm = 0.1m diameter = 7mm = 0.007m radius = d/2 = 0.0035m Young modulus = 2*10^11 Nm^-2 force constant k = ? extension e = Fl/YA but from f=ke; e = f/k therefore f/k = Fl/YA making k the subject of the formula k = YAF/Fl the F above cancels the F below therefore, k = YA/l; (Area = pie x r^2) k =(2*10^11 x 3.14 x 0.0035 x 0.0035) ÷ 0.1 k = 7.693 x 10^7 Nm^-1 |
| Re: UI 2018/2019 Admission Thread by Aceed: 1:30pm On Jul 24, 2018 |
Nltaliban:Because they are not too hard for the Taliban ![]() me still dey wait the remaining ooo |
| Re: UI 2018/2019 Admission Thread by Nltaliban(m): 1:33pm On Jul 24, 2018 |
Aceed:I'll try. You people should not tag me with physics questions again. I'm not good at physics and no longer do physics as a course here in ui |
| Re: UI 2018/2019 Admission Thread by Obamahmud: 3:56pm On Jul 24, 2018 |
Pls! I want to study industrial chemistry,my jamb score is 213 and my subject combination is bio,chem,phy,eng.Am I good to go? |
| Re: UI 2018/2019 Admission Thread by Aceed: 4:01pm On Jul 24, 2018 |
Nltaliban:Abeg no vex....exam dey door and aspirants here don't answer questions as quick as before or maybe as quick as I expected I con say make I tag unayou try sef ajet, Isiss dem no even answer but I understand they are in a very demanding department... What course do you study? |
| Re: UI 2018/2019 Admission Thread by Eben331: 4:04pm On Jul 24, 2018 |
Nltaliban:But your residual knowledge on Physics will still be useful to us here. You can still assist us in solving some questions you still know. |
| Re: UI 2018/2019 Admission Thread by Eben331: 4:07pm On Jul 24, 2018 |
Obamahmud:Check through this link on that - https://ui.edu.ng/content/advertisement-course-requirements |
| Re: UI 2018/2019 Admission Thread by Nltaliban(m): 5:38pm On Jul 24, 2018 |
Aceed:Ikr. Most guys here rarely visit nairaland. I'm in the faculty of veterinary medicine. No problem. I'll send the answers later |
| Re: UI 2018/2019 Admission Thread by Nltaliban(m): 5:40pm On Jul 24, 2018 |
Eben331:no problem man |
| Re: UI 2018/2019 Admission Thread by Eben331: 7:20pm On Jul 24, 2018 |
(1) A bullet of mass of 20g is fired from a gun of mass 1kg. If the velocity of the bullet is 250m/s, calculate the backward velocity of the gun. (2) A ball P of mass 0.25kg loses one-third of its velocity when it makes a head-on collision with an identical ball Q at rest. Q moves off with a speed of 2m/s in the original direction of P. Calculate the initial velocity of P. PLS SHOW YOUR WORKINGS! |
| Re: UI 2018/2019 Admission Thread by DrBESTJC(m): 9:22pm On Jul 24, 2018 |
Eben331:I always post corrections to my questions naa........Abi you no de see am ? |
| Re: UI 2018/2019 Admission Thread by DrBESTJC(m): 9:40pm On Jul 24, 2018 |
DrBESTJC: |
| Re: UI 2018/2019 Admission Thread by shiffynaani(m): 9:55pm On Jul 24, 2018 |
PLS I COULDN'T FIND NECESSARY DOCUMENTS REQUIRED BY OOU. PLS AUX, WHERE DO THEY LIST THEM? ITZ NOT IN THEIR SCREENING FORM EITHER |
| Re: UI 2018/2019 Admission Thread by BiafranDel: 10:36pm On Jul 24, 2018 |
A given amount of gas occupies 10dm*3(10dm cube) at 4 atm and 273degree celsius. The number of moles of the gas present is [molar volume of a gas at stp is 22.4dm*3] A0.89mole B1.90mole C3.80mole D5.70mole |
| Re: UI 2018/2019 Admission Thread by BiafranDel: 10:40pm On Jul 24, 2018 |
The pressure of a gas which occupies a volume of 500cm3 at temperature 27degree celsius is 900mmhg. what is the pressure of the gas at a temperature of minus -48degree*C if the volume is reduced to 250cm3. A500mmhg B750mmhg C1350mmhg D400mmhg |
| Re: UI 2018/2019 Admission Thread by Nltaliban(m): 10:44pm On Jul 24, 2018 |
Aceed:
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| Re: UI 2018/2019 Admission Thread by Orezy5(m): 10:58pm On Jul 24, 2018 |
BiafranDel:Volume, v = 10dm³ Pressure, P = 4 atm Temperature = 273°C Convert to Kelvin: 273 + 273 = 546K number of moles, n = ? Gas constant, R = 0.08205 PV = nRT n = PV/RT n = 4 x 10/0.08205 x 546 n = 0.89 mol |
| Re: UI 2018/2019 Admission Thread by Orezy5(m): 11:03pm On Jul 24, 2018 |
BiafranDel:V1 = 500cm³ P1 = 900mmHg T1 = 27°C = (273 + 27)K = 300K V2 = 250cm³ P2 = ? T2 = -48°C = (273 + - 48)K = 225K Applying the general gas equation, P1V1/T1 = P2V2/T2 P2 = P1 x V1 x T2/V2 x T1 P2 = 1350mmHg |
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